The Math Book 9 9786055485511
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I
MATHBOOK - 10
TÜRK EĞİTİM DERNEĞİ YAYINLARI Copyright © 2012, Türk Eğitim Derneği (TED) İktisadi İşletmesi. Her hakkı saklıdır. Kitabın tamamı ve/veya herhangi bir kısmı hiçbir yolla kopyalanamaz, çoğaltılamaz, yayımlanamaz ve dağıtılamaz. Yayıncı Sertifika No: 11687
The Math Book 9 ISBN: 978-605-5485-51-1 5. Baskı Ankara, 2016
Sipariş: TED İKTİSADİ İŞLETMESİ Tel: 0312 489 85 90 www.tedisletme.com.tr [email protected]
Tasarım: YURDAGÜL REKLAM AJANSI G.M.K. Bulvarı Denizer İş Merkezi No: 52/15 Maltepe / Ankara Tel: 0312 232 58 15 • 0532 542 85 07
Baskı: Dumat Ofset (Sertifika No: 14021) Bahçekapı Mah. 2477 Sokak No: 6 Şaşmaz - Etimesgut / Ankara Tel: 0 312 278 82 00 (pbx) Faks: 0 312 278 82 30 [email protected]
III
Contents CHAPTER 1: NUMBERS AND ALGEBRA 1. SETS
..................................................................................................................................................................... 7
1.1. Set
...............................................................................................................................................................................7
1.1.1. Definition and Notations..........................................................................................................................................7 1.1.2. Types of Sets.............................................................................................................................................................8 1.2. Set Operations..................................................................................................................................................................10 1.2.1. Union and Intersection...........................................................................................................................................10 1.2.2. Complement and Difference..................................................................................................................................11 1.2.3. Cartesian Product...................................................................................................................................................19
2. EQUATIONS AND INEQUALITIES................................................................................................................................. 29 2.1. Real Numbers...................................................................................................................................................................29 2.1.1. Irrational and Real Numbers..................................................................................................................................29 2.2. First Degree Equations and Inequalities...........................................................................................................................31 2.2.1. First Degree Equations in One Unknown................................................................................................................31 2.2.2. Simple Inequalities.................................................................................................................................................35 2.2.3. First Degree Inequalities With One Unknown........................................................................................................37 2.2.4. Absolute Value.......................................................................................................................................................39 2.2.5. First Degree Equations and Inequalities With Two Unknowns...............................................................................47 2.3. Exponents and Radicals....................................................................................................................................................63 2.3.1. Exponents...............................................................................................................................................................63 2.3.2. Radicals..................................................................................................................................................................72 2.4. Applications of Equations.................................................................................................................................................86 2.4.1. Ratio and Proportion..............................................................................................................................................86 2.4.2. Problems................................................................................................................................................................94
IV 3. FUNCTIONS.............................................................................................................................................................. 107 3.1. Function..........................................................................................................................................................................107 3.1.1. Definition of Function...........................................................................................................................................107 3.1.2. Linear, Identity and Constant Functions...............................................................................................................115 3.2. Graphs of Functions........................................................................................................................................................121 3.3. Types of Functions..........................................................................................................................................................134
CHAPTER 2: GEOMETRY 4. TRIANGLES............................................................................................................................................................... 144 4.1. Congruency....................................................................................................................................................................155 4.1.1. Angles of Triangles................................................................................................................................................155 4.1.2. Congruent Triangles.............................................................................................................................................161 4.1.3. Angle-Side Relations.............................................................................................................................................187 4.2. Similarity of Triangles.....................................................................................................................................................193 4.2.1. Similarity of Trianges............................................................................................................................................193 4.2.2. Triangle Proportion Theorem...............................................................................................................................201 4.3. Bisectors, Median and Altitude......................................................................................................................................209 4.3.1. Angle Bisector......................................................................................................................................................209 4.3.2. Median.................................................................................................................................................................223 4.3.3. Altitude.................................................................................................................................................................230 4.3.4. Perpendicular Bisector.........................................................................................................................................231 4.4. Right Triangle and Trigonometry....................................................................................................................................233 4.4.1. Pythagorean Theorem..........................................................................................................................................233 4.4.2. Trigonometric Ratios in Right Triangle..................................................................................................................240 4.4.3. Unit Circle and Trigonometric Ratios....................................................................................................................244 4.4.4. The Law of Cosines...............................................................................................................................................251 4.5. Area of a Triangle............................................................................................................................................................253 4.5.1. Area of a Triangle.................................................................................................................................................253 4.5.2. The Law of Sines...................................................................................................................................................271
V 5. VECTORS ................................................................................................................................................................. 273 5.1. Vectors and Operations with Vectors.............................................................................................................................273 5.1.1. Vectors..................................................................................................................................................................273 5.1.2. Operations with Vectors.......................................................................................................................................280
CHAPTER 3: STATISTICS AND PROBABILITY 6. STATISTICS............................................................................................................................................................... 286 6.1. Measures of Central Tendency and Spread....................................................................................................................286 6.1.1. Measures of Central Tendency.............................................................................................................................286 6.1.2. Measures of Spread..............................................................................................................................................288 6.2. Graphs ...........................................................................................................................................................................294 6.2.1. Graphical Tools.....................................................................................................................................................294 6.2.2. Scatter Plot...........................................................................................................................................................300 6.2.3. Box and Whisker Plot...........................................................................................................................................306
7. PROBABILITY........................................................................................................................................................... 311 7.1. Probability......................................................................................................................................................................311 7.1.1. Basic Definitions and Computing Probabilty........................................................................................................311
6
CHAPTER 1
NUMBERS AND ALGEBRA
7 1. SETS 1.1. Set 1.1.1. Definition and Notations
Set
A set is a well-defined collection of objects or elements. Sets are usually denoted by capital letters such as A, B or C. Let the set A is given as A = {a, b, {a, b}}
Element Note of Interest: The modern study of set theory was initiated by Georg Cantor and Richard Dedekind in the 1870s. Set theory is generally taken to have been the work of a single man, Georg Cantor, who developed a basic discipline that has deeply affected the shape of modern mathematics. Remark! In roster form,
If ‘a′ is one of the objects of the set A, we say that ‘a is an element of A′ or ‘a belongs to A′, denoted by a Œ A. If it is not true that a is an element of A, we write a œ A.
The number of the elements of the set A is denoted by n(A) ; here n(A) = 3
Notations Roster From (The List Method) In the roster form of the set, bracets, { }, are used to enclose the members of the set when listed. The members are separated by comma.
If we are talking about the set of vowels in the Turkish alphabet, we may denote this set as V = {a, e, ı, i, o, ö, u, ü}
1. repeating elements are only written once 2. the order of the elements doesn’t change the set The Set Builder Notation
The set notation {x: x has property P} is intended to contain exactly those elements having the property P. Here, (:) or (|) stands for “such that”
Ex(1)
The following sets are given by set builder notation. Write them in roster form. E = {x| x is an even natural number less than 8} = {…………….……….} E = {a: 0 ≤ a ≤ 6, a = 2k, k Œ Z} = {…………………….}
8 Venn Diagram Note of Interest: Around 1900, it was discovered that Cantorian set theory gave rise to several contradictions, called paradoxes. Bertrand Russell and Ernst Zermelo independently found the simplest and best known paradox, the Russell’s paradox, “the set of all sets that are not members of themselves.”
Sets can be represented by means of a diagram called the Venn diagram, where the elements of the set are represented by a point in a closed plane figure. E
2 0
6 4
1.1.2. Types of Sets Equal Sets
Ex(2)
Two sets that contain exactly the same elements are said to be equal sets. A = B if and only if (x ΠA and x ΠB)
Determine whether the following sets are equal or not. A = {x: –2 < 3x ≤ 12, x Œ Z } B = {x: x2 < 25, x Œ Z } C = {x: x! < 100, x Œ N}
Empty (Null) Set Ex(3)
The empty (null )set, denoted by Ø or { }, is the set which contains no elements. Determine whether the following sets are empty or not. A = {x | x2 < 0, x Œ Z } B = {x | 2 < x < 3, x Œ N} C={Ø}
Universal Set
Finite and Infinite Sets
The universal set U is a set which has the property that all sets under consideration are contained in it.
Note that although called the universal set, U is not uniquely defined unless precisely stated. Certainly any set containing U could be used as the universal set. Sets whose number of elements can be counted are finite. A set is infinite if it is not finite.
Ex(4)
Determine whether the following sets are finite or infinite. A = {x: –5 < x < 7, x Œ Z} B = {1 ,2, 3, 4, …}
C = {x | 0 < x < 1, x ΠR}
9 Subset Proper Subset
Given any two sets A and B, if every element of A is also an element of B, then A is said to be a subset of B, and is written as A Õ B or B È A (B contains A) . A subset of a given set that is not the set itself is called a proper subset. A is properly contained in B is written as A ⊂ B or B ⊃ A (B properly contains A).
Property A Õ A, ∅ ⊂ A, A ⊂ U (A Õ B and B Õ A) Ş A = B (A ⊂ B and B ⊂ D) Ş A ⊂ D Pascal's Triangle 1
= 1 = 20
1 +1
=2 = 2
1 +2 + 1
= 4 = 22
1+3+3+1
= 8 = 23
the subset of A with no element:
the subsets of A with 1 element: {1}, {2}, {3} = 3
= 1
∅
the subsets of A with 2 elements: {1, 2}, {2, 3}, {1, 3}
=3
the subsets of A with 3 elements: {1, 2, 3} = 1 + + ––––––––––––––––––––––––––––––––––––––––––––––––– –––––– ∅, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, {1, 2, 3}
= 8
The proper subsets of A:
∅, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}
1 + 4 + 6 + 4 + 1 = 16 = 2 .. .
= 7
If n(A) = n then the number of subsets of A is 2n
the sum of the numbers on every line gives the total number of the subsets Note :
the number of proper subsets of A is 2n – 1
Ex(5) A = {x : –2 < x < 4, , x Œ N} is given. Find the number of the
a) elements of A there are
b) subsets of A
i) a ≤ x ≤ b Ş b – a + 1 a0
f0 f 0
7
x
125 Graphing Functions
The graph of a function y = f(x) is the collection of all points (x, y) that satisfy (x,y) ∈ f. We graph functions in exactly the same way that we graph equations. 1. Construct a table of values 2. Plot enough points to recognize a pattern 3. Connect the points with a line or a curve to obtain the graph.
Ex(6)
Draw the graph of y = f(x) = x – 2, x y
y
Note: Generally, the values of the independent variable (generally the x-values) are placed on the horizontal axis; the values of the dependent variable (generally the y-values) are placed on the vertical axis.
Ex(7)
x
x-intercept: y-intercept: Domf(x) = Rngf(x) =
Draw the graph of f(x) = x – 2 for –1 ≤ x < 4 x y
y
x
x-intercept: y-intercept: Domf(x) = Rngf(x) =
126 Ex(8)
Draw the graph of f(x) = 1 – 2x for –2 < x ≤ 1 y
x y
x
x-intercept: y-intercept: Domf(x) = Rngf(x) =
The Rate of Change and the Slope of a Line Rate of Change
A rate of change is a rate that describes how one quantity changes in relation to another quantity. If x is the independent variable and y is the dependent variable, then rate of change =
change in y change in x
Let us determine the rate of change of a linear function. 1. Consider the linear function f(x)= 2x+1
1 1 1 1
x
y
0
1
1
3
2
5
3
7
4
9
y 2
2:1=2
2
2:1=2
constant (positive)
2
2:1=2
rate of change
2
2:1=2
x
0
Positive rate of change 2. Consider the linear function f(x) = 4 – 3x
2 1 1 3
x
y
0
4
2
–2
3
–5
4
–8
5
–17
y –6 –6 : 2 = –3 –3 –3 : 1 = –3 constant (negative) –3 –3 : 1 = –3 rate of change –9 –9 : 3 = –3
0
x
Negative rate of change
127 3. Consider the linear function f(x) = 5
1 1 1 1
x
y
0
5
1
5
2
5
3
5
4
5
y 0
0:1=0
0
0:1=0
0
0:1=0
0
0:1=0
constant (zero) rate of change
x
0
Zero rate of change As it can be seen from the above example, a linear function has a constant rate of change. Ex(9)
The graph below shows a company's yearly profits for the past 10 years, where x is the year and f(x) is the profit (in millions of dollars). 60 56 (million 52 dollars) 48 44 40 36 32 0
1
2
3
4 5 6 (year)
7
8
9 10
Use the graph to answer the following questions a) How many times did the company's yearly profit have a positive rate of change? b) How many times did the company's yearly profit have a negative rate of change? c) During which year(s) did the company show no change in its profit? d) During which year(s) did the company show the greatest rate of change in its profit? Slope
Note: The slope of a linear function is the coefficient of the independent variable x.
The constant rate of change of a linear function is also called the slope. Slope is a number that tells how steep the line is.
The rate of change (or slope) of the linear function f(x) = mx + n can be found by; y −y f (x 2) − f (x 1) mx 2 + n − (mx 1 + n) = =m Rate of change: x 2 − x 1 = x2 − x1 x2 − x1 2 1
128 Ex(10) Note:
Find the rate of change, or the slope of the following lines. a) b) y y l2
l1
"Uphill" "Downhill" Horizontal Vertical
3
2 –1
Positive Slope
Negative Slope
Slope = 0
2
x
x
Slope is Undefined
c)
d)
y
l4
y
l3 2 x
0
Ex(11)
–3
0
x
Find the rate of change, or the slope of the following linear functions. a) f(x) = 3x – 6
b) f(x) = 2 – x
c) 2y = x + 1
d) x + 3y = 5
e) y – 2 = 0
y
Ex(12)
According to the given figure, find k. C
Note: The slope is the same for any two points on a straight line
B A (–2, –1)
(7, k)
(2, 3) y
129 Mathematical Modeling
When a mathematical structure is used to describe and study a real world problem, such a structure is called a mathematical model for the original problem. If the structure is linear, then the resulting equation is called a linear model.
Ex(13)
Suppose an art object purchased for 10,000 tl is expected to appreciate (increase) in value at a constant rate of 1000 tl per year. Write an equation to determine the value of the art object after t years. What will be its value 3 years from the purchase date? Draw the graph of the function.
Ex(14)
You spent 1200 tl on a computer and it depreciates (decreases in value) at a constant rate of 200 tl per year to 0 tl. Write an equation to determine the value of the computer after t years. What will be its value 5 years from the purchase date? Draw the graph of the function.
Linear Cost, Revenue and Profit Functions
The cost function C(x), is the total cost of manufacturing x units of the product. The revenue function R(x), is the total payment received from the sale of x units of the product. The profit function P(x), is the total profit realized from the manufacturing and sale of the x units of product. Profit = Revenue – Cost Suppose a firm has a fixed cost of F TL, a production cost of c TL per unit, then the cost function is given by C(x) = cx + F If the firm has a selling price of s TL per unit, then the revenue function is given by R(x) = sx and the profit function P(x) for the firm is given by P(x) = R(x) – C(x) where x denotes the number of units of the commodity produced and sold.
130 The breakeven point is the point where two linear functions intersect. In marketing, this point represents the point where products neither make a profit nor incur a loss. The breakeven point in the graph is the point, where the total revenue line crosses the total costs line (i.e. costs and revenue are the same). Everything below this point is produced at a loss, and everything above it is produced at a profit. Total sales Break – even point
Profit
TL
Total costs Variable costs Fixed costs
Loss Total units
Ex(15)
A little chef wants to start her own cupcake stand. It costs her 3 tl to make each cupcake, plus she has to pay a 5o tl renter's fee for her stand. She sells the cupcakes for 5 tl per cupcake. She wants to know how many cupcakes she has to sell before she can start making a profit. Find the cost function, the revenue function, the profit function. Find the breakeven point and thus the number of cupcakes she has to sell to start making profit. Draw their graphs and comment on the graph.
Ex(16)
A shoe manufacturing company has a monthly fixed cost of 15,000 tl, a production cost of 20 tl per unit, and a selling price of 50 tl per unit. Find a) the cost function, the revenue function b) the number of shoes that has to be sold before the company can start making a profit c) the profit function and the profit when 100 shoes are sold d) the number of shoes that has to be sold to make a profit of 6000 tl.
131 Piecewise Defined Functions and Their Graphs
A piecewise defined function is a function defined by at least two equations (‘pieces’), each of which applies to a disjoint subset of the domain.
Ex(17)
f: [a, b] Æ R and c Œ [a, b] f (x ) = )
g (x), a ≤ x < c h (x), c ≤ x < b
The function f is piecewise defined as f (x) = ( a) Find f(0) + f(2) + f(3)
2x − 1 , x < 2 x 2 − 1, x ≥ 2
b) Find a if f(a) = 8
Ex(18)
Draw the graphs of the following piecewise defined functions. x, x < 2 a) f (x) = ( x − 2, x ≥ 2
− 1, x < 0 x − 1, x ≥ 0
b)
f (x ) = (
c)
Z − x, x < 1 ] f (x ) = [ 4, 1 ≤ x < 3 ] + x 2, x ≥ 3 \
132 Absolute Value Function Absolute value function is the function y = |f(x)| defined as
|f(x)| = )
Ex(19)
− f (x) , f (x) < 0 f (x), f (x) ≥ 0
Define the following absolute value functions as piecewise and draw their graphs. a) f(x) = |x|
b) f(x) = |x + 3|
c) f(x) = |2 – x| + x
Ex(20)
Define f(x) = |x| + |x + 2| as piecewise and draw its graph.
133 Ex(52) Note: To graph
Given the graph of y = f(x), draw the graph of y = |f(x)| on the same set of axes. a)
y
b) y = f(x)
c)
y
y = |f(x)| from y = f(x); Any part of the graph on or above the x-axis stays the same;
y
y = f(x)
y = f(x)
x
x
x
reflect any part of the graph below the x-axis above the x-axis.
Graphs of f(x) = xn Ex(21)
Draw the graphs of the following functions. a) f(x) = x x y –1 0 1
y
x
b) f(x) = x2 x y –2 –1 0 1 2
y
x
134 c)
f(x) = x3 y
x y –2 –1 0 1 2
x
d)
1 f (x) = x , x ≠ 0 y
x y –2 –1 0 1 2
x
3.3. Types of Functions One to One (1-1) Function
Let f: A → B be a function If for all x1, x2 Œ A, f(x1) = f(x2) Ş x1 = x2 or x1 ≠ x2 Ş f(x1) ≠ f(x2)
then f is a one to one (1-1) function (injective function). A function is one to one if each element in the domain corresponds to exactly one element from the range
Ex(22)
Determine whether the following functions are 1-1 or not. f A
B
A
g
B
135 Ex(23) Note:
Determine whether the following functions from R to R are 1-1 or not. y
If any horizontal line cut the graph at most one point, then the function is 1 – 1, else not. (horizontal line test)
Ex(24)
y
g
f
y
x
x
y
h x
r x
Write the intervals where f is 1-1. y
–4
2 –2 –3
3
5
x
Ex(25)
f: R → R, f(x) = 2x + 1 is given. Is f 1-1?
Ex(26)
f: R → R, f(x) = 2x2 + 1 is given. Is f 1-1?
Onto Function
Let f: A → B be a function. If is onto (surjective function) iff, for any y Œ B, there exists at least one x in A such that f(x) = y. In other words ; If f(A) = B, then f is onto.
(Range=Codomain)
136 Ex(27)
Ex(28)
Determine whether the following functions are onto or not. A
t
A
h
B
B
y
f
Note:
A
n
B
B
y
g x
x
y
y
r
Note:
x
A 1-1 and onto function is called a bijective function.
Ex(29)
m
Write the types of the following functions from R to R. y
If any horizontal line intersects the graph at at least one point, then the function is onto; else into.
A
x
k x
Which of the following functions is/are onto? a) f: Z → Z, f(x) = 2x + 1 b) f: N → N, f(x) = x + 2 c) f: Z → Q, f(x) =
x+1 3
d) f: R → R, f(x) = x
h
137 REVIEW EXERCISES - I 1. Determine whether the graphs of the following equations define functions from R to R.
a) y
b)
3. According to the graph of f(x), find y
y = f(x)
y
x
3 0 1
x
f (0) + f (1) f (3)
3
x
–4
c) y
d)
y
x
x
2. Find the Domain and Range of the given functions.
a)
b)
y
y
f
2
2
d)
y
–3
–1
1
Dom f = Rng f =
3
1 –3
–1
3 –3
1 x
–4 Dom f = Rng f =
2
y = f(x)
6
–1
5
–5 f
2 1
Dom f = Rng f =
c) y
x
4
Dom f = Rng f =
y
f x
1
4. According to the graph of y = f(x), find f (0) + f (3) + f (5) f (–3) – f (–1)
x
5 y = f(x)
x
138 5. The graphs of f(x) and g(x) are given.
7. According to the graph of y = f(x), find
Find f(0) + f(–6) + g(2) + g(0) + g(5)
y 4 –6
y 3
y = f(x) 2
y = f(x)
1
–5 –4
1
–2 0
x
0 2
1
3
–1
6
7
x
y = g(x)
a) Dom(f) :
b) Rng(f) :
c) x-intercepts :
d) y-intercept :
e) sum of the roots of the equation f(x) = 0 :
f) the value(s) of x satisfying f(x + 1) = –1 :
g) the number of x satisfying f(x) = 1 :
h) the intervals where f(x) > 0 :
i) the intervals where f(x) < 0 :
6. According to the graph of the function y = f(x) find the values of a, b, c, d if y f(x) –3
–2
1
1 –1
2
x
–3 –4
a) f(a) = 1 Ş a =
b) f(b) = 0 Ş b =
c) f(c) = –1 Ş c =
d) f(d) = –3 Ş d =
e) f(e) = –4 Ş e =
8. f: [–2, 3) Æ R, f(x) = 3 – x is given. Find the range of y = f(x) and draw its graph.
139 9. According to the given tables of some values of the functions, find the rate of change and determine whether they are linear or not. Find the rule for the linear function(s).
a)
b) x
y = f(x)
x
y = f(x)
1
4
1
–1
2
3
2
1
3
1
4
5
4
2
6
9
c)
d) x
y = f(x)
x
y = f(x)
1
1
2
3
3
3
4
3
7
7
7
3
13
13
11
3
11. Find the slope of the following linear functions.
a) f(x) = 2x – 2
b) f(x) =
c) f(x) = 3 – 4x
d) 5y = 1 – x
e) x + 2y – 3 = 0
x –1 3
12. Draw the graphs of the following functions.
a) f(x)=2x – 2
b) f(x) = 3 – x
x c) f (x) = 2 + 1
d) f(x) = –2
10. Find the slope of the following lines.
a)
b)
y
2
1
c)
x
1
0
0 –1
0
x
2
d)
y
y 3
y
x
–1
x –2
140 13. Suppose you find a new job at a company. The starting salary is 20,000 TL per year, with a 600 TL annual (yearly) raise. Write an equation to determine your salary after t years. What will be your salary 6 years from the day you started working?
15. A manufacturer has a monthly fixed cost of 150,000 tl and a production cost of 18 tl for each unit produced. The product sells for 24 tl per unit.Find a) the cost function, the revenue function, the profit function b) Compute the profit (loss) corresponding to production levels of 22,000 and 28,000 c) How many units must the company produce and sell before the company can start making a profit d) How many units must the company produce and sell if they wish to make a profit of 30,000 tl?
14. A bicycle has an original value of 1000TL and is to be depreciated linearly over 5 years with a 300TL scrap value. What is the rate of depreciation of the bicycle?
Find an equation giving the value of the bicycle at the end of year t. What will be the value of the bicycle at the end of the second year?
141 REVIEW EXERCISES - II 2 – x, x < 3 is given. Find f(10) – f(100). 2 , x≥3
5. Draw the graphs of the following piecewise defined functions. x + 2, x < 1 a) f (x) = ( x − 1, x ≥ 1
1.
f (x ) = '
2.
x – 3, x < 2 f (x) = * 2x, 2 ≤ x < 6 is given. Find f(2) + f(6) + f(0). x + 2, x ≥ 6
b) f (x) = (
x − 3, x < 2 − 1, x ≥ 2
3.
f (x ) = )
–x 2, x even is given. Find f(4) – f(-3). x 3, x odd
c) f (x) = (
− 1, x < − 1 1, x ≥ − 1
4.
f (x ) = (
3x + a, x < 5 is given. If f(3) = 14, then find f(a). ax – a, x ≥ 5
d) f (x) = (
2 − x, x < 0 x + 1, x ≥ 0
e) f (x) = *
1 − x, x < − 1 x, − 1 ≤ x < 2 3, x ≥ 2
142 6. Define the following absolute value functions as piecewise and draw their graphs. a) f(x) = |x + 2|
b) f(x) = |3 – x|
c) f(x) = |x – 1| + x + 1
x d) f (x) = x , x ≠ 0
e) f (x) =
| x | + 3x 2
f) f(x) = |x + 1| + |x – 1|
g) f(x) = |x + 2| – |x|
7. Determine whether the following functions are one to one or not. a) f : N → N f(x) = 2x + 1
b) f : Z → Z
f(x) = 5x – 3
c) f : R → R
f (x) =
d) f : R → R
f(x) = x2 + 1
e) f : R+ → R f(x) = 2 – x2
x+1 2
8. f: {–1,0,1,2} Æ {–5, –2,1,4}, f(x) = 3x – 2 is given. Is f an onto function?
9. Determine whether the following functions are onto or not. a) f : N → Z f(x) = x + 4
b) f : Z → Z
f(x) = 2x – 1
c) f : Z → R
f (x) =
d) f : R → R
f(x) = 3x – 2
3x + 1 2
10. Determine whether the following functions are one to one, onto, both or neither.
a) f : Z → Z
f(x) = 2x + 5
b) f : R → R
f(x) = 3x + 1
c) f : R → R
f(x) = x2 + 2
143
CHAPTER 2
GEOMETRY
144 4. TRIANGLES
Point
A point represents position and has no length, no width, and no height or thickness. It is represented by a dot, which is generally labelled with a capital letter. A point is the most fundamental or basic element in geometry. All geometric figures consist of points. C
A N
Line
A line has length only, with no width, and no height or thickness. It is a set of connected points that indefinitely extends in one dimension in both directions. A line contains collinear points, which are points that lie in the same line. A line can be labelled by two of its points that are represented by capital letters or another label for a line is a lower case letter. ↔ line AB or AB B A
line d
d
Postulates Pertaining to Lines one line postulate
One and only one straight line can be drawn between any two points. B A
line has two points postulate
two points on line postulate
line intersection postulate
A line contains at least two points.
Through any two points there is one and only one line (two points determine a line).
Two straight lines can intersect each other at one and only one point.
145 Parallel Lines
Parallel lines are straight lines that lie in the same plane and do not intersect even at extended lengths. ¬ m
parallel line postulate
¬ // m
Through a given point not on a line, there is one and only one line parallel to that line. parallel line throught point line
Transversal
A transversal is a straight line that crosses two or more parallel or non-parallel lines. t
a b
transversal line
Ray
A ray has an end point at one end and extends indefinitely in the other direction. Rays are named by the letter representing the endpoint and any other point on the ray. A
B Æ AB
C
Angle
A
Æ AC
If two straight rays or segment meet at their endpoints, an angle is formed. The point where the endpoints meet or lines cross is called the vertex of the angle, and the sides of the angle are called the sides.
Measure of an Angle
The measure of an angle is the separation between the sides in degrees or radians. sides
Note: D R 180 = π 1o = 60’ , 1’ = 60’’
The angle on the left can be called
B
BA∑C, CA∑B or A
A
a vertex
C
BA∑C = [AB » [AC m(BA∑C) = a
146 Angle Bisector
An angle bisector is a line, ray or segment that divides an angle into two congruent angles. A
The bisector of AOB is the ray OC.
C
O
m(AO∑C) = m(CO∑B)
a a
B
Each point lying on the angle bisector is equidistant to the arms of that angle. Aʹ A C
O
Angle Bisector Postulate
a a
B
|CA| = |CB|
Cʹ
|CıAı| = |CıBı|
Bʹ
Any angle has one and only one bisector.
Types of Angles acute angle
Angles smaller than 90o are called acute angles. B
O
right angle
0° < a < 90°
a A
A right angle measures 90o and is often identified using a square at the interior of the vertex. A m(AO∑B) = 90°
O
obtuse angle
B
Angles larger than 90o but less than 180o are called obtuse angles.
A
90° < a < 180° a O
B
147 straight angle
A 180o angle, also called a straight angle, has its sides lying in a straight line. a = 180° A
O
B
Pairs of Angles Congruent Angles
Congruent angles have the same measure. A b
a B
Adjacent Angles
E
C
F
Adjacent angles are two angles that have a common vertex and a common side
A
AO∑B « BO∑C = [OB
B b a
O
Vertical Angles
a=b
D
AO∑B and BO∑C are adjacent angles
C
Vertical angles are the angles that have the same vertex, but are across the intersection point of any two intrsecting lines. D b 0
A a
a
B
b
m(AO∑C) = m(BO∑D) = a
are vertical angles
m(DO∑A) = m(CO∑B) = b
C
Ex(1)
B
D
C
A, O, D are collinear
40° x A
According to the given figure, find x.
25° O
E
148 Ex(2)
E
A, O, B are collinear
D
[OC and [OF are angle bisectors
C
F
m(FO∑D) = 100°
A
O
B
m(EO∑C) = 95°
are given. Find m(EO∑D)
Complementary Angles
If the sum of any two angles equals 90o, then these two angles are called complementary angles. A a + b = 90°
B b
Supplementary Angles
a
O
C
If the sum of any two angles equals 180o, then these two angles are called supplementary angles. B
a + b = 180° a
A
b O
C
Ex(3)
The complement of an angle is 15 less than two times of the angle. Find the angle.
Ex(4)
The sum of the degrees of the complement and supplement of an angle is 3 times of the degree of the angle itself. Find the degree of this angle.
149 parallel transversal postulate
If two parallel lines are cut by a transversal, the corresponding angles are congruent. d1 // d2 d
a
c
d1
b
h
e
g
d2
f
Corresponding angles : = aV eU= , bV W f= , cU gW= , dV hV Alternate interior angles = : bV hV= , cU eU
Alternate exterior angles: = aV gW= , dV W f
Vertical angles : = aV cW = , bV dV= , eU gW= , W f hV
Consecutive interior angles : m ( bV) + m ( eU) = 180° m ( Uc ) + m (V h ) = 180°
a ) + m (W f ) = 180° m ( dV) + m (gW) = 180° Consecutive exterior angles : m ( V Results
1. d1 A
B x
x = y
d1 // d2 D
y C
2.
d2 A
B
d1
x
x + y = 180° d1 // d2
y
D
d2
C
3. A
d1
x a
d2
y
C
4.
B
x y
z C
d1 // d2 a = x+y
d1
A B
d1 // d2 d2
x + y + z = 360°
150 5.
C
y b
E
Ex(5)
d1
A a
x
B
d1 // d2
D
z
x+y+z = a+b
d2
A
B
[BA // [CD, 45°
x = ? D
C
105°
x E
Ex(6) [BC // [ED A 70° C
m(BK∑ E) = x = ? B D
x K
55° F
E
A
Ex(7)
According to the given figure, find x.
50°
F 45° B
110°
x F
D 70°
35° C
151 A
Ex(8)
d1
30° 50° C
d1 // d2
B
According to the given figure, find x.
135° 100°
D
x
d2
E
A
Ex(9)
d1
d1 // d2 B
120° x
According to the given figure, find x.
D d2
C
C
Ex(10)
60°
B
[AD // [ML
According to the given figure, find x.
A D F
x E
L M
152 REVIEW EXERCISES - I D
1.
x
B
A
A, B, C are collinear.
C
y
5. An angle is 15 more than half of the complement of it. Find this angle.
x + y = 140° is given. Find x.
E
6. If the ratio of three angles is 3:4:15 and the first and the third angles are supplementary, then find the complement of the second angle. A
A, O, C are collinear,
120°
[OC is the angle bisector of BO∑D.
2.
O
B
C E
m(EO∑B) = 90° Find m(EO∑D)
7.
D
A
C
F
AB // DE
60°
According the given figure, find x.
B
5x
–1
0°
K
D
G E
130° H
3.
E
[OB is the angle bisector of AO∑C.
D C
B
40° O
x A
[OC is the angle bisector of AO∑E,
[OD is the angle bisector of BO∑E. Find x.
K
8.
A
4. If the ratio of the degrees of two complementary angles is 2:3, then find the measure of each angle.
3x + 7° C
AB // CD
B
E
According to the given figure, find m(AE∑F)
2x – 9° F
G
D
153 12. B
9.
A
165° C
[AB // [DE
According to the given figure, find m(CD∑E)
80°
[BA // [DE
C
B
A
2x E
D D
E
13. 10.
[AB // [DE
B
A
x
A 3x
E
x
65° C
120°
x
d1
d1 // d2
14. A
According to the given figure, find x.
F
B
[AB // [CD
According to the given figure,
150°
B C
find m(BA∑E).
E
140°
130° A
According to the given figure, find x.
E
43°
11.
D
C
E D
[BA // [CD // [EF
B
According to the given figure, find x.
C
d2
According to the given figure, find x.
x
3x
F D
154 B
15.
A
19.
60°
F A
25° E
150°
100° D
x
80°
70° B D
C
[BA // [CE // [FK
B
20. A
C
[BC // [EF
B
According to the given figure, find x.
x
According to the given figure, find x.
F
E
50°
C
L
E
A
16.
According to the given figure, find x.
G
According to the given figure, find x.
x
C
[AB // [ED
D D
x F
50°
E
K
21. A
B
17.
85° D
H
C
According to the given figure, find D m(EC∑D).
F
[BA // [CD // [EF
60° C
[BC] // [DE]
[AB // [CD
E
According to the given figure, find m(DC∑G)
B
A
150° E
G
F
A
22. 18.
A
F
d1
120° B
D 100° C
E
d2 d3
d1 // d2 // d3 According to the given figure, find m(AB∑C).
d1
E
B
x
120° C
D
50°
F
d2
d1 // d2 According to the given figure, find x.
155 4.1. Congruency 4.1.1. Angles of Triangles Triangle
Triangle is a three sided polygon that contain three angles. A triangle is formed by three line segments that join three non-collinear points. (non-collinear points are points that are not in a straight line)
A
AB∆C = [AB] » [AC] » [BC] c
b
B
a
C
Perimeter of the triangle ABC is a + b + c. The sum of the interior angles in a planar triangle is 180o.
A a
a + b + c =180o
B
b
c
A
Proof
B
Exterior Angle of a Triangle
C
C
The exterior angle of a triangle is the angle formed between the extension of one of the sides of the triangle and the outside of the triangle. Each exterior angle is a supplement of an adjacent interior angle.
A
exterior angle
Aʹ
Cʹ
Bʹ B
C
156 The measure of an exterior angle of a triangle is equal to the sum of the measures of the two non-adjacent opposite interior angles. a+b=d
A a
B
c
b
, d>a d>b
d C
Proof A
C
B
The sum of the exterior angles of a triangle is 360o.
A
e
a f
b
d
c C
B
A
Proof
B
C
d + e + f = 360o
Results
157 A
1.
x
z
t
y
x+y+z = t
C
B
2. A
x
B O
y=
x+z 2
y
a a
z
C
According to the given figure, find x.
A
Ex(1)
50°
E x
D
35°
60°
B
ABC is a triangle
A
Ex(2)
Find x.
x + 15°
B
C
5x – 40°
2x + 15° C
Ex(3) B
According to the given figure,
2x F x
C
x
E
3x
Find x. D 3x
A
158 Ex(4)
A
In the given figure m(DC∑B) = m(EC∑D) = m(ED∑A)
E
75°
Find m(BD∑C)
D
B
C
Ex(5)
According to the given figure,
40°
Find x. a a
x 110°
x
Ex(6)
B
According to the given figure, Find x.
D 5y
A
E 6y
140° C
159 REVIEW EXERCISES 1. D a
According to the given figure, find x, y, a, b
85°
x
C b y 40°
A
4.
A
x
3x + 20°
105°
D
4x – 30°
B
C
B
5. 2.
A
A
According to the given
130°
According to the given figure, find x
3x
figure, find m(AC∑B).
E 2x
B
3.
According to the given figure, find x
A 50°
C
E
6.
D
x – 20°
2x + 10° C
According to the given figure, find x
A
B B
20°
B
According to the given figure, find x
C
290°
70°
30°
x D
25°
C
160 11. A
7.
According to the given figure, find x
45°
3x
F
K
54° C
D
50°
144°
2x
L 70°
2x B
According to the given figure, find x
A
B
E
D
x C
8.
E
12.
A
D
50° F
110° C
A
E
According to the given figure, find x F
x
120° B
y
x
D
C
160°
B
K
In the figure, y – x = 40° is given. Find x.
9. 13.
F A
B
150°
According to the given figure, find x
80° C
E x
D
F
140°
A
E
120°
25°
C
B
According to the given figure, find x
D
N
14. A
10.
85° B
According to the given figure, find x
x
C
D
a A
E B
x
150°
C
2a
According to the given figure, find x
100°
x D
161 4.1.2. Congruent Triangles Congruence of Triangles
Two triangles are congruent if i)
their corresponding angles are equal, and
ii) their corresponding sides are equal A c B
D b
f C
a
W) W) = m ( D m (A W m ( B ) = m ( EW) m ( CW) = m ( FW)
E
and
e F
d
_ a = db & , DEF & b = e ` + ABC b c=fa
AB∆C is said to be congruent to DE∆F and denoted by AB∆C ≅ DE∆F
SSS Criterion for Congruence
If three sides of one triangle are equal to three sides of another triangle, then the triangles are congruent. D
A
B
C
Ex(1)
E
F
AB∆C ≅ EF∆D
A
F
B
x
|AE| = |DC| |AF| = |EC| |EF| = |ED|
65°
E
C
D
mFE∑D = 65° Find x.
162 SAS Criterion for Congruence
If two sides and the included angle of one triangle are equal to two sides and included angle of another triangle, then the triangles are congruent. (An included angle is the angle formed by the two given sides) A
D
B
C
Ex(2)
E
F
AB∆C ≅ DE∆F A
In triangle ABC,
% % m (BAD) = m (EDC )
Find AE
14
|AD| = |DE| E
|AB| = |DC| = 14 cm 8 |EC| = 8 cm B D 14 C
ASA Criterion for Congruence
If two angles and the included side of one triangle are equal to two angles and included side of another triangle, then the triangles are congruent. A
A
B
C
E
F
AB∆C ≅ DE∆F
E
Ex(3)
In the figure 30°
|AB| = |EC| is given. Find m(BC∑D)
D 40° C A
30°
x 40°
B
163 Ex(4)
D A 3
|AB| = 3 cm |AE| = 1 cm
1 E
m(BA∑C) = m(DE∑C)= m(DC∑B) = 90°
B
Ex(5)
In the given figure
|BC| = |DC| are given Find |DC|.
C
In the figure,
A
|AD| = |EC|, |DB| = |DE|, D
|BC| = |CD| are given. x
Find x.
55° B
E
C
A
Ex(6)
ABC is an equilateral triangle, A, C and D are collinear. |BE| = |CD|
m(CE∑D) = 10° Find m(BA∑E) B
E
10°
C
D
164 Congruent Right Triangles 1) Hypotenuse - Leg Theorem
If the hypotenuse and one leg of one right triangle are congruent to the hypotenuse and one leg of another right triangle, then the right triangles are congruent.
E
C
A
2) Hypotenuse - Angle Theorem
D
B
If the hypotenuse and one acute angle of one right triangle are congruent to the hypotenuse and one acute angle of another right triangle, then the right triangles are congruent.
E
C
A
3) Leg -Angle Theorem
B
D
F
If one leg and one acute angle of one right triangle are congruent to one leg and one acute angle of another right triangle, then the right triangles are congruent. E
C
A
4) Leg - Leg Theorem
F
B
D
F
If two legs of one right triangle are congruent to the two legs of another right triangle, then the right triangles are congruent. E
C
A
B
D
F
165 Ex(7)
D
C
F
ABCD is a square E
m(FB∑E) = 90°
A
Show that |AE| = |CF|, |EB| = |BF|
B
Ex(8)
A
D
ABCD is a square find m(AF∑E).
F E
C
B
Special Triangles I. I. Right-Angled Triangle:
A right triangle or right-angled triangle is a triangle in which one angle is 90o. In a right triangle, the side opposite to the right angle is called the hypotenuse, and the two sides that meet to form the right angle are called legs. The hypotenuse is always the longest side. B a
c2 = b2 + a2 hypotenuse c
C
b
A
legs
Properties
1.
B
A
D
C
In a right triangle, the median from the right angle to the v is one - half of the length of the hypotenuse. If m(A) = 90° and |BD| = |DC|, then | | | AD | = BC 2
166 2. 90° – 45° – 45° triangle A
B
a
a
45°
45°
C
a§2
3. 30° – 60° – 90° triangle A
a
a§3
60°
30°
B
C
2a
A
Ex(9)
BAC is a right triangle [BA] ^ [AC]
E
|BD| = |DC| = |EC| m(AB∑C) = 40°
B
40°
D
Ex10)
C
In the given figure,
A
[AB] ^ [CE]
|AE| = |BF| = |FC|
E D
m(AB∑C) = 48°
x
Find m(AD∑C) = x
48° B
Find m(AC∑E)
F
C
167 A
Ex(11)
In triangle ABC,
[DA] ^ [AC]
|DC| = 2.|AB| 25°
m(AC∑ B) = 25° B D C
Find m(BA∑D)
A x
Ex(12) z
In triangle ABC,
y
Find x, y and z.
30° B
C
4§3
Ex(13)
D
According to the given figure, find |AD|.
8§3 C 60° A
18
B
A
Ex(14)
According to the given figure, find |BC|.
2 22,5° B
C
168 II. Isosceles Triangle
An isosceles triangle is a triangle with (at least) two congruent sides. A
vertex
|AB| = |AC| ğ m(B) = (C)
B
Properties
C
base
A
1)
B
i) ii)
a/2
H
a/2
C
|AB| = |AC| Ş [AH] = nA = Va = ha
|AB| = |AC| AH ^ BC
iii) |BH| = |HC|
iv) m(BA∑H) = m(CA∑H) If any two of the above conditions are satisfied, then the rest are also satisfied.
2) Isosceles-right triangle: A
m(BA∑C) = 90° |AB| = |AC|
|AH| = |BH| = |HC| 45° B
45° H
C
169 3) 120° – 30° – 30° triangle A
m(BA∑C) = 120°
120°
a
|AB| = |AC|
a
30°
30°
B
C
a§3
4) If |AB| = |AC|, then
A
D
A
F
D
F
E B
E
|CD| = |BF|
C
B
C
|DE| = |FE| |BE| = |CE|
5) If |AB| = |AC|, then A
A
K
H
K
H
L
L
B
C
B
|BL| = |CL|
|BK| = |CH|
|HL| = |KL|
6) If |AB| = |AC|, then A
E
A
N
E
M
B
|BN| = |CE|
C
N M
C
B
|EM| = |NM| |BM| = |CM|
C
170 A
Ex(15)
ABC is an isosceles triangle, |AC| = |BC|
20°
m(AD∑B) = 50° m(DA∑C) = 20° Find m(BA∑D)
50° B
D
C
In the figure,
A
Ex(16)
|AD| = |BD| |AE| = |EC|
m(DA∑E) = 50° B D E
C
Find m(B∑AC)
A
Ex(17)
In the figure, |AB| = |AC|
35°
m(BA∑D) = 35°
E B
|AD| = |DC|
C
25°
x
m(BC∑D) = 25° Find x.
D
A
Ex(18)
According to the given figure, Find m(AD∑E) E
B
D
C
171 Ex(19)
A
In the figure, [AD] is angle bisector |BD| = |DC|
m(AD∑C) = 110°
D
110°
Find x.
x B
C
A
Ex(20)
ABC is a right angled triangle, [ED] ^ [BC]
E
|AE| = 2 cm B
D
C
A
Ex(21)
|BE| = 4 cm
Find m(AB∑C)
In the figure, |AD| = |AC|
B 3
D
18
|BD| = 3 cm
|DC| = 18 cm
C
m(DA∑C) = 90°
are given. Find |AB| A
Ex(22) F
m(BA∑C) = 120°
120° 4
B
In the figure, |AB| = |AC|
x D
E
[DE] // [AB] C
[DF] // [AC]
|BC| = 6§3 cm |DF| = 4 cm
are given. Find |DE| = x
172 A
Ex(23)
ABC is an isosceles triangle |AB| = |AC|
H
A, C and D are collinear.
3
[CH] ^ [AB]
|CH| = |BH| + |DC|
C
B
|CH| = 3 cm are given. Find |BD|
D
III. Equilateral Triangle
An equilateral triangle is a triangle with all sides congruent. A 60°
a
a 60° B
Properties
1)
60° C
a
A
C
H
B
ha = h b = h c = V a = V b = V c = n A = n B = n C = h
2)
A 30° 30°
h = a a a 3 2
60°
B
a/2
60°
H
a/2
C
a 3 2
a.h Area = 2 a2 3 h2 3 = = 4 3
173 A
Ex(24)
ABC is an equilateral triangle. |AB| = |BD|
D
m(AB∑D) = 10° Find m(AC∑D) = x
x
10° B
C
ABC is an isosceles right triangle, ADC is an equilateral triangle,
A
Ex(25)
Find m(DB∑A) = x.
D x B
C
A
Ex(26)
ABC is a right triangle
D
[DE] ^ [BC]
|AC| = |BE| = |AE| Find m(BD∑E). B
E
Ex(27)
C
A F
8
ABC is an equilateral triangle
|AD| = 8 cm, |FB| = 10 cm D 10
B
x
E
C
Find |BE|.
174 Ex(28)
ABC is an equilateral triangle.
A
[PE] // [AB]
E
[PF] // [BC]
3 F
[PD] // [AC]
1 P
|PF| = 1 cm
2 B
D
C
|PD| = 2 cm |PE| = 3 cm
Find |AB| = x A
Ex(29)
ABC is an equilateral triangle. |AB| = 6 cm
B
D
C
|AD| = 2§7 cm
Find |BD|
A
ABC is an equilateral triangle.
Ex(30) D
|AD| = |DB| , |DE| = 5 cm,
|EF| = 2 cm are given.
F E
B
C
D
Ex(31)
12
ABC is an equilateral triangle. x
|AE| = |EC|
E
Find |DE| = x
A
B
Find |AF|
H
|BH| = 12 cm
C
175 REVIEW EXERCISES - I 1.
A
4. A
D
B
C
Given AB∆C ≅ FE∆D
Find |DF| + |EF|
E
F
D
6
8
2
C
x B
D
E
C
E
|BD| = |EC|
m(DA∑E) = 70°
B
Find x
In the given figure,
|AB| = |AE|, m(BA∑D) = m(DA∑C)
3.
E A
In the given figure, |AC| = |CE|
B
A
|AB| = |AC|
Find |EF|
In triangle ABC,
70°
|BE| = 2 cm
E
B
A
|AE| = 6 m
F
5. 2.
|AB| = |BC|
9
6
In the given figure,
C
D
C
m(BD∑A) = 2 . m(ED∑C) are given. Find m(ED∑C)
A
6.
ABC is an isosceles triangle. |AB| = |AC|
|DE| = 6 cm
Find |AB|
|BD| = 10 cm
D
D
B
K
|DK| = 3 cm
E
C
|KC| = 5 cm Find |BE|
176 7. C
A
80°
9.
|AB| = |AC|,
|AD| = |AE|
80°
In the figure,
D Find x.
F x
B
A D E
|AD| = |BE|
B
ABC is an equilateral triangle.
A
E F
B
D
F
C
|AE| = |BD|
Find x.
x C
|AE| = |BF|
m(AC∑B) = 36° Find x.
A
10.
|AC| = |BC|
x
E
8.
ABC is an isosceles triangle
D
E
|DC| = 10 cm
B
ABC and ADE are equilateral triangles.
C
Find |BE|
177 REVIEW EXERCISES - II A
1.
A
4. E x
50° B
C
D
In triangle ABC,
m(BA∑C) = m(DE∑C) = 90°, |BD| = |DC|
60°
B
According to the given figure, find |BC|.
m(AB∑C) = 50°. Find m(AD∑E) = x
D
18
C
ABC is triangle
A
5. ABC is a right triangle
A
2.
|AD|=|BE|=|EC|
D 70°
B
m(AB∑C) = 70°
x E
C
A
3.
ABC is a right triangle
m(AC∑B) = 45°
E
45° B
6.
|ED| = 3§2 cm
4§2
3§2
C
C
|AE| = |EB| = |CD|
E
Find m(DE∑C) = x
15°
Find m(ED∑B) 60° B
60° C
D
A
|DC| = 4§2 cm Find |EB|
According to the given figure, find |BC|.
m(AB∑D) = 60°
right
[ED] ^ [AC]
D
a
1
D
B
178 A
7.
x
6
45°
30°
B
C
8.
A
x
105°
8§2
In triangle ABC,
m(AB∑C) = 45°
m(AB∑C) = 135°
m(AC∑B) = 30°
|AC| = 6 cm
B
C
A
E
B
D
C
B
m(BA∑C) = 105°
Find |AB| = x
ABC is a right triangle
12.
m(AC∑B) = 20°
20°
135°
|AC| = 8§2 cm
|BD|=|DC|=2.|DE|
Find m(EA∑C)
m(AC∑B) = 30°
x
11. A
[AE] ^ [ED]
10
In triangle ABC,
m(AB∑C) = 45°
45°
9.
10. A
Find |AB| = x
In triangle ABC,
30°
C
|AC| = 10 cm Find |AB| = x
In triangle ABC, 30°
m(BA∑C) = 30°
x
m(AC∑B) = 15° B
D
60°
A
C
|BC| = 6§2 cm Find |AC| = x
In the given figure, m(DA∑B) = 60°
7§3
6§2
C
15°
m(AB∑C) = 60° 60° 18
[DC] ^ [CB]
B |AB| = 18 cm
|DC| = 7§3 cm
Find |AD|
179 13. A 6 D
75° 135° B
According to given figure, |BC|.
the find
Note: 90° – 15° – 75° triangle A
2§2
h
C 75° B
15°
H
C
4h
A
14.
ABC is a right triangle
4
15°
B
|AB| = 4 cm C
15. A 8 E
30°
B
D
ABC is a right triangle
[AD] ^ [DE], |BD| = |DC|
m(ED∑C ) = 30°, |AE| = 8 cm
Find |BC|
C
Find |AC|
16.
BAC is triangle
A
|BC| = 16 cm
m(AB∑C) = 15°
15° B
a
C
Find A(AB∆C)
right
180 REVIEW EXERCISES - III A
1.
In triangle ABC, |AD| = |BD| = |AC|
m(AC∑B) = 72°
Find m(AB∑C) = x
B
72°
x
D
In triangle ABC, |AB|=|BD|=|AC|
40°
m(DA∑C) = 40°
B
3.
Find m(AB∑C). C
D
In triangle ABC,
A
|BD| = |AD|,
x B
In the figure,
x
|AB| = |BD| = |BC|
D
m(CA∑D) = 20°
D
E
m(AC∑D) = 10° B
Find m(BA∑C) = x C
A
C
A
2.
4.
E C
|ED| = |EC|,
m(DB∑A) = m(DA∑E) = 40°. Find m(AD∑E) = x
A
5.
In the figure, D |AB|=|AC| = |AD|
35°
E
m(BD∑C) = 35° Find m(BA∑C)
B
C
A
6.
In triangle ABC, |AE| = |ED| = |EC|
x
|BD| = |DC|
E
m(AC∑B) = 65° Find m(BA∑D) = x
65° B
D
C
181 A
7. D
11.
[AC] ^ [BC]
[ED] ^ [DC]
ABC and EDC are right triangles
B
|AD| = |BD| = |DE|
C
m(BA∑C) = 55°
E
Find m(BC∑E)
8.
|AB| = |AC|
D
[CD] ^ [AB]
E
F
3
[BE] ^ [AC]
|FE| + |FC| = 6 cm
B
C
|DB| = 3 cm
Find |BC|
ABC is an isosceles triangle
A
10
|AB|=|AC|=10 cm
10
B
In triangle ABC,
A
D 1 C
15
|BD| = 15 cm
Note: If |AB| = |AC|, then A
|DC| = 1 cm Find |AD|
|PD| + |PE| = |CH|
H
ABC is an isosceles right triangle
A
9.
B
8
D B
[AB] ^ [AC]
x
D
2
C
E
P
C
|AB| = |AC|
|BD| = 8 cm |DC| = 2 cm
Find |AD| = x
A
12. 10.
D
x
A
30° B 6
E
9
30°
C
ABC is an isosceles triangle
In triangle ABC,
m(DE∑B) = 90°
m(AB∑C) = 30°
[PD] ^ [AB]
m(AC∑B) = 30°
m(BA∑C) = 30°
|BE| = 6 cm
D
B
Find |AD| = x
|EC| = 9 cm
|AB| = |AC|
30°
[PE] ^ [AC] 4
2 P
E C
|PD| = 2 cm
|PE| = 4 cm Find |AB|
182 REVIEW EXERCISES - IV A
1.
In triangle ABC,
110°
D
m(AC∑B) = 60°
D
m(AE∑D) = 110°
Find m(BA∑C) = x
60° B
C
ABC is an equilateral triangle
A
E B
m(BC∑D) = 145° Find m(BD∑C) = x
C
ABC is an equilateral triangle
F
E
D
B
|DB| = |BE| = |AF|
4
C
|FC| = 4 cm Find |DE|
E
B
A
In the figure
DEF and CFB are equilateral triangles. DE Find CF
F
A
5.
ABC is an equilateral triangle.
G F
|EF| = 6 cm,
E
A
3.
x
m(BA∑D) = 70°
D
30°
2.
C
|BE| = |BC| = |DC|
x
E
4.
B
3|AF| = 2|FC| are given.
D
C
A
6.
ABC is an equilateral triangle. |BD| = 2 cm
x
|DC| = 6 cm
B
2
D
Find |ED|.
6
C
Find |AD| = x
183 ABC is an equilateral triangle.
A
7.
|BD|=|DE|=|EC|,
B
D
E
C
|AB| = 6 cm are given. Find |AE|.
ABC is an equilateral triangle
A
9.
F
5
1
[PF] ^ [AB]
3 B
D
C
A
ABC is an equilateral triangle
|AD| = |DB|
D
E
F
B
Find |BE|
|CD| = 6§3 cm C
A
E F
B
P
h
D H
C
D 2
[PD] ^ [AB]
P
[PF] ^ [BC]
Note: ABC is an equilateral triangle with one side a, a 3 then | PD | +| PE | +| PF | = h = 2
ABC is an equilateral triangle
A
|PE| = 5 cm Find |AB|
10.
|AE| = 2|EC|
|PD| = 3 cm |PF| = 1 cm
8.
[PE] ^ [AC]
P
[PD] ^ [BC]
E
B
F
C
|BF| = |FC|
|PD| = 2 cm
A(AB∆C) = 48§3 cm2 Find |PF|
184 REVIEW EXERCISES - V In triangle ABC,
A
1.
|AB| = |AC| = |BD|
m(DB∑C) = 24°
D
Find m(AB∑D)
24°
B
40°
E
50°
B
5.
|AB| = |AC|
|CE| = |CB|
m(EC∑B) = 50°
B
56°
|CF| = |CE|
m(BA∑C) = 100° B
E
C
C
In triangle ABC,
D
x
[DE] ^ [AC]
Find m(DE∑F)
m(AB∑C) = 56°
m(BC∑A)= x
|AB| = |BD|
E
|AE| = |ED| = |DC|
Find m(AE∑D) = x D
C
In triangle ABC,
A
Find
ABC is a right triangle
x
6.
|AE| = |EC|
C |BA| = |BD|
A
B
Find m(AB∑E)
E
|BD| = |BE|
F
100°
D
ABC is an isosceles triangle
A
m(BA∑C) = 40°
3.
In triangle ABC,
A
C
A
2.
4.
[ED] is angle bisector
D
B
E
C
|AD| = |DC|
|BE| = |EC|
Find m(BA∑C)
185 7.
9.
A
D
B
E
ABC is a right triangle
[AB] ^ [BC]
Find m(AC∑B)
30°
[AB] ^ [BC]
|AD| = |DC| = |DE| m(CD∑E) = 30°
x
A
Find m(CB∑E) = x
B
|AB| = |BD| = |DE| = |EF| = |FC|
D
D
C
8. A
E
F
ABC is a right triangle
C
10.
[BD] ^ [BC]
|BD| = |BC|
x B
In triangle ABC,
C
|AB| = |DC|
Find m(DB∑A) = x
ABC is an isosceles triangle
A
D
F
1
|AB| = |AC|
E
[BE] ^ [CD]
m(AB∑E) = m(A∑CD)
5
B
C
|FE| = 1 cm
|BD| = 5 cm Find |BC|
186 11.
E
A
In triangle ABC
30°
m(BA∑C) = 30° H
B
x
C
[BH] ^ [AC]
|AE| = |AH|
D |AB| = |AC| = 8 cm
|CD| = 4 cm
Find m(ED∑B) = x
A
13.
[DE] ^ [AB]
E
Find |AE| + |AF|
In triangle ABC, |AC| = |BC|
2 D
Find |DC|
3 B x
B
In triangle ABC
|AC| = |AD| = |AE| = |BD|
Find m(BD∑E) = x
C
3
E
D
|DF| = 3§3 cm
A
|ED| = §3 cm
3§3
§3
B
A
[DF] ^ [AC]
F
14.
12.
ABC is an equilateral triangle
D
C
C
187 4.1.3. Angle-Side Relations In a triangle, the longest side is opposite the largest angle, the shortest side is opposite the smallest angle. A b
c
B
a
C
W) > m ( BW) > m ( CW) + a > b > c m (A
Ex(1)
In triangle ABC,
A
m(AB∑C) = 75° c
b
75°
55°
m(BC∑A) = 55° Order the sides of ABC.
B
a
C
A
Ex(2)
80°
50° 65° 60° B
C
D
Order the sides in the figure.
188 Triangle Inequality Theorem
The sum of the lengths of any two sides of a triangle is always greater than the length of the third side.
Proof
A
|b – c| < a < b + c
c
b
B
a
Ex(3)
|a – b| < c < a + b
C
In triangle ABC,
A
|AB| = 4 cm |AC| = 6 cm
6
4
B
|a – c| < b < a + c
Find the integer values that
x
|BC| = x can take
C
Ex(4)
A 3a
5a
B
In the given triangle find an interval for a.
16
C
A
Ex(5)
In the given figure, find the possible integer values of x.
6 14
B
x
D 3
8
C
189 A
Ex(6)
In the triangle ABC,
7
W) < m ( CW) m (A
5
B
2x–3
Find the sum of the integer values of x.
C
Note: A
A b
c
c
A b
b
c B
a
C
B
a
m(B) = 90°
m(B) < 90°
b2 = a2 + c2
b2 < a2 + c2
8
6
B
a
C
m(B) > 90° b2 > a2 + c2
If m(A) < 90° , then find the possible integer values of |BC|
A
Ex(7)
C
C
B
Ex(8)
A E
5 B
Find the integer value of x in the given figure.
12 D
x 4
11 C
190 REVIEW EXERCISES 1.
In triangle ABC,
A
m(B) = 48°
D
Order the sides of ABC.
48° B
62°
a
A
m(C) = 62°
b
c
4.
B
According to the given figure, find the longest side
50°
C
75° C
A
2.
In the given figure,
35°
m(BA∑D) = 35°
m(AD∑B) = 75°
B
75° D 55°
25°
5.
D
m(DC∑B) = 25° Order the sides of ABC
E
C
A
Find the shortest side in the figure.
85° 55° 35° C 75°
80°
B
3.
A
In triangle ABC,
39°
m(A) = 39°
m(B) = 51°
b
c
6.
A
51° B
a
Simplify
|a – c| – |b – c| – |b – a|
4
7
C B
x C
D
Find the possible integer values of x in the given figure.
191 7.
|AB| = 8 cm
14
9
In triangle ABC,
A
10.
A
12
8
|AC| = 12 cm
B
2x–1
m(B) > m(A)
C
In triangle ABC, find the number of the possible integer values of x.
B
x
Find the number of the integer values of |BC| = x.
11. 8.
A
x+2
2x
B
6
In triangle ABC, x is an integer.
Find the number of the possible values of x.
C
9.
According to the given figure, find the possible integer values of x.
A 4
8 x
B
C 5
2 D
In triangle ABC,
A
m(C) < m(A) x+3
C
B
5
2x–1
|AC| = 5 cm C
|AB| = x + 3 cm
|BC| = 2x – 1 cm
Find the sum of the integer values of x.
192 12.
In triangle ABC,
A
|AB| = 3 cm
3
|BC| = 4 cm
x
m(AB∑C) < 90° B
4
C
In triangle ABC,
A
3
|AD| = |AC| = 3 cm
|BD| = 4 cm
3
B
4
D
C
In the figure,
A 12
5 x
B
D
9
Find the minimum integer value of |AB|
C
m(BC∑D) < 90°
In triangle ABC,
A
m(BA∑C) < 90°
4
12
|BD| = |DC|
|BC| = 9 cm
Find the integer value of |BD| = x
15.
|AB| = 5 cm m(BA∑D) > 90°
given
|AD|=|DC|=12 cm
12
Find the possible integer values of |AC| = x.
13.
14.
|AB| = 4 cm B
D
C
|AC| = 12 cm
Find the possible integer value of |AD|
193 4.2. Similarity of Triangles 4.2.1. Similarity of Triangles Similarity of Triangles
Two triangles are similar if i) their corresponding angles are equal, and ii) their corresponding sides are proportional D A
B
C
E
F
We say that AB∆C is similar to DE∆F and denote it by writing AB∆C ~ DE∆F
If AB∆C ~ DE∆F, then by definition m(A) = m(D) m(B) = m(E)
and
AB = DE
BC = EF
CA = k FD
m(C) = m(F) where k is the similarity ratio If the similarity ratio k = 1, then the triangles are congruent.
AAA Criterion for Similarity
If the corresponding angles of two triangles are equal, then the triangles are similar. D A
B
C
E
m(A) = m(D)
m(B) = m(E) m(C) = m(F)
& ~ DEF & & ABC
F
194 Ex(1)
A
According to the given figure, find x + y.
3
E
y 6
4 B
x
D
3
C
A
Ex(2)
According to the given figure find |AD|
D
6 3 B
E
5
C
A
Ex(3)
In the figure [DE] // [BC],
9
|AD| = 9 cm, |BD| = 3 cm
E
D
Find |BC|.
3
B
C
Ex(4)
C B
x 8
9
E
6
D
A
Ex(5)
According to the figure find |EC|.
In triangle ABC,
A 6 10
B
m(AB∑D) = m(AC∑B)
D
|AB| = 10 cm
x
C
|AD| = 6 cm
Find |DC| = x
195 In triangle ABC,
E
Ex(6)
A
m(EA∑C) = m(AD∑B)
8
|AB| = 8 cm
4
B
D
3
|DC| =3 cm
C
|AD| = 4 cm Find |BD|
SAS Criterion for Similarity
If one angle of a triangle is equal to one angle of the other triangle and the sides containing these angles are proportional, then the triangles are similar. D A
B
C
E
F
AB W) and = W) = m ( D m (A DE
AC & ~ DEF & = k & ABC DF According to the given figure
A
Ex(7)
2
4 E
Find |BC|.
D
3
10
2 B
C
Ex(8)
In the given figure,
E
|AC| = 10 cm
9 A
10 5
15
C
D
6
|AB| = 5 cm |BC| = 6 cm |CE| = 9 cm
B
|CD| = 15 cm Find |ED|
A
Ex(9)
x 16
D
12
15
B
20
C
According to the given figure, find x
196 Ex(10)
A 12
B
In the given figure, |AB| = 12 cm |AD| = 6 cm
6
9
D
|BD| = 9 cm
C |DC| = 7 cm
7
are given Find |AC|
SSS Criterion for Similarity
If the corresponding sides of two triangles are proportional, then the triangles are similar. D A
B
C
AB = DE
E
BC = EF
F
AC & ~ DEF & = k & ABC DF
AB∆C ~ DF∆E is given.
D
Ex(11) A 6
9 7
B
C E
F
A
Ex(12)
In the given figure, 16
25
15
|AB| = 25 cm
E
20 12
B
Find |DE|
18
D
|BD| = 15 cm
x C
|AD| = 20 cm |AE| = 16 cm
|DE| = 12 cm Find |EC| = x
197 Properties of Similar Triangles
1) In two similar triangles, the ratios of the •
corresponding angle bisectors,
•
corresponding altitudes,
•
corresponding medians,
•
perimeters
are equal to the similarity ratio ‘k’.
2) If k = 1, then the triangles are congruent.
ABC and DEF are similar triangles
A
Ex(13)
D
[AP], [DK] are angle bisectors 2|AP| = 3|DK| B
P
C E
K
F
|EF| = 6 cm Find |BC|
Ex(14)
In triangle ABC,
A
D
|AF| = 2 cm, |FG| = 4 cm
E
F
P(AD∆E) = 12 cm Find P(AB∆C)
B
G
C
198 REVIEW EXERCISES 1.
3
A
[AB] // [CD]
B
According to the figure,
2 E
Find |AD| . |CD|
6
3
D In the given figure
E
[AE] // [DC]
8
m(EA∑C)=m(EB∑D) = 90°,
A
6
B
12
C
|BC| = 12 cm
2.
According to the given figure find x.
A 4
6
Find |DC|
6.
In triangle ABC,
A
D
[DE] ^ [BC]
x
2
[AB] ^ [AC]
E
E
B
C
B
10
D
6
C
A
8
B
6 E
4.
9
D
x
E 6
B
4
F
m(BA∑E) = m(AE∑D)
|AB| = 8 cm
|BD| = 21 cm
C
In the figure
8.
[DE] // [BC],
|AE| = 9 cm and
Find x.
|BF| = 4 cm,
|EC| = 6 cm
|BD| = 10 cm Find |AE|
In triangle ABC,
A
m(AB∑C)=m(DA∑C)
x B
9
D
|BD| = 9 cm
C |DC| = 3 cm
3
Find |AC| = x
Find |CD|
A
7.
D |ED| = 6 cm
C
In the given figure,
|AE| = |EC|
|DC| = 6 cm
3.
|AB| = 6 cm |AE| = 8 cm
D
C
5.
In triangle ABC,
A
m(DA∑B) = m(BC∑A) 4
3
B
2
D
Find the perimeter of AB∆C.
C
|BA| = 3 cm
|AD| = 4 cm
|DB| = 2 cm
199 C
9.
E
D
6
4
6
x
F
B
C
A
10. D
E
4
3 6
F
B
H
C
In triangle ABC,
12.
m(BD∑C) = m(BE∑C)
|AD| = |BD| = 6 cm
Find |EC| = x
|AE| = 4 cm
A 6 B
D
3
E
In triangle ABC,
13.
A
|DE| = 4 cm
|BC| = 12 cm
|EF| = 3 cm
Find the perimeter of AB∆C.
C x
|DF| = 6 cm
In the given figure,
4
In triangle ABC,
2
|AD| = |DB| = 4 cm |AE| = 2 cm
14
B
|AB| = 6 cm
Find |EC| = x
E
4
m(AC∑B) = m(AE∑C)
|BD| = 3 cm
5
D
[AB] // [CE]
C
|DE| = 5 cm
|EC| = 14 cm Find |BC|
A
11.
F
3
x
6
14.
E
D B
15
C
In triangle ABC, m(DA∑C) = m(FC∑B) = m(EB∑A)
|AC| = 6 cm, |BC| = 15 cm, |FE| = 3 cm, Find |ED| = x
In triangle ABC,
A E
5
|AE|=|BD|=5 cm
21
10
B
5
D
25
C
|EB| = 10 cm
|DC| = 25 cm |AC| = 21 cm Find |ED|.
200 15.
D
8
6
12
A
18.
In the given figure,
C
B
18
[DC] // [AB] |DC| = 8 cm
9
a
B
Find |BC|
3x
18
|AB| = 12 cm, |AC| = 9 cm, |BC| = 3x
12
B
15
D
|AB| = 18 cm
x
10
C
Find m(DF∑E) = a
3 D
|AD| = 10 cm
2
F 6 B
4 E
|AE| = |FK| = 4 cm |DE| = 2 cm
10
4 8
|AD| = 3 cm
K
x
|BF| = 6 cm
C
|BK| = 8 cm
Find |AC| = x
In triangle ABC,
A
|BD| = 12 cm
F
|DE| = 6 cm, |EF| = 8 cm, |DF| = 2x
|DC| = 15 cm
8
In triangle ABC,
E
In triangles ABC and DEF,
A
C
19. 16.
2x
6
60°
|AD| = 6 cm
|AB| = 18 cm
D
50°
12
|AC| = 12 cm
A
|EC| = 10 cm Find |KC| = x
20.
D
A
17. D
A
B
16
12
9 15
C
E
12
20
F
According to the given figure, name the similar triangles.
K
C
E
N
B
In similar triangles ABC and DEF,
[AK] and [DN] are medians respectively.
F
5|AK| = 3|DN|
If the perimeter of AB∆C is 15 cm, then find the perimeter of DE∆F.
201 4.2.2. Triangle Proportion Theorem If a line is drawn parallel to one side of a triangle, then the other two sides of the triangle are divided proportionally. A
6DE @ // 6BC @ & D
AD AE = DB EC
E
B
C
The converse of the above result is stated as follows: If a line divides any two sides of a triangle in the same ratio, then the line is parallel to third side of the triangle.
A
Ex(1) 6
[DE] // [BC] |AD| = 6 cm
x
|DB| = 4 cm
E
D
6
4 B
C
A
Ex(2)
|EC| = 6 cm
are given. Find |AE| = x
[DE] // [BC]
3|AD| = 4|DB| D
B
|AC| = 21 cm
E
C
are given. Find |EC|
202 A
Ex(3)
[DE] // [BC] [EF] // [AK]
K
8
According to the given figure, find x.
x
E
D 4
8
B
F
C
A
Ex(4)
[DF] // [BC]
[DE] // [BF]
6 E 4
D
|AE| = 6 cm F
|EF| = 4 cm x
B
are given. Find |FC| = x C
Theorem
A B
C
D
[AD] // [BE] // [CF]
E
AB DE = BC EF
AB DE = AC DF
F
Ex(5) 5
d1
8
d1 // d2 // d3 Ş x = ? d2
15
and
x d3
203 Theorem
A
D
[DE] // [BC] AD = AB
E
B
DE BC
C
Ex(6)
A
In triangle ABC,
3
[DE] // [BC]
6 D
|AD| = 6 cm
E
|DB| = |DE| = 3 cm
3 B
x
Find |BC| = x
C
A
Ex(7)
F
In the given figure,
5
B
E
3
C
15
A
Ex(8)
F
Find x.
D
E
21
[EF] // [BK], [ED] // [BC]
2
x
K
D
AE = AC
In the given figure,
B
x
20
[AB] // [EK] // [DC] Find x.
K
C
204 D
Ex(9)
In quadrilateral ABCD,
C
[DC] // [EF] // [AB]
3
Note: In a triangle, the midline, joining the midpoints of two sides is parallel to the third side and half as long. Conversely, the line joining points on two sides of a triangle, parallel to its third side and half as long is a midline.
6
x
E
|DE| = 3 cm
F
|EA| = 2 cm
2 A
11
|DC| = 6 cm
B
|AB| = 11 cm Find |EF| = x
A M
B
x Midline 2x
N
C
Ex(10)
In triangle ABC,
A
[BE] // [ND]
E
F 6 B
[BF] = 6 cm
D N
|BN| = |NC| |AF| = |FN|
C
Find |ND|
Ex(11)
A
In triangle ABC,
4
|AF| = |FD|
E
F
|AE| = 4 cm
10
|EC| = 10 cm B
6
D
C
|BD| = 6 cm Find |DC|
205 Theorem
B
A
[AB] // [DE]
C
AB = DE D
AC = CE
BC CD
E
In triangle ABC,
A
Ex(12)
[ED] // [BC]
6 E
3|FC| = 5|FE|
D F
Find x.
x B
C
Ex(13)
In the given figure,
B
A
[AB] // [DF] C
|AC| = |GF| = 2|CG| G
|FE| = 4 cm Find |DE|
D
Ex(14)
E
D
A
Note: E
B
10
C
z
x D
F
[AB] // [CD] // [EF] 1 1 1 & x = y+z
B
F
In the given figure, [AB] // [EF] // [CD]
E
15
x
y A
F
C
Find |EF|
206 REVIEW EXERCISES 1.
A
D
8
10
5.
[DE] // [BC]
|AD| = 10 cm
E
B
C
y
2
6
x
2.
A
E
D
Find x + y
d3
are given.
B
C
In triangle ABC,
[DE] // [BC]
3|AE| = 2|EC|
|AB| = 30 cm
3.
A
K
E
D
F
B
In triangle ABC,
A
[DE] // [BC] x
D
|AD| = 2|DB|
E
|BC| = 15 cm B
15
Find |DE| = x
C
Find |AD|
7. In the given figure,
[DE] // [BC],
[EF] // [AK]
3|AD| = 4|DB|
In triangle ABC,
A
[DE] // [BC] E
D
|BC| = 3|DE|, |AE| = 2 cm
B
C
Find |EC|
|FK| = 12 cm
C
Find |CK|
4.
In the given figure,
A
A
8.
F
3 D
E
5 B
C
ADEF is a parallelogram,
9
[DE] // [BC],
9
d1 // d2 // d3
|EC| = 3 cm
6.
4
Find |DB| = x
In the given figure,
d1 d2
|AE| = 8 cm
3
x
In triangle ABC,
[FE] // [DC]
|FD| = 3 cm
x
|BD| = 5 cm
Find |AF|
F
D
B
2 E
|AF| = 9 cm
3
|FC| = 3 cm C
|FE| = 2 cm Find |DB| = x
207 9.
In the given figure,
13.
[DE] // [BC]
[EG] // [AF]
|DE| = 8 cm
|BC| = 12 cm
|EG| = 6 cm
Find |AF| = x
A
x
8
D
F
E
6
B
12
G
C
A E
E
A
8
x
F 6
B
D K L 3
M
C
11.
B
F
E
x
[LM] // [ED]
B
Find x.
|BC| = 30 cm
K
Find |FK| = x
B
C
30
12.
In triangle ABC,
A
[EF] // [BC]
D
E
x
|BF| = |FD| 5|AD| = 3|DC|
F
|BC| = 20 cm B
20
C
Find |EF| = x
C
|AK| = |KC| |ED| = 8 cm
In the given figure,
[KL] // [AE],
2|AE|=6|EK|= 3|KC|
F
A
|BE| = 4|AE|
Find |FK| = x
[FK] // [BC],
[DE] // [FK] // [BC] D
D
[FK] ^ [BC]
x
In the given figure,
In triangle ABC,
A
[ED] ^ [BC] K
14. 10.
In triangle ABC,
9
14
x
[AB] // [CD]
C
2
|AB| = 14 cm
y
O
|AO| = 9 cm
3
|OD| = 3 cm
D
|OC| = 2 cm
|BO| = x
|CD| = y
Find x . y
15.
A 6
B
10 y
In the given figure,
F 8
[AF] // [BE] // [CD]
E
12
|AF| = 10 cm x
C
|AB| = 6 cm D
22
|BC| = 12 cm
|FE| = 8 cm
|CD| = 22 cm
Find x + y
16. A
In triangle ABC,
[DB] ^ [BC]
x
D
|AD| = |DC|
3
|BC| = 6 cm
B
6
C
|BD| = 3 cm Find |AB| = x
208 A
17.
K
D
x
9
E
In triangle ABC,
21.
[BK] and [CK] are angle bisectors
[DE] // [BC]
6
|AD| = 9 cm
B
30
C
|DB| = 6 cm
|BC| = 30 cm
Find |AE| = x
18.
In triangle ABC,
A
|DK| = 3 cm
E
D 3
|KC| = 5 cm
x
K
5
[DE] // [BC]
4
B
A
[AB] // [DE]
|BK| = 4|KF|,
B
Find |EC| = x
A E
Find |KB| D
22.
A K
D
B
C
|BF| = 24 cm
B
C
A
23.
B
K
[AB] // [CD] x Find y
1 F 5
C
x
E
y
D
Note: Menelaus Theorem
A
Find |BC|
A
In the given figure, K
[EK] // [BA]
|FK| = 2|FE| B
6
D 2 E
F
In the given figure,
4
x
C
Find |FC| = x
K
B
24.
A
[MK] « [CK] = {K}
then | KB | . | CM | . | AL | | KC | | MA | | LB | = 1 C
In triangle ABC,
3|AE| = 2|EC|
E
F
3|BD| = 4|DC|
6
x
K, B and C are collinear
M
L
In the given figure, [DE] // [BC], |AD| = 3 cm |BD| = 9 cm, |KE| = 8 cm |DK| = 4 cm, |AC| = 16 cm Find |KL|
[DF] // [BC] 5|DE| = 4|DF|
20.
E L
[DE] // [BF], F
D
C |FC| = 5 cm
In the given figure,
F
K
|EF| =|FD|
E
given
|AE| = 4 cm
C
19.
In the figure,
B
|FE| = 6 cm D
C
Find |BF| = x
209 4.3. Bisectors, Median and Altitude 4.3.1. Angle Bisector Angle Bisector
Angle bisector of a triangle is a line, segment, or a ray that bisects the angle and is extended from any of the three angles to the side opposite the angle. An angle bisector divides the angle in half, resulting in two congruent angles which are half the measure of the vertex angle that is bisected. A
The angle bisector divides the angle into two congruent triangles B
Constructing the bisector of a given angle: B
N
C
Given: Angle A Construct: The bisector of angle A Procedure: 1) Begin with angle A.
B
2) Using a compass set with any radius and its center point on point A, draw an arc that intersects each side of angle A. Label the intersection points B and point C.
B
3) Using a compass with point B as the center and a radius greater than half the distance between points B and C, draw an arc beyond the BC arc. Then using the same compass radius and the center on point C, draw an arc that intersects the arc made from point B. Label where the arcs intercept as point D.
A
C
A
C
D
A
C
B
A
D
D
C
• Note that length AB = length AC, and the distance between points B and D is equal to the distance between points C and D. • Note also that two congruent triangles ABD and ACD, are formed (SSS postulate), with angle BAD corresponding to and being congruent to angle CAD. this confirms that line AD is the bisector of angle BAC.
C
B
A
4) Draw a line between points A and D using a straightedge. Therefore, line AD is the bisector through angle A.
210 Properties
1) Each point lying on the bisector of an angle is equidistant to the arms of that angle. A
[OC angle bisector ğ |CA| = |CB|
C
O
B
[OC angle bisector
A
C
O
[OA ^ [CA]
ğ OA∆C ≅ OB∆C
[OB ^ [CB]
B A
Ex(1)
According to the given figure, find x. 17
5
C
B x
5 D
2) Angle bisectors of a triangle intersect at a constant point. This point is the center of the inscribed circle (incircle) and it is called the incenter.
|AN| = nA ,
A
K
I
B
N
|BM| = nB ,
|CK| = nC ,
M
I: incenter
C
N I-incenter N N
N
211 Ex(2)
In triangle ABC,
A
[AP] and [BP] are angle bisectors
2
[HP] ^ [AB]
x
H
|AH| = 2 cm
P
3 B
7
|BH| = 3 cm
C
|BC| = 7 cm
Find |AC| = x 3) Two exterior angle bisectors and the interior angle bisector of the third angle intersect at a point and it is called the excenter. A
B
EA: excenter of AB∆C
C
EA
A
EB: excenter of AB∆C
EB B
Aı
C
EB
C
EA
I B
A
EC
EA, EB, EC: excenters of AB∆C
212 A
Ex(3)
80°
D
50°
According to the given figure find x.
x
40°
B
70° C
4) i)
A
W) m (A % m ( BIC ) = 90° + 2
I
b B
a a
b
C
A
ii)
B
a
b
a
W) m (A % m (BKC ) = 90° − 2
C b
K
iii)
A T
2a
B
Ex(4)
W) m (A % m (BTC ) = 2
a b
a a
C
b
In triangle ABC,
A 100°
[BD] and [CD] are angle bisectors
D
B
m(BA∑C) = 100° C
Find m(BD∑C)
213 A
Ex(5)
In triangle ABC, [BD] and [CD] are angle bisectors
x
m(BD∑C) = 120°
D
Find m(BA∑D) = x
120° B
C
Ex(6)
A
In triangle ABC,
70° B
[BD] and [CD] are exterior angle bisectors C
m(BA∑C) = 70° Find m(BD∑C)
D
Ex(7)
In triangle ABC,
A
D
80°
[BD] is interior angle bisector, [CD] is exterior angle bisector
E
C
m(BA∑C) = 80°
B
Ex(8)
A
E
Find m(BD∑C)
In triangle ABC, [AF], [BE] and [CE] are angle bisectors m(BF∑C) = 120° Find m(BE∑C)
F 120° B
Ex(9)
C
D
A E D 96° x B
C
According to the given figure, find x.
214 Angle Bisector Theorem
The angle bisector of an angle in a triangle divides the opposite side in the same ratio as the sides adjacent to the angle. A
AB BN = AC NC
nA
B
Ex(10)
N
C
Fill in the missing ratios in the given triangles. A
A
4
B
2
...k
N
...k
...k
C
...k
B
N
8
A
A 12
...k
...k
3
B
N
C
6
...k
C
B
...k
15
N
6
A
Ex(11) 4
B
According to the given figure, find x.
6
2
Ex(12)
N
x
C
A D
B
In triangle ABC,
9
12
C
|AB| = 14 cm Find |BD|
C
215 Ex(13)
In triangle ABC,
A
[AH] ^ [BC] D
|BH| = |HC|
x
B
B
H
Ex(14)
2|AD| = 3|BD|
15
|EC| = 15 cm
C
Find |DE| = x
A
In the figure,
N
9
B
|BD| = |DC|, |AB| = 9 cm are given.
D
Length of the Interior Angle Bisector
C
Find |BN|.
A
|AN|2 = b . c – m . n
b
c
B
m
Ex(15)
n
N
C
A
Intriangle ABC,
12
6
[AN] is angle bisector
x B
|AB| = 6 cm, |AC| = 12 cm N
E
|BN| = 4 cm Find |AN| = x
Ex(16)
In triangle ABC,
A
[AD] is angle bisector 8 4 B
D
6
|AB| = 8 cm |AD| = 4 cm, |AC| = 6 cm C
Find |DC|
216 Exterior Angle Bisector
Bisector of an exterior angle of a triangle divides the opposite side into segments which are proportional to the adjacent sides. A
EC = EB
n'A
B
Ex(17)
C
AC EC = or AB AC
EB AB
E
Fill in the missing ratios in the given triangles
A
A
5
... k
3
B
... k
Ex(18)
C
... k
E
A 4
4
C
2
E
According to the given figure, find x.
3 2
B
B
... k
x
C
E
A
Ex(19)
6
E
x
According to the given figure, find x.
9
B
N 3
C
217
Length of the Exterior Angle Bisector A c
|AE|2 = m . (m + n) – b . c
b
B
n
C
m
E
A
According to the given figure, find |AE| = x.
Ex(20) x
4
3
E
B
Ex(21)
2
C
In triangle ABC,
A
[AD] is the exterior angle bisector
6§2
6
|AB| = 6 cm
4 B
x
C
D
|AC| = 4 cm |AD| = 6§2 cm Find |BC| = x
218 REVIEW EXERCISES 1. A
2x–1
D
x+5 B
C
In the given figure,
5.
A
[BD] is angle bisector
x
|AD| = 2x – 1 cm
|DC| = x + 5 cm
Find |DC|
E
In triangle ABC, [KE] ^ [AB] F
[KF] ^ [AC] m(AB∑C) = 44°
K 66°
44° B
D
C
A D
20
15
ABC is a right triangle, [BD] is angle bisector
B
E
C
[DE] ^ [BC]
|AB| = 20 cm
|DE| = 15 cm
Find |BD|
3.
A
6.
I is the incenter. D
I
|NC| = 2|BN|
Find m(AC∑B)
N
B
x
A
D 10
B
[BD] is angle bisector
6
x
In the given figure,
3 C
[BA] ^ [AD] |AD| = 6 cm |DC| = 10 cm
|BC| = 3 cm
Find |AD| = x
I is the center of the incircle of triangle ABC.
A
I
32°
B
8. 4.
C
x
C
2
Find |BC| = x
B
|AB| = |AD| + 5 is given.
In the figure,
In the figure,
7.
A
m(IB∑C) = 32° m(IC∑B) = 28°
28°
C
ABC is a right triangle
I
I is the center of the incircle of ABC
E 6
4
B
Find m(BA∑I) = x
A 2 D
|EK| = |FK| Find m(DA∑C) = x
2.
m(AC∑B) = 66°
x
[DE] // [BC] C
|AD| = 2 cm
|DB| = 4 cm
|EC| = 6 cm
Find |BC| = x
219 A
9.
E
ABC is a right triangle
D
12. A
D 85°
x
[BD] and [CE] are angle bisectors
x
B
C
Find m(DE∑C) = x
B
A
B
According to the given figure, find x.
A x
C
According to the given figure, find x.
13.
10.
E
80°
In triangle ABC,
D
K
70°
x
[AK] and [CK] are exterior angle bisectors m(AK∑C) = 70°
C
E
Find m(DB∑E) = x
I 130° C
B
A
14.
According to the
E given figure, find F
A
In triangle ABC,
110°
B
[AD] and [BD] are angle bisectors
D C
|AB| =|AC|
m(AD∑B) = 110°
Find m(DA∑C)
C
15. A
In triangle ABC,
K x
50° B
m(A)
5x
B
11.
4x
D
C
D
[BD] is interior angle bisector [AD] is exterior angle bisector m(AC∑D) = 50° Find m(AD∑B) = x
220 16.
In triangle ABC,
A
D
50°
x
[CD] is bisector
80°
C
B
A
20.
angle
According to the given figure, find x.
9
2x–3
m(BA∑C) = 80° B
m(DA∑B) = 50°
x
N
6
Find m(BD∑C) = x
21.
A x+1
In triangle ABC,
17.
A
D
x
[AD] is exterior angle bisector
B
C
|AB| = |AD|
m(BD∑C) = 32°
Find m(AD∑B) = x
D
A
m(BC∑A) = 20°
60° 60°
m(AC∑D) = 80°
20°
B
Find m(AD∑B)
80° C
According to the given figure, find x.
A x
B
N
N
2
C
C
angle
|AB| = 8 cm |AC| = 12 cm |BC| = 20 cm
Find |NC|
In triangle ABC,
A
[AD] ^ [BE]
6 9
E
F
x
5
4
[AN] is bisector
12
B
In triangle ABC,
8
23. 19.
C
A
8
According to the given figure, find the perimeter of AB∆C
m(BA∑C)=m(CA∑D)=60°
In the given figure,
2x
B
22. 18.
N
6
[BD] is interior angle bisector
C
B
D
|AF| = |FD| |AE| = |DC| = 6 cm
6 C Find |EC| = x
221 24. A
ABC is a right triangle [AD] is angle bisector
|BD| = 3 cm
|DC| = 5 cm
B
3
D
5
C
Perimeter
D
|BN| = 9 cm,
Find |AD|
N
C
9
4
D
C
of 29.
|CN| = 6 cm B
B
AB∆C = 45 cm
According to the given figure, find |AD|.
A
6
Find |AB|
A
25.
28.
In triangle ABC,
A
[AD] is bisector
6
4
|AD| = 4 cm
|AC| = 6 cm B
x
D
3
C
4
B
D
[DF] is angle bisector in triangle DBC
15
6 E
[AF] is angle bisector in triangle ABC,
A x
F
|BE| = 6 cm
C
|DE| = 4 cm
|DC| = 15 cm
Find |AD| = x
30.
In triangle ABC, A is divided into three equal angles.
A
|BD| = |DE|
B
D
E
C
|EC| = 2|DE| AC Find AB
In triangle ABC,
31. 27.
In triangle ABC,
A
x
B
D
E
|DC| = 3 cm Find |BD| = x
26.
angle
C
A
[AD] is exterior angle bisector
6
m(AB∑C) = m(EA∑C),
m(BA∑D) = m(DA∑E)
2|AB| = 3|AE|
|BC| = 4 cm
|DE| = 6 cm
Find |DB| = x
Find |AC| = x
3 D
x
|AC| = 6 cm B
4
C
|AB| = 3 cm
222 32.
A
8
In triangle ABC,
36.
[AE] is exterior angle bisector
|AN| = §3 |AE|
is given
|AC| = 5 cm
5
|AB| = 8 cm
B
x
C
Find |BC| = x
33.
10
E
|CE| = 10 cm
[AN] is exterior angle bisector
2|AB| = 3|AC|
|BN| = 18 cm
B
C
34.
Find x
x B
N
N
37. 7
Find |BC|
B 5 N 2 C
B
x
3 C
8
x
D
A
E x+3
B
35.
A
x
According to the given figure, find x.
According to the given figure, find x.
A
E
E
A
38.
C
In triangle ABC,
A
In the figure,
A
5
3
B
N
According to the given figure, find x.
C
x
x+1
C
4
According to the given figure, find |AE|.
E
223 4.3.2. Median Median of a triangle is a line segment joining a vertex to the midpoint of the opposing side. A
Constructing the median of a given triangle: C
B
D
C
Given: Triangle ABC Construct: The median of the side [BC]. Procedure:
1) Begin with triangle ABC. B
A
2) With the compass point on vertex B, set the compass width to any medium setting.
C
3) Draw an arc on each side of the line BC. 4) Without changing the compass width, place the compass point on vertex C, B
A
C
and make two more arcs so they intersect with the first two. 5) Draw a line between the points where the arcs cross. This will bisect the triangle side, dividing it into two equal parts. Label this point F. 6) Draw a line between F and A-the vertex opposite.
B
A
7) Therefore AF is the median of the side [BC].
C
The other two medians can be constructed in a similar way
B
A
C F A
B C F
B
A
224 Properties
1) In triangle ABC, the line segment [AD] joining the vertex A to the midpoint D or the opposite side [BC] is the median to the side [BC] and is denoted by Va. A
c
b
Va
B
a
D
C
2) The medians Va, Vb and Vc are concurrent. Their intersection point is the centroid (center of mass) of the triangle. A centroid E
F G B
D
C
3) A 2x
F
G
2z
x
B
1 GD = 3 AD 2 AG = 3 AD
E
y
2y
z
D
C
G is the centroid of the triangle
A
Ex(1)
ABC. Find |BG| + |GD|
E
8 3
G B
D
4)
C
In a right triangle, the median from
A
the right angle to the hypotenuse is one - half of the length of the hypotenuse m(A) = 90° ğ |BD| = |DC| = |AD| B
D
C
225 Ex(2)
ABC is a right triangle,
A
G is the centroid |BD| = 2x – 1 cm
G B
2x–1
D
x+7
Find |AG|
C
A
Ex(3)
|DC| = x + 7 cm
In triangle ABC, [AD] and [BE] are medians
3
B
|BF| = 3 cm
E F
|FD| = 2 cm
x
2 D
C
are given. Find |CF| = x
G is the centroid of AB∆C.
A
5)
• [FE] // [BC] BC • |FE| = 2
3m K m
F
G B
Ex(4)
E 2m D
C
A
In triangle ABC, E and D are midpoints of
K
E
[AB] and [BC] respectively. [EF] // [BC]
F
|AD| = 24 cm
L B
D
• |AK| = |KD| AD • |KG| = 6
|CE| = 18 cm C
|BC| = 28 cm
Find the perimeter of the triangle EGK.
226 A
Ex(5)
G is the centroid of the triangle ABC. [CG] is angle bisector
10
|AC| = 10 cm
G
B
x
D
Find |BD| = x
C
A
Ex(6)
G
15
B
G is the centroid of ABC,
x
E
[BG] is the angle bisector of AB∆C.
24
D
C
[GD] // [AC]
Find |BE| = x
227 REVIEW EXERCISES 1.
According
A
B
4
figure, find
E
|AF|.
2
1 D
ABC is a right
A
to the given
2 F
5.
4
triangle.
F
4
E
[AD] and [BE] are medians.
C
B
D
C
2.
In the given
A
figure,
E
F
|AD| = 6 cm
Find |BC|
C
6.
6
|GE| = 3 cm
G
B
D
C
Find |AG| + |BG| + |CF|
3.
A
|CG| = 8 cm
G is the centroid of AB∆C. Find |AG|
G 8
B
G
A
is
the
A
7.
triangle ABC.
E
G
B
[GF] // [BC]
D
C
|EF| = 4 cm Find |AB|
D
B
E
C
B
30° 4
E
C
|BE| = 4 cm Find |AC|
Find |NE| = x.
4 G 3
ABC.
G
E
centroid of
D
|AB| = 24 cm
A
G is the
A
ABC. |GN| = |NB|
G
N
8. 4.
G is the centroid of the right triangle
centroid of the F
|AF| = 4 cm
B
D
In triangle ABC, G is the centroid
|GE| = 4 cm, |GD| = 3 cm. Find |BG|
C
228 ABC is a right triangle. G is the centroid
A
9. E
2 G
B
D
C
13.
|AG| = |BC|
of AB∆C.
|BE| = 9 cm
|GE| = 2 cm
Find |BC|
D
G is the centroid of the right triangle ABC.
A
G x
B
|BH| = 8 cm H
C
|HC| = 10 cm Find |GH| = x
are given. Find |BC|
G
C
G is the centroid of the triangle ABC.
A
11.
E
G
L
B
N x D
A
14.
5
B
18
F
G B
D
In the triangle ABC
|GN| = |NC|
|AE| = 2|EC|
|AL| = 24 cm
Find |LD| = x
D E B
|AD| = 3 cm |BC| = 5 cm Find |AC|
C
A
10
x E 3
Find |GD| = x
B
C
Find |AG| = x
C
A
|AH| = 18 cm
x
|BC| = 16 cm
m(BA∑C) = m(DA∑C)
E
G
15.
G is the centroid of ABC. H
|DG| = 5 cm
C
A
m(BA∑G) = m(CA∑G)
16
16. 12.
G is the centroid of the triangle ABC
x
D
|DG| = 2 cm
E
B
10.
G is the centroid of ABC.
A
D
C
In the given figure, [ED] // [AC]. Find |AC| = x
229 A
17.
D
G is the centroid of ABC. F
E
19.
D is the midpoint of [BC].
|AD| = 6 cm
|AB| = 5 cm
Find |AC|.
|AC| = 6 cm
|ED| = |EG|,
G
B
C
B
D
Find |AD|
G is the centroid of ABC.
A 3§2 E
E is the midpoint of [AB]
G 45°
B
D
C
Find |AG|
Note: Median Length, Apollonius' Theorem: A
c
B
b
va
D a
a2 2Va2 + 2 = b 2 + c 2
C
E
C |BD| = 2 cm
A
20.
18.
In triangle ABC,
A
G is the centroid of the triangle ABC.
6 4
3
|AG| = 6 cm
D
|EG| = 4 cm
G B
x
C
|GD| = 3 cm Find |BC| = x
230 4.3.3. Altitude The altitude of a triangle is the segment from a vertex perpendicular to the opposite side. h
h
h
acute
right
obtuse
The three altitudes drawn from the three angles of any triangle meet at a single point (are concurrent), which may or may not be inside the triangle. The orthocenter of a triangle is the point where the altitude lines are concurrent. A
A ha
ha hb
B hc
C hc
B
C orthocenter
Constructing the altitude of a triangle:
hb
Given: Triangle ABC Construct: The altitude to the side [BC]. Procedure:
1) Start with the vertex A. C
C
2) Place the compass on A. 3) Set the compass width to a approximately 50% more than the distance to
B
A
B
A
C
the line [BC].
4) Draw an arc across the line on each side of A, making sure not to adjust the
C
compass width in between. Label these points P and Q B
A
B
C
A C
5) At this point, you can adjust the compass width. Recommended: leave it as is. 6) From each point P, Q, draw an arc below the line so that the arcs cross. 7) Place a straightedge between A and the point where the arcs intersect. Draw
B
A
B
A
the perpendicular line from A to the line, or beyond if you wish. 8) This line is perpendicular to the first line and passes through the point A. It also bisects the segment PQ (divides it into two equal parts). Therefore this line is the altitude to the side [BC].
231 Ex(1)
D is the orthocenter of the triangle ABC.
A
m(BA∑C) = 45°
45° E
|BE| = 6 cm
x
Find |DE| = x
D
B
C
4.3.4. Perpendicular Bisector A perpendicular bisector of the side of a triangle is a line, ray or segment that is perpendicular to and bisects that side at its midpoint. A
B
D
C
The perpendicular bisectors of the sides of a triangle are concurrent and intersect at a point that is equisdistant from the three vertices of the triangle. A
B
C
The perpendicular bisectors of the sides of a triangle intersect at the center of a circle that circumscribes the triangle and has the distance from the center to each vertex as its radius (dashed). A
B
C
232 Perpendicular Bisector / Endpoints Theorem
If a point lies on the perpendicular bisector of a segment, then the point is equidistant from the endpoints of the segment.
Endpoints/ Perpendicular Bisector Theorem
If a point is equidistant from the endpoints of a segment, then it lies on the perpendicular bisector of the segment.
point A
point B
perpendicular bisector
Constructing the perpendicular bisector of a triangle
Given: Triangle ABC Construct: The perpendicular bisector of the side [BC]. Procedure:
1) Start with a line segment [BC].
2) Place the compass on one end of the line segment. 3) Set the compass width to a approximately two thirds the line length. 4) Without changing the compass width, draw an arc above and below the line. 5) Again without changing the compass width, place the compass point on the the other end of the line. Draw an arc above and below the line so that the arcs cross the first two. 6) Using a straightedge, draw a line between the points where the arcs intersect. 7) This line is perpendicular to the first line and bisects it. Therefore this line is the perpendicular bicector of the side [BC]. A
Ex(1)
x 7
E B
20
[ED], [EF] are the perpendicular bisectors of [BC] and [AC].
F
|BD| = 20 cm |ED| = 15 cm are given.
D
C
|EF| = 7 cm Find |AF| = x
A
Ex(2)
O is the center of the circumscribed circle. [OD] ^ [BC] |OD| = 2 cm, |BD| = 5 cm
x 2 B
O
D
Find |AO| = x C
233 4.4. Right Triangle and Trigonometry 4.4.1. Pythagorean Theorem Pythagorean Theorem
In a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the right sides. C
a2 = b2 + c2
a
c
A
b
B
Converse of Pythagorean Theorem: If the sum of the squares of two sides of a triangle is equal to the square of the third side, then the triangle is a right triangle. m (W A ) = 90° + a 2 = b 2 + c 2
Most common Pythagorean triangles: 5k
3k
5k
4k
12k 25k
17k
8k
7k 24k
15k
Ex(1)
According to the given figure, find x.
A 4
3
B
13k
x
C
234 A
Ex(2)
[AC] // [DB]
25
|AD| = 25 cm x
D
|DB| = 7 cm
7
|BC| = 30 cm B
30
C
Find |AC| = x
Note: A A a
a
d
B
c c
c a2 + c2 = b2 + d2
C
A
Ex(3)
[AD] ^ [BC]
5
|AB| = 5 cm
x
|DC| = 6 cm
C
B 4
|BD| = 4 cm
6
Find |AC| = x
D
Ex(4) E
3
ABC is a right triangle
A
|DB| = |DC|
8
|AE| = 3 cm
9 B
d
B
D b
b
D
|EB| = 9 cm D
C
|AC| = 8 cm Find |DE|
235 Euclid’s Theorems
A c
i) h2 = p . k b
h
ii) b2 = k . a iii) c2 = p . a
B p H k C 144444244443 a
iv) a . h = b . c
Proof
A
Ex(5)
According to the given figure find x
4 8
B
Ex(6)
H
x
C
A
According to the given figure find x
x
B
3
Ex(7)
H
12
C
According to the given figure find x
A x
B
3
H
6
C
236 A
Ex(8)
According to the given figure find x 4
3
x
B
H
C
Ex(9)
A
ABC and ADE are right triangles |BD| = |DC|
10 6 B
E
|AE| = 10 cm D
C
|BE| = 6 cm Find |DE|
237 REVIEW EXERCISES 1.
According to the given figure, find x.
A 4x+1
5.
2x
2.
According to the given figure, find a2 – b2
A 2 D
b
3
6
|AB| = |BD| |AC| = 6 cm
D
2
|DC| = 2 cm Find |BC|
A
4.
C
|AB| = 3 cm |CD| = 6 cm
3 B
C 6 D
|BC| = 12 cm Find |AD|
Find |BC| = x
C
ABC is a right triangle
A
|AC| = 4§5 cm
|AD| = 6 cm
4§5
|BD| = 4 cm
6
Fİnd |DC| = x B
ABC is a right triangle
A
B
x
B
C
|BD| = 5 cm
a
|AB| = 8 cm
D 5
C
6.
3.
8
B
B
|AD| = |DC|
15
ABC is a right triangle
A
4
D x
C
A
7.
[AE] ^ [BC]
|AB| = 4 cm
8.
5
4 3 B
D
Find |DC| = x
C
E
ABC is triangle 18 E x
B
a
|BD| = |DC|
|DB| = 3 cm
x
A
|AC| = 5 cm
D
|AE| = 18 cm
6 C
|EC| = 6 cm Find |DE| = x
right
238 A
9.
6
Note:
|AE| = |EC|
Carnot's Theorem
|AD| = 6 cm
E
ABC is a right triangle
|DB| = 2 cm
F
|BC| = 10 cm
D 2 B
B y
F
E
B
A
1
4 x
D
C
z
D
10.
55°
E
P
13. A
x2 + y 2 + z 2 = a 2 + b 2 + c 2
c
a
Find |DE|
C
10
A
x
b
3
C
E 6
P
B
3
D
x
C
According to the given figure, find |DC| = x
ABC is a right triagle
|BD| = |DC|, |AB| = |ED| m(ED∑B) = 55°. Find m(AC∑B) = x
14.
11.
18
12
B
D
6
ABC is a right triangle. A, B and E are collinear
A
C
x
|AD| = |DC|
ABC is a right triangle
A y
4
[AH] ^ [BC]
|AB| = 4 cm B
2
H
x
C
Find |HC| = x and |AC| = y
|AC| = 18 cm |AB| = 12 cm
E
|BE| = 6 cm
Find |DE| = x ABC is a right triangle
15. A 12. 5
m(DA∑C) = 90°
A
B
|BH| = 2 cm
D
3
m(AB∑C) = 2.m(AC∑B)
x
|AB| = 5 cm C
|AD| = 3 cm Find |AC| = x
[BD] ^ [AC]
D
[DH] ^ [BC]
|BH| = 2 cm
B
2
H
8
C
|HC| = 8 cm Find |AB|
239 16. A
3
ABC is a right triangle
12
6
|AD| = 3 cm
E
|DC| = 12 cm H
B
17.
C
According to the given figure, find |BC|.
A
|DH| = |BH| = |HC|
D
20.
2
Find |AB| + |BC|
D
B
C
According to the given figure, find x.
A H 15
x
B
21. B
C
20
According to the given figure, find x.
D x E 6
4 A
18.
According to the given figure, find x.
A 12 5
C
x
B
H
C
22. A
6
19.
A
6
B
x
H
y
C
In the figure,
B
9
D
x2 + y2 = 97 is given.
ABC is an isosceles triangle
|AB| = |AC|, m(BA∑D) = 90°
Find |BC|.
|BD| = 9 cm, |AD| = 6 cm
Find |DC| = x
x
C
240 4.4.2. Trigonometric Ratios in Right Triangle The trigonometric functions sine, cosine, tangent and cotangent can be defined according to the ratios of the sides of a right triangle. opposite sin θ = hypotenuse = adjacent A cos θ = hypotenuse = opposite b hypotenuse c tan θ = adjacent = a b opposite adjacent a q cot θ = opposite = b B a C adjacent sin θ tan θ = cos θ cos θ cot θ = sin θ Ex(1)
x and y are acute angles
A
y 4
B
x
C
3
Find a) sinx
b) cosx
c) tanx
d) cotx
e) siny
f) cosy
g) tany
h) coty
Ex(2)
The figure is consisting of identical squares. a
Find sina, tanq q
b c a c
241 Ex(3)
0 < a < 90° 2 If cosa = 3 , then find the value of sina . tana
Trigonometric Values of 30o,45o and 60o
Given equilateral triangle ABC, and isosceles right triangle DEF, determine the trigonometric function values for 30o, 60o and 45o D
A
Note: A pattern for the values of sine function : q sinq
0
30°
0 2
1 2
45°
2 2
60°
3 2
90°
4 2
Reverse the order to get cosine values of the same angles.
C
B
q
30°
E
45°
F
60°
sin cos tan cot
Ex(4)
Find the value of cos60o + sin245o + cot230o – tan45o
Ex(5)
When walking along a sidewalk, you notice that by turning 30° to your left you see a large building. After walking another 40 m, you now notice that the building is directly to your left. How far is the building when you looked the second time?
242 REVIEW EXERCISES 1.
ABC is triangle
A 3
2
right
|AB| = 2 cm
5.
The figure is consisting of identical squares.
q a
Find cosa + tanq
|AC| = 3 cm
x B
a
C
Find the value of sinx . cosx + tanx
6. 2.
A
ABC is a right triangle [BA] ^ [BC]
a
m(BA∑C) = a
|AC| = 6 cm 2 cos α = 3
B
C
Find tana + cotq
find tana
3.
ABC is a right triangle
A x
[CD] ^ [AB]
|AB| = |AC| = 8 cm D
|BC| = 6 cm
Find sina
a
B
C
m(AB∑C) = 37° 15
C
|BC| = 15 cm
Find |AB| = x
(sin37° = 0,6)
8.
4. 0 < a < 90° If tana = 2, then find the value of sina . cota
8
In triangle ABC,
A
In triangle ABC,
A
[BA] ^ [AC]
37° B
q
a
7.
The figure is consisting of identical squares.
|AB| = |AC| = 8 cm
8
|BD| = 3 cm
a B 3
B
7
C
|DC| = 7 cm Find tana
243 ABC is an isosceles triangle
A
9.
|AB| = |AC| 5 tanA = 12
Find cotC
C
B
10. Find the value of
sin 45° + cos 45° cos 30°
11. Find the value of sin230o + cos230o +sin245o + cos245
12. Find the value of
13. Find the value of
tan 60° . cot 60° cos 2 45°
sin 30° + cos 60° cos 30° + sin 60° + tan 60° cot 2 30°
14. You are walking in a park and you notice a tree in front of you. To see the top of the tree you have to look up at an angle of 20°. If you reach the bottom of the tree after walking 10 m, find the height of the tree. (Assume your eye level is 1,4 m above the ground.) (tan20° = 0,36)
244 4.4.3. Unit Circle and Trigonometric Ratios By using a right triangle, we can only define the trigonometric functions of acute angles. To be able to work with the trigonometric functions of any angle, the trigonometric functions will be defined by using the unit circle. Unit Circle
The circle with a radius of 1 unit, centered at the origin of a rectangular coordinate system is called the unit circle. y B(0,1) C(–1,0)
A(1,0)
O
x
D(0,–1)
Sine and Cosine Functions
Letq be the angle measured from the positive x-axis to the terminal side, and P(x,y) be the point on the unit circle where the terminal side intersects the unit circle. y 1
–1
Note:
1 P(x,y) 1 y q x x H 1
O
O
y x
P(x,y)
H 1
x
–1
| OH | x In triangle POH, cos θ = | OP | = 1 = x
y
–1
1 q
So, the coordinates of the point P can be written as P(cosq, sinq) That is, the first coordinate (abscissa) of a point on the unit circle is cosq and the second coordinate (ordinate) is sinq. y
–1
P(x, y) be a point on the unit circle. x = cosq, y = sinq since
x 2 + y2 = 1
cos2q + sin2q = 1
y | PH] sin θ = | OP | = 1 = y
1 y q x
P(x,y)=(cosq,sinq)
245 Ex(1)
Tangent and Cotangent Functions
Find the value of sin220o+sin250o+ cos220o+cos250o
Let us draw the line x = 1 to the unit circle. y 1 P –1
O
q
A
B H 1
x
–1 x=1
| AB | | AB | In triangle AOB, tan θ = | OB | = 1 = | AB | So, tanθ is the ordinate of the point A; where the ray [OP cuts the line x=1, i.e the length of the line segment [AB].
y 1 P
–1
O
q
A tanq B x H 1
–1 x=1
| PH | sin θ In triangle POH, tan θ = | OH | = cos θ So, sin θ tan θ = cos θ
246 Let us draw the line y=1 to the unit circle. y 1 D
y=1
P –1
q
O
q
C
x
H 1
–1
| CD | | CD | In triangle DOC, cot θ = | OD | = 1 = | CD | So, cotθ is the abscissa of the point C; where the ray [OP cuts the line y=1, i.e the length of the line segment [DC]. y cotq
1 D
x=1
P –1
q
O
q
H 1
C
x
–1
| OH | cos θ In triangle POH, cot θ = | PH | = sin θ So, cos θ cot θ = sin θ
Ex(2)
1 A c 1, 3 m 2 2 –1
O
a
–1
1
According to the given unit circle find sina =
tana =
cosa =
cota =
247 Ex(3)
sin x . cot x + cos x . tan x sin x + cos x
Simplify
y 1
Trigonometric Ratios of Obtuse Angles P(x,y)
y q
–1 x
O
1
x
Let P be a point on the second quadrant of the unit circle, so q be an obtuse angle Since x < 0 and y > 0, then cosq < 0 and sinq > 0
–1
y=1
y 1
D P
q –1
O
1
x
Since tanq is the ordinate of the point E, tanq< 0 cotq is the abscissa of the point D, cotq < 0
E
–1
x=1
Note: Quadrant II (–, +) only sine positive
y
So,
Quadrant I (+, +) All positive 1
x
–1
O
0° < q < 90° sinq > 0 cosq > 0 P tanq > 0 cotq > 0
y
q
1
x
–1
Ex(4)
90° < q < 180° sinq > 0 cosq < 0 tanq < 0 P cotq < 0
1
q
–1
O
–1
Determine the signs of the following a) sin 80°
e) sin 150°
b) cos 110°
f) cos 170°
c) tan 75°
g) tan 95°
d) cot 10°
h) cot 160°
y
1
x
248 Ex(5)
According to the given unit circle find
1 3 1 Bc− 2 , 2 m
a
1
–1
sina =
tana =
cosa =
cota =
–1
Ex(6)
A 2
According to the given figure, find sina . cosa
1
a B
Ex(7)
Reference Angles y Quadrant II Reference angle q′
Quadrant I
q
x
C
3 x ∈ (90° , 180°), sinx = 5 is given. Find the values of cosx, tanx and cotx
To find the trigonometric values of angles in quadrants other than the first quadrant, reference angles are used. The reference angle q¢ for an angle q is the smallest positive angle between the terminal side of q and the x – axis. For instance, the reference angle for 120° is 60° To find the value of trigonometric function of any angle q, first determine the function value for the associated reference angle q¢. Then, depending on the quadrant in which it lies, prefix the appropriate sign to the function value. y
Reference angle : 180° – q
1 P'(–x,y) –1
On the unit circle, the triangles
P(x,y)
OPH and OP¢H¢ are congruent, then
180– q Hı
q
O
q
H 1
x
sin(180° – q) = sinq cos(180° – q) = –cosq
–1
therefore
tan(180° – q) = –tanq cot(180° – q) = –cotq
249 Ex(8)
Trigonometric Values of Some Obtuse Angles
Find the value of cos120° + cos60° + sin40° – sin140°
Given the unit circle, determine the trigonometric function values for the following angles. 90° y
(0,1)
(1,0)
180° (–1,0)
x 0°
(0, –1) 270°
q
0°
30°
45°
60°
sin cos tan cot
Ex(9)
Find the value of
cos 120° – cot 135° sin 120° . tan 150°
90°
120°
135°
150°
180°
250 REVIEW EXERCISES 6.
sin x cos x 1. Simplify ` cos x + sin x j . sin 2 x
4 3 Ba − 5 , 5 k
a
1 – sin 2 x 1 – cos 2 x + cos 2 x sin 2 x
x
O
2. Simplify
According to the given unit circle find
y
tana =
cosa =
cota =
D
7.
10
C
1 3 6 Ac 3 , 3 m
–1
a
–1
According to the given unit circle find
|AD| = 6 cm
6
ABCD is a rectangle |AE| = |EB|
3.
sina =
|DC| = 10 cm
x A
E
B
Find tanx
sina = 1
tana = cosa = cota =
8.
The figure is consisting of identical squares. a
Find sina . cosa + tanq
9. Find the value of
sin 150° – sin 180° cot 120° . sin 90°
10. Find the value of
sin 135° – cos 135° tan 180° + cos 0°
4. If sinx – cosx = 0 , then find the value of tanx.
2 sin x – cos x 1 5. If sin x + 3 cos x = 2 , then find the value of tanx.
q
1 11. 90o < x < 180o , sin x = 3 is given. Find the value of 3cosx – cotx + tanx
12. 0 < x < 180o , 3sinx + 2cosx = 0 is given. Find the value of sinx – cosx.
251 4.4.4. The Law of Cosines By the Law of Cosines and the Law of Sines, it will be able to work not only with right triangles, but also with all sorts of triangles. The Law of Cosines
In a triangle ABC with sides of lengths a, b and c, A c B
a2 = b2 + c2 – 2bc . cosA b
a
b2 = a2 + c2 – 2ac . cosB C
c2 = a2 + b2 – 2ab . cosC
Proof Note: In a right triangle where m (W A ) = 90° , since cos90° = 0, The law of cosines simplifies to give a2 = b2 + c2 (The Pythagorean Theorem)
The Law of Cosines applies when 1) two sides of a triangle and the included angle (SAS) or 2) all three sides of a triangle (SSS) are known
Ex(1)
W) = 30o, b = 5cm, c = §3 cm are given. Find a. In triangle ABC, m ( A
Ex(2)
In triangle ABC, a = 3§2 cm, b = æ10 cm, c = 4 cm are given. Find m ( BW) .
Ex(3)
Two ships leave harbor at the same time. The first sails N 15o W at 25 knots (a knot is one nautical mile per hour). The second sails N 45o E at 20 knots. Find the distance between the ships after 2 hours.
252 REVIEW EXERCISES 1. In triangle ABC, a = 2 cm, b = 4 cm and m(C) = 60° are given. Find c
6.
A
D
2
5
3
K
x
5
4 B
2. A
According to the given figure, find |CD| = x
C
In triangle ABC,
2
m(AB∑C) = 150°
x
|AB| = 2 cm
150°
§3
B
C
|BC| = §3 cm
7.
Find |AC| = x 4
x
3
3. 4
|BC| = æ37 cm Find |AC| = x
æ37
C
8.
In triangle ABC,
A
|AB| = 3 cm
3
According to the given figure, find |ED| = x
A 3 E
8
4.
C
|AB| = 4 cm
x
B
5
2 D
m(BA∑C) = 120°
120°
B
In triangle ABC,
A
According to the given figure, find |AC| = x
A
B
6
4
x
C
4
D
|AC| = 5 cm
5
|BC| = 6 cm Find cos(BA∑C)
6
B
C
9. In triangle ABC, a2 = b2 + c2 – bc is satisfied. Find m(A)
5.
According to the given figure, find |DE| = x
A D
1
2 x
3 B
E 1
4
C
bc 10. In triangle ABC, b + c − a = a + b + c is satisfied. Find m(A).
253 4.5. Area of a Triangle 4.5.1. Area of a Triangle 1) If one side of the triangle and the altitude to this side are given
A 3
4
Ex(1)
a
C
hc
4
hb
4
B
C
a
B
c
1
hc
b
4
ha
2
hb
b
4
ha
c
4
A
a.h a b.h b c.h c & A ( ABC = ) = 2 2 = 2
In ABC,
A
[AH] ^ [BC] |AC| = 5 cm
5
B
Ex(2)
2
H
|BH| = 2 cm
3
C
|HC| = 3 cm
Find A(AB∆C)
A
ABC and AEC are right triangles, 4
B
|AB| = 4 cm
8 x
D
C 3
E
|AD| = 8 cm |EC| = 3 cm
Find |DC| = x
254 Ex(3)
In ABC,
A
[AD] is angle bisector
10
B
3
D
[AB] ^ [BC]
|AC| = 10 cm
C
|BD| = 3 cm
Find A(AD∆C)
Ex(4)
According to the given figure, find A(AD∆C).
A
17
17
B
16
Ex(5)
D
Ex(6)
|AB| = |AC| = 13 cm, |BC| = 10 cm are given. Find |CH|.
10
C
In triangle ABC,
A
[AH] ^ [BC]
Note:
D
H
In the figure, 13
B
A
|AD| = 3 cm
D
AD . BC A (ABDC) = 2
C
C
A H
B
6
|BC| = 8 cm
Find A(ABDC) B
H
C
255 Result 1
If ABC is an isosceles triangle, |AB| = |AC|, then A
E
|PD| + |PE| = hb = hc
H
hb
D
B
P
C
Proof
A
Ex(7)
ABC is an isosceles triangle, |AB| = |AC|, find A(AB∆C)
30° E
F 6
3 B
Result 2
D
C
If ABC is an equilateral triangle then A F D
B
Proof
|KE| + |KF| + |KD| = |AH| = h
K E H
C
256 A
Ex(8) Note:
E
The height of the equilateral triangle with one side a is a 3 equal to 2
ABC is an equiletaral triangle If |OD| + |OE| + |OF| = 6§3 cm, then find A(AB∆C).
F
O
D
B
C
The area of the equilateral triangle with one a2 3 side a is equal to 4
2) If two sides and an included angle of a triangle are given A
c
b
B
a
C
A
Ex(9)
& ) = 1 b.c sin A A ( ABC 2 1 = 2 a.c sin B 1 = 2 a.b sin C
According to the given figure, find A(AB∆C).
8 30° B
C
6
Ex(10)
A D
Find x. F
12 B
A(DB∆C) = A(AB∆E) is given.
x
4
E
6
C
257 In triangle ABC,
A
Ex(11)
5|AE| = 2|AB| E
3|DC| = |AD|
D
A(AE∆D) = 12 cm2
B
C
Ex(12)
A 3 D
3
According to the given figure & A ( ABC ) find A (DFCE)
6 E
2 F
3
B
Find A(AB∆C).
C
3) If three sides of a triangle are given A
c
& ) = u (u − a) (u − b) (u − c) A ( ABC
b
B
a
where 2u = a + b + c
C
Ex(13) A 9
B
Find A(AB∆C).
5 6
C
258 Ex(14)
A
7
D
6
Common Altitude - Triangle Area Relationship
5
x
B
According to the given figure, find x.
C
A
&) A ( ABD a &) = b A ( ADC
h B14243D14243C a b
a.h a 2 = b .h b 2
&) A ( ABD = & A ( ADC )
A
b.S
a.S B
a
D
b
C
A
Ex(15)
In the given figure, A(AB∆D) = 9 cm2
B
D
C
2|BD| = |DC|
Find A(AB∆C)
259 Ex(16)
In the given figure ,
A
A(AB∆C) = 3 . A(AB∆D) 5
B
Ex(17)
5
Find |AD|.
2 D
C
A
In the given figure, |BD| = |DC|
E
B
D
Ex(18)
|AE| = |EC| & A ( ABC ) & =? A (EDC )
C
In AB∆C,
A
|EC| = 5|AE|, |DC| = 3|BD| are given. &) A ( ABC Find & A ( ADE )
E
B
D
C
Note: If G is the centroid of the triangle ABC then, A
S
A
S
E
S
S S C
B
S
G D
F
S
S
S G
S
S
D
S
B
A
C
B
S
C
260 A
Ex(19)
G is the centroid of AB∆C [DG] // [BC] A (ADGC) Find A (DBCG) G
D
B
C
Note: In triangle ABC, [AD] is the angle bisector. Since the altitudes to the triangles ABD and ADC are the same; & A ( ABD) c m &= = n A ( ADC ) b A b
c
B
m
D
n
C
A F b
c E B
h
h m
D
n
C
A b
c
b.S
c.S B
D
C
261 Ex(20)
In the given figure,
A
[AD] and [DE] are angle bisectors
E
B
|DC| = 2|AD|
D
2|AC| = 3|AB|
C
A(AB∆C) = 25 cm2 Find A(AD∆E)
Note (1) In triangle ABC, I is the center of incircle. Since the altitudes to the triangles ABD, ADC and BDC are the same; A
A c
b
I
c
K
h
F I
b
h
h B
C
a
B
E
C
a
A c
c.S I
b
b.S
a.S B
C
a
(2) Let O be the center of the incircle of the triangle ABC, and r be the radius of the incircle. A c
K
A F b
r
r Or
B
E
b
c O
a
C
a .r b .r c .r a+b+ck & A ( ABC ) = 2 + 2 + 2 = a .r 2 = u .r
B
a
r C
262 Ex(21)
I is the center of the incircle of the triangle ABC.
A
7
6
Ex(22)
|AB| = 7 cm
5
I
B
|BC| = 6 cm
|AC| = 5 cm & A ( IBC ) Find & A ( ABC )
C
A
ABC is a right triangle |AB| = 5 cm
12
5
B
C
|AC| = 12 cm
& Find the radius of the inscribed circle of ABC .
A
Common Base - Triangle Area Relationship
D h2
h1
& A ( ABC ) h 2 & = A (DBC ) h 1
B14444244443C a
a.h 2 h2 2 = a.h 1 h1 2
& A ( ABC ) = & A (DBC )
In triangle ABC,
A
Ex(23)
[AE] ^ [BC] [DF] ^ [BC]
D
|DF| = 4 cm |AE| = 14 cm
B
E
F
C
A
Ex(24)
5 E
A(ABDC) = 15 cm2 & Find A (DBC ) .
ABC is a right triangle [AB] // [ED] |AE| = 5 cm
|DC| = 4 cm
Find A(BD∆E) B
D
4
C
263 Similar Triangle - Area Relationship
D A
f
c
e
b
B
a
C
E
AB∆C ~ DE∆F and k =
d
F
BC then EF
&) A ( ABC a j2 = `= k2 & d A (DEF )
Ex(25)
D
A
A(DE∆F) = 10 cm2
6 6
Find A(AB∆C)
E 4
B
9
C
In triangle ABC,
A
Ex(26) Note:
D
S 3S 5S
[AB] and [AC] are divided into 4 equal pieces. If the sum of the areas of the shaded regions is 20 cm2, then find A(AB∆C).
E
F
G
K
7S 9S
F
N
º
B
C
Ex(27)
In triangle ABC,
A
[KL] // [DE] // [BC] K D B
|AK| = |KD| = 2|DB|
L
A(KDEL) = 36 cm2
E
Find A(DBCE) C
264 REVIEW EXERCISES
|AB| = 10 cm
|AH| = 6 cm
10
6
5.
In triangle ABC,
A
1.
Find A(AB∆C) B
H
3
[AD] ^ [BC]
|HC| = 3 cm
In triangle ABC,
A
|BC| = 10 cm
|AC| = 8 cm
C
[BE] ^ [AC]
E
B
|AD| = 5 cm
C
D
6.
2.
ABC is triangle
A 4
30°
B
3.
C
right
4
[ED] ^ [AB]
|AB| = 4 cm
Find A(AB∆C)
C
|AB| = 6 cm
[AC] ^ [BC]
|AE| = 3|EB|
ABC is a right triangle
8.
|AD| = 13 cm
|DC| = 7 cm
Find A(AD∆C)
In triangle ABC,
C
E
A
7
3
7.
D
B
|BE| = 3 cm
Find |AC|
|BD| = 5 cm
5
2
B
|DE| = 2 cm
D
8
E
C
A
In triangle ABC, [AC] ^ [BC]
A
m(AC∑B) = 30°
3
A
B
a
According to the given figure, find A(ABCD)
D 13
4.
Find |BE|
|AC| = 8 cm |BD| = 6 cm
B
6
D
C
ABC is a right triangle
A
|AE| = |EC|
E
6 B
Find A(BE∆D)
D
|AB| = 6 cm 8
C
|DC| = 8 cm Find A(DE∆C)
265 A
9.
10
D
4
E
B
C
14.
[AB] ^ [BC]
[AH] ^ [BC],
[DE] // [BC]
|AC| = |BC| = 10 cm
|AB| = 10 cm
|AB| = 12 cm
|DE| = 4 cm
Find |AH|
A
Find A(AC∆E)
B
H
15. A
10.
In AB∆C,
|HC| = 4|BH|,
A(AB∆C) = 20 cm2 are given
In triangle ABC,
In triangle ABC,
B
H
C
C
In the figure,
A
|BC| = 12 cm,
E B
H
C
Find |BC|. In AB∆C,
A
16. A
11.
In the figure,
|AB| + |AC| = 12 cm
B
13
C
Find the perimeter of (AB∆C)
C
D
B
17.
In AB∆C,
12. C
A
c
B
Find A(AB∆C)
A
[AD] is an angle bisector
[AC] ^ [BC|
B
A
In the figure,
|AD| = |AC| = 10 cm
|BD| = 5 cm
|DC| = 12 cm
B
D
C
Find A(AB∆D)
|AB| = |AC|
30°
[ED] ^ [AB] [DF] ^ [AC]
|DC| = 4 cm
A
|BD| = 8 cm
Find A(AB∆D)
13.
C
D 4
8
18.
ABC is a right triangle
b – c = 14 cm, a = 26 cm are given.
a
b
Find A(AB∆C).
H
4
A(AB∆C) = 30 cm2
A(ABEC) = 48 cm2 are given. Find |AE|.
E B
x
1 D
m(BA∑C) = 30°
F
A(AB∆C) = 16 cm2 C
|ED| = 1 cm Find |DF| = x
266 A
19.
§6
45° x
B
|KD| = 3 cm
C
D
According to the given figure, find x.
A
|KE| = 2 cm
K B
24.
|KF| = 1 cm
E
F
ABC is an equilateral triangle
2 C
D
Find A(AB∆C)
20.
ABC is triangle.
A
an
equilateral
|BD| = 1 cm
|DC| = 3 cm
Find A(AB∆D)
B 1 D
3
C
A
25.
In the figure, |BD| = 8 cm,
B
E
60°
|AD| = 14 cm are given.
C
Find A(AB∆D)
D
21. 4§3 30°
According to the given figure, find A(ABCD)
A 5§3 C
3
B
4§3
26.
60°
D
22.
B 3 E
27. A C B
Find A(AB∆C)
4
3|AB| = 2|BC| is given.
D
Find A(CD∆E)
6
In triangle ABC,
A
C
B
12
23.
|DE| = 15 cm
Find cosx.
20 x
|BE| = 3 cm
8
15
A(AB∆C) = 15 cm2,
C
|DC| = 8 cm
D
|AD| = 9 cm
9
A
In triangle ABC,
A
3|AD| = 2|DB|
E
|EC| = 3|AE|
D
A(DBCE) = 6 cm2 Find A(AD∆E)
E
B
C
267 28.
In triangle ABC,
A
| DE | |AD| = |EC| = 4
D
A
32.
A(BCEF) = 15 cm2
F E
E
10
2|AF| = 3|BF| B
According to the given figure, find A(AD∆E).
D
H
8
C
Find A(AB∆C) C
B
33. A 29.
In triangle ABC,
A
|AD| = 2|FD| = 2|BF|
E
30°
B
C
D
C Find A(AB∆C)
B
Note:
34.
A z c
Find A(AD∆C)
A(DFCE) = 36 cm2
F
A(AB∆D) = 6 cm2
5|AE| = 3|EC|
D
In the figure
E b
F
p x
D
K L C
B
C
m
F
D
y
B
E
& ) x.y.z + m.n.p A (DEF = a.b.c A (& ABC)
n
[AC] is divided into 5 equal parts. D is the midpoint of [AB]. & A ( ADF ) Find A (BCLD)
A
a
35. A A
30.
In triangle ABC,
F B
|AF| = 4|BF| B
C
D
3
D
C
A(DE∆F) = 28 cm2 Find A(AB∆C)
31.
A
4
8 B
H
C
In the figure,
|AB| = 8 cm, |AC|=4 cm, |BC| = 6 cm
are given. Find |AH|
In triangle ABC
A
36.
A(DE∆F)
6
2|AC| = 5|AE|,
F
E
|BD| = |DC|
E
According to the given figure, find
6
2|AD| = 3|DC| E
|AB| = 3|BE|
D
A(AB∆C) = 30 cm2 B
C
Find A(BD∆E)
268 37.
A
D
F
B
E
C
38.
42.
|AD| = |DC|
|BE| = 2|EC|
|BF| = 3|FD|
A(DE∆F) = 2 cm2
Find A(AB∆C)
|EC| = 9 cm
Find A(AB∆C)
4
B
5|DC| = 3|BC|
D
E B
C
D
5|BD| = 2|DC| & A (DEC ) Find & A ( ABD)
|AE| = 4 cm
C
A
43.
13
39.
According to the given figure, find A(AD∆C) 15
14
B
D
7
C
In triangle ABC
A
|BC| = 5|BD| F
|AE| = 3|EC|
E B
C
D
|AD| = 3|AF| & A (DEF) = 6 cm2
44.
40.
Find A(AB∆C)
E
|AE| = |EF| = |FC| F
B
B
In triangle ABC
A
C
D
6 30
C
4|BD| = 3|DC| & A ( ABC ) Find & A (DEF )
According to the given & A (EBD) figure, find & A ( ADC )
A 3
6
15
A
D
G is the centroid of the triangle ABC. |AB| = |AC|
E
|BG| = 5 cm
G
E 1 B
D
45. 41.
According to the given figure, find A(AB∆C)
A 7
[DE] ^ [AC]
9
[AD] ^ [BC]
E
|AE| = 3|EC|
In triangle ABC
A
In triangle ABC
A
In triangle ABC
C
|BD| = 4 cm B
D
C
Find A(DCEG)
269 In triangle ABC,
50.
[AD] is the angle bisector
|AB| = 8 cm
|AC| = 10 cm
A(ABC) = 27 cm2
Find A(AD∆C)
46.
A
B
C
D
47.
In the given figure,
A
[AE] ^ [CE]
B
[CB] // [AD] |AE| = 10 cm
C
|CE| = 15 cm
E
D
Find A(ABDE)
In triangle ABC,
A
In triangle ABC
[BE] is the angle bisector
51.
|AB| = 12 cm
|BD| = 8 cm
|DC| = 4 cm
|DF| = 2 cm
A(AB∆C) = 45 cm2
|AC| = 10 cm
Find A(AB∆E)
12
E
8
B
48.
D 4
C
In triangle ABC,
A 13
B
D
B
C
A
E
B
|AC| = 14 cm
|BC| = 15 cm
C
Find A(DB∆C)
Find A(AB∆E)
According to the given & A ( ABD) figure, find & A (BCD)
15
5
[DF] ^ [AC]
E
52.
|AB| = 13 cm
C
15
[DE] // [BC]
F
[BD] and [CD] are angle bisectors
14
D
A
D
F
53. A 49.
ABC is a right triangle
A
[AB] ^ [AC]
[EF] // [BC]
|AE| = 4 cm
|FC| = 7 cm
Find A(DE∆F)
E
4
F 7
B
In the figure,
D
C
E 10
6 B
C
D
given
[AB] ^ [BD] [ED] ^ [BD] |AB| = 10 cm |ED| = 6 cm & A ( ACD) Find & A (ECD)
270 54.
According to the given
A E
3
4
figure, find D
B
2
6
D
C
55.
16
F E
E
D
|AE| = |EC| A(BCED) = 12 cm2
B
Find A(AB∆C)
C
A
60.
F A
A2
75
A1
A2
G
D
C
|AF| = |FD| = |DC|
E
B
A(GEDF) = 75 cm2 Find A2 – A1
61.
A 20
57.
Find A(FBCK) 62.
G is the centroid of AB∆C.
A
E B
D
A(AB∆C) = 36 cm2 C
Find A(BDGE)
D
[EG] // [BD],
G
In triangle ABC
A
58.
B
12
A(AD∆E) = 18 cm2
C
B
E
According to the given figure, find A(AE∆D)
2|AD| = 3|FB| = 6|DF|
K
F
D
[DE] // [FK] // [BC]
E
D
C
In triangle ABC
A
Find A3
[FG] // [DE] // [CB].
D
A1 + A2 = 42 cm2
In the figure,
C
5|BD| = |BC|
E
A1 B
56.
[AB] // [DE],
A3
C
|AD| = |DB|
According to the given areas, find A(AF∆E)
40
In triangle ABC
A
[DE] // [BC]
E
B
A (BCDE) & A ( ABC )
In triangle ABC
A
59.
[DE] // [BC]
E
|BD| = 3|AD|
K B
C
A(DK∆E) + A(BK∆C) = 51 cm2
Find A(AB∆C)
271 4.5.2. The Law of Sines The Law of Sines
In a triangle ABC with corresponding sides a, b and c, A
a b c = = B sin C sin A sin
b
c a
B
C
Proof
A
Ex(1) x 45° B
Ex(2)
Ex(3)
According to the given figure, find x. 4 30°
C
W) = 60° are given. Find m ( CW) . In triangle ABC, a = 2§3 cm, b = 2 cm, m ( A
In triangle ABC, a = 2§3 cm, b = 2 cm, c = 2 cm are given. Find the interior angles of the triangle ABC
272 REVIEW EXERCISES A
1. x
According to the given figure, find x.
60° 45°
B
4§3
2.
C
According to the given figure, find x.
A 135°
x
30° B
5. In triangle ABC, c = 2 cm, m(C) = 30o, m(A) = 105o are given. Find b.
6. The perimeter of the triangle ABC is 26 cm.
If 2sinA = 3sinB =4sinC is satisfied for ABC, then find |BC|
C
2§2
3. In triangle ABC, m(A) = 30o, c = 4cm , a=2cm are given. Find m(C)
7.
14
D 6 B
4.
ABC is an acute angled triangle.
A 5§2
B
45° 5§3
According to the figure, find m(A). C
A
E 10 6§2
45° x
C
According to the given figure, find x.
273 5. VECTORS 5.1. Vectors and Operations with Vectors 5.1. 1 Vectors Vector
A vector in the plane is a directed line segment. B
Point of Interest The word vector is derived from the Latin ‘vehere’ meaning ‘to carry’. Note: A quantity which is specified by just its magnitude is called a scalar (e.g. distance, speed).
head
B
A
A
line containing A and B
tail
line segment AB
B
A
directed line segment AB
If the points of the line segment are ordered so that they proceed from A to B, a directed line segment from A to B is obtained, which is denoted by A∂B. In a directed line segment A∂B, A is called tail and B head.
A vector is a quantity that has magnitude and direction (e.g. displacement, velocity ) and sense. direction
v
u
direction
u
direction
u same direction but different sense different direction and sense
Components of a Vector
v = v1 + v2
v2
vertical component
horizontal component
v1
Whenever a vector v can be written as a sum v = v 1 + v 2 of two nonparallel vectors (the line segments representing the vectors are nonparallel), the vectors v 1 and v 2 are said to be the components of v .
274 Ex(1)
Sketch the components of the given vectors y
u v z
Ex(2)
A kite string exerts a pull (|F| = 12) on a kite and makes a 45o angle with the horizontal. Find the horizontal and vertical components of F. 45°
F
y
Representing Vectors in the Cartesian Plane
vy
y - component
v = vx + vy
x - component x
vx
If v is a vector with tail at the origin and head at P(a, b), then v can be represented as OP = v = (a, b) y
y P(4,3)
P(a,b) b O
v
a v = ^a, b h
x
3
v
O
4 v = ^4, 3 h
x
275 Ex(3)
Represent the following vectors in component form. y u= u
v
x
z
y
v= y= z=
Position Vector
If the tail of the vector is at the origin, then the vector is called a position vector. y
B(x2, y2)
P(a,b)
(y2 – y1)
v
O
Note: AB = B − A BA = A – B
A(x1, y1)
(x2 – x1)
x
The position vector of AB is the vector v . Let A(x1, y1) be the tail and B(x2, y2) be the head of the vector AB . Then the position vector v of AB can be found by
AB = –BA v = B – A = (x 2 – x 1, y 2 – y 1)
276 Ex(4)
Find the position vectors u and v of the vectors AB and CD where A, B, C, D are the points, A(–1, 2), B(3, 3), C(3, 1) and D(5, 3), both by sketching and finding components y
B
D
A C
x
O
Ex(5)
The points A(–5, 7) and B(6, 12) are given. Find the position vectors of AB and BA
Ex(6)
Magnitude of a Vector
Find the coordinates of the point A where AB = (5, –4) and B is the point (–3, 7)
The magnitude or norm of the vector AB , is the length of the line segment AB and is denoted by | AB |. Let AB be a vector defined from the point A(x1, y1) to the point B(x2, y2) y y2
B y2 – y1
y1 O
A x1
x2 – x 1 x2
x
Then the magnitude of AB , is defined by AB = AB = (x 2 − x 1) 2 + (y 2 − y 1) 2
277 If A = ^ x, y h is a position vector, y y
A
O
x
x
A = x2 + y2
Ex(7)
Ex(8)
Given A = ^3, 2h and B = ^ 2, –4h , find a)
A
b)
2A
c)
B
d)
− 3B
e)
AB
f)
BA
For the points A(m, 1) and B(4, –3), |A∂B| = 5 is given. Find the sum of the values of m.
278 Unit Vector
A unit vector is a vector which has a magnitude of 1.
Zero Vector
The vector which is of zero magnitude is called the zero vector, and is denoted by O . Unlike other vectors, zero vector has no direction.
Ex(9)
Ex(10)
1 A = a 3 , b k is a unit vector. Find the values of b.
Find the unit vector with same direction and sense as the vector A = (4, –3)
Note: The unit vector with same direction and sense as A A is , the unit vector A with same direction and opposite sense as A is A − A Equal Vectors
Two vectors are equal or the same if they have the same magnitude and direction. B A
C
D
A∂B = C∂D = E∂F
F
E Let A = (x 1, y 1) and B = ^ x 2, y 2h be two vectors then A = B + x 1 = x 2 and y 1 = y 2 Ex(11)
A = (m + 1, 1 – n) and B = (n + 3, 2m – 6) are given. If A = B then find m + n.
Ex(12)
A(3, 4), B(1, –2), C(2, –5) are given. If AB = CD , then find the coordinates of D.
279 REVIEW EXERCISES - I 1. Given the points A(3, 2), B(2, 1), find the position vectors of A∂B and B∂A.
2. Given the points A(1, –2), B(0, –1), C(–2, –5), find A∂B and B∂C.
3. Find the coordinates of the point A where BA = (–2, 1) and B is the point B(1, 5).
4. Given the vector AB = (1, –3) and B is the point B(–3, –5), find the point A.
5. Given the points A(3, 1) and B(3, –3), find |AB|.
6.
7.
1 A = a a, 2 k is a unit vector. Find the values of a.
A = (–3, 4) and B = (–4, k) are given. If A∂B is a unit vector, then find k.
8. A(3 + m, m +n) and B(n + 1, 4) are given. If AB is a zero vector, then find 2m + n.
9. Given A = (m + 2,3), B = (5, n – 1) and A = B . Find m + n.
10. A(5, 2), B(3, m) and C(n, –3) are given. If A∂B = C∂A, then find m + n.
280 5.1.2. Operations with Vectors Operations with Vectors (Geometric) Scalar Multiples of a Vector
A vector is multiplied by a positive real number by multiplying its magnitude by the number. A vector is multiplied by a negative number by reversing the vector’s direction and multiplying the magnitude by the number’s absolute value. u
Note: A∂B = –B∂A
−u
2u
Addition of Vectors a) Head-to-Tail Rule:
To add vector v to vector u, move vector v so that its tail touches to the head of u. The sum u + v is the vector from the tail of u to the head of v.
Note: A∂B + B∂D = A∂D
tail
head v
u
head
u
=
+
A∂B + B∂D + D∂C + C∂A = A∂A
v
u+v
tail b) Parallelogram Rule
To add vector v to vector u, move vector v so that its tail touches to the tail of u. Draw a representative of v from the head of u and a representative of u from the tail of v. The sum u + v is the vector from the tail of u to the head of the representative of v.
v
u
u
u+v
=
+
v
Note that, u – v is the vector obtained by adding –v to u. –v –v
u
=
+
u
–v +
u−v
u
u−v
u
=
–v
281 Ex(1)
The vectors u , v and z are given in the figure. Sketch
z y
a)
v u
u + v
f) u + v + z
b)
u + y
g) u − y
c)
v + z
h) v − z
d)
1 2y + 3 v
i)
3 5 u − v + 2z
282 Ex(2)
For the rectangle ABCD, find the resultant vector of the following operations D
C
A
B
a)
A∂D + A∂B
b)
B∂C + D∂C – A∂C
c)
A∂C – A∂D + B∂A
Operations on Vectors (Algebraic) Scalar Multiples of a Vector
Let A = ^ x 1, y 1h be a vector and k ΠR be a scalar then,
k . A = (k . x1, k . y1)
Addition of Vectors
Let A∂ = (x1, y1) and B∂ = (x2, y2) be two vectors, then y y1 + y 2
y (x2, y2)
y2 B y1
(x2, y2)
y2
B
B (x1, y1)
A
y1 A
x2
x1
x
x2
A + B = (x 1 + x 2 , y 1 + y 2)
A − B = x 1 + y 1 + (− x 2) + (− y 2)
= (x1 – x2 , y1 – y2)
(x1 + x2 , y1 + y2)
A+B
(x1, y1) x1
x1 + x 2
x
283 Ex(3)
Given A = (3, 2) and B = (2, 4) , find the vectors a) 2A
b) A + B
c) 3A − 2B
Ex(4)
Given A = (2, − 2) and B = (− 1, 3) , find C if A + C = 2B
Ex(5)
For the vectors u = (3, 9), v = (2, 7) and z = (2, 4), find m + n for m, n ∈ Z satisfying m . u + n . v = z
284 REVIEW EXERCISES - II 1. ABCD is a rectangle, |CE| = |EB|. Find the resultant vector of the following operations D
C E
A
4.
A
B
B
D
In AB∆C, |BD| = 2|DC|. Find AD in terms of AB and AC .
a) E∂A + 2B∂E + D∂C
A
b) A∂E + B∂E + B∂A
Note: | DB | m = n In AB∆C, if | DC |
c) D∂C – 2B∂E + B∂A 1 d) A∂E – 2 A∂D + B∂A
C
B
m
D
n
C
then AD =
n . AB + m . AC m+n
5. Given A = (–2, 2) and B = (4, 1) find 2. ABCD is a rectangle |DE| = |EC|. Find D
E
C
a) 2A
b) A+B c) 3A + 4B d) A
A
B
a) A∂B + A∂D + E∂D – A∂E
b) A∂E – E∂B = k . B∂C Ş k = ?
3.
D
A−B
f)
6.
Given A = (1, 2), B = (m, 5) and A – 2B = (4, –8). Find m
7. For the vectors , u = (2,1), v = (–1, 3) and z = (7,0),
A
B
e) A+B
find m + n for m, n ∈ Z satisfying m . u + n . v = z
C
8. A∂B = (–2, 5) and A∂C = (3, –1) are given. Find |B∂C|
In AB∆C, [AD] is the angle bisector.
9. A∂ + B∂ = (–6, 1)
| AB | 2 | AC | = 3
3A∂ – 2B∂ = (–3, 13)
B∂D = (2, 6) is given
Find C∂D
are given. Find |A∂B|
10. Find a vector of magnitude 2 units which has the same direction and opposite sense with the vector A = (6, 8)
285
CHAPTER 3
STATISTICS AND PROBABILITY
286 6. STATISTICS 6.1. Measures of Central Tendency and Spread Statistics is the development and application of methods to collect, analyze and interpret data. Medical, biological, social sciences, economics, finance, marketing research, government are some of the areas that use statistics.
6.1.1. Measures of Central Tendency The three most common measures of central tendency are the mean, the median, and the mode. Mean ( x )
The mean is the sum of the value of each observation in a data set divided by the number of observations. x=
x 1 + x 2 + x 3 + ... + x n n
Ex(1)
Find the mean of 5, 7, 8, 10, 15.
Ex(2)
The mean of the Physics marks of 4 students is 60. If the mean of the Physics marks of 3 of them is 54, then find the mark of the 4th one.
Mode
The mode is the most frequently occuring value in a distribution.
Ex(3)
Find the mode of the following data 3, 2, 5, 1, 7, 4, 5, 2, 5.
Ex(4)
Find the mode of the following data 2, 4, 8, 15, 20.
Remark! If no number occurs more often than the other numbers, then a set of data has no mode.
Ex(5) Remark! A set of data may have more than one mode.
Find the mode of the following data 4, 5, 5, 7, 8, 8.
287 Median
Median is the middle value in distribution when the data is arranged in order. If the number of the data is n+1 i) odd, then the median is the a 2 k th term n n ii) even, then the median is the mean of the “ ` 2 j th ” and the “ ` 2 + 1j th ”term.
Ex(6)
Find the median of the following data a) 4, 7, 8, 10, 15 b) 9, 7, 4, 8, 15, 10
Ex(7)
Find the mean, mode and median of the following data a) 1, 4, 4, 6, 7, 10, 12, 14, 24
b) 18, 28, 22, 26, 23, 19, 18, 20
c) 14, 10, 45, 38, 60, 14, 23, 35, 68, 50
Ex(8)
The following six integers are arranged from smallest to largest 1, x, 3, y, 14, z . The mode is 1, the median is 5 and the mean is 7. Find the values of x, y and z.
Ex(9)
If the mean of the given scores is 4, then find the value of a. score number of students with this score a) Find the mode b) Find the median
1
2
3
4
5
6
7
4
5
a
10
17
5
1
288 6.1.2. Measures of Spread The measures of central tendency are not always adequate to describe data. For example, two data sets can have the same mean but they can be entirely different. Thus to describe data, the extent of variability must be known. This is given by the measures of spread. Range, interquartile range, and standard deviation are the three commonly used measures of spread. Range
Ex(1)
Quartiles
The range of a set of data is the difference between the highest and the lowest value.
Find the range of the data 12, 23, 30, 32, 43, 57
The quartiles divide a group of observations into four quarters. Q1 (25th percentile) : The lower quartile. 25% (one quarter) of the values are less than or equal to it. Q2 (50th percentile) : The median. 50% (one half) of the values. Q3 (75th percentile) : The upper quartile. 75% (three quarters) of the values are less than or equal to it.
Interquartile Range (IQR): Q3 - Q1 It describes the middle 50% of the observations. Large IQR means that the middle 50% of observations are spaced wide apart.
Ex(2)
Find the quartiles of the following data a) 3, 3, 5, 6, 8, 9, 12, 14, 19, 20, 24
b) 20, 23, 23, 26, 27, 28
c) 147, 150, 154, 158, 159, 162, 164, 165
d) 10, 12, 13, 15, 19, 19, 24, 26, 26
289 REVIEW EXERCISES 1. From 5 tests, a student received the following scores; 72, 86, 92, 63, and 77. What test score must the student earn on the sixth test so that the mean score for all six tests will be 80?
4. A survey of 100 households asked how many bicycles there were in each household. The results are given below.
No of Bicycles
Frequency
0 1 2 3 4
5 70 21 3 1
2. The maths quiz marks of a group of 10 students are given below.
10, 12, 6, 13, 19, 25, 15, 8, 15, 17
Find
a) the median mark of the students
Calculate the mean number of bicycles per household.
b) the mean mark of the students
c) the mode
5. In a class of 20 students, the Chemistry marks for the last exam are shown in the table. Marks
50
60
70
80
90
100
Number of students
1
2
4
7
4
2
A new student joins the class and the mean math mark of the group of 11 students become 15.
Calculate
d) Find the math mark of the student who has joined the group
a) the mean
b) the median
c) the mode
3. Find the mean, median, mode of the following data
a) 2, 3, 3, 6, 8
b) 12, 8, 13, 5, 11, 5
c) 4, 8, 10, 4, 7, 6, 4, 10, 5, 6
290 Ex(3)
The following are the grades which 12 students get in a physics test 46, 68, 77, 30, 51, 17, 34, 55, 10, 72, 49, 85.
Find a) the range and the quartiles
b) the interquartile range (IQR)
Ex(4)
The times of arrival of 15 trains were recorded, and the number of minutes that each train was late is recorded in the table minutes late
0
1
2
3
4
5
number of trains
1
4
3
1
0
6
Find a) the median (Q2)
b) the lower quartile (Q1)
c) the upper quartile (Q3)
d) the interquartile range (IQR)
291 Ex(5)
The heights of the 40 players in a rugby match are given in the table height (cm)
170
175
180
190
195
number of players
5
15
10
8
2
Find a) the mean, mode and median
b) the minimum height in the top 25% of the players.
Standard Deviation
The sample standard deviation(s) shows how much dispersion exists from the mean. A low standard deviation indicates that the data tend to be very close to the mean; high standard deviation indicates that the data are spread out over a large range of values.
To find the standard deviation of the data 2, 4, 4, 4, 5, 5, 7, 9
i) find the mean :
x=
2 + 4 + 4 + 4 + 5 + 5 ++7 + 9 =5 8
ii) find the difference of each data from the mean, take the square of each, then find their sum x−x
2 – 5 = –3 4 – 5 = –1 4 – 5 = –1 4 – 5 = –1 5 – 5 = 0 5 – 5 = 0 7 – 5 = 2 9 – 5 = 4
(x − x ) 2 9 1 1 1 0 0 4 16 32
32 i) divide the sum to 7 ((number of data) –1) : 7 ii) take the square root :s =
32 = 2, 14 7
292 Ex(6)
Find the standard deviation of the following data a) 12, 8, 11, 10, 13, 6
b) 10, 16, 7, 5, 17
Ex(7)
The mean of the numbers 3, 6, 7, a, 14 is 8. Find the standard deviation of the set of numbers.
Ex(8)
Two machines A and B are used to pack biscuits. A sample of 10 packets was taken from each machine and the mass of each packet, measured to the nearest gram, was noted. Find the standard deviation of the masses of the packets taken from the sample from each machine. Comment on your answer. A (mass in g)
196 198 198 199 200 200 201 201 202 205
B (mass in g)
192 194 195 198 200 201 203 204 206 207
293 REVIEW EXERCISES 1. 14, 3, x, 7, 22, y, 9, 6, z are given. Find x, y and z given the following information. i) mode = mean = median = 9,
5. The population below is listed in ascending order,
5, 6, 7, 7, 9, 9, r, 10, s, 13, 13, t.
The median of the population is 9.5, the upper quartile is 13. a) Write down the value of
ii) 22 is the maximum value of the data,
iii) the range is 21
iv) z < x < y.
i) r
2. Consider the four numbers a, b, c, d with a ≤ b ≤ c ≤ d where a, b, c, d ∈ Z. The mean of the four numbers is 4. The mode is 3, the median is 3 and the range is 6. Find the values of a, b, c and d.
ii) s
b) Find the value of t if the mean of the population is 10.
6. The mean of the data 1, 3, 7, 9, a, 13, 21 is 9.
Find the standart deviation of this data.
7. A sample of two different batteries was tested to see how long they last
3. For the data set
6, 4, 7, 5, 3, 4, 2, 6, 5, 7, 5, 3, 8, 9, 3, 6, 5
find the quartiles and the interquartile range.
hours
1
2
3
4
5
6
Battery A
10
20
30
90
80
70
Battery B
50
40
55
45
50
60
Find the sample standart deviation of each battery and choose the reliable one. Explain your answer.
4. The following table shows the marks of the History multiple choice test. Marks
10
20
30
40
50
60
70
80
90
100
Number of students
1
3
2
5
7
4
1
2
2
2
Find
i) mode, median and mean
ii) the lower and upper quartiles
8. Two streetball teams compare their successful shots for the last 8 matches. I
23
17
31
25
25
19
28
32
II
9
29
41
26
14
44
38
43
a) Find the standart deviation of successful shots of each team
b) Which team is more consistent?
294 6.2. Graphs 6.2.1. Graphical Tools Once a research question has been developed and the data is collected, the next step is to classify and organise the data. There are two types of data, dependant upon on their possible values: qualitative (categorical) and quantitative (numerical). Quantitative data are further divided into discrete and continuous. DATA
Qualitative (Categorical)
Quantitative (Numerical)
Discrete
Continuous
A qualitative data places an individual into one of several groups or categories. For example, • The data gender has two possible values, male and female. • The data major has numerous values such as Mathematics, Biology, Physics, Economics, Chemistry... A quantitative data takes numerical values for which arithmetic operations (such as adding and averaging) make sense.
Quantitative data is divided into discrete and continuous :
• Discrete quantitative data takes on values which are spaced, i.e, for two adjancent values, there is no value that goes between them. We obtain them by counting.
The number of children is discrete. It takes on integer values, for example there cannot be 2.5 kids in a family.
• Continuous quantitative data take values in a given interval. For any two values of the data, we can always find another value that can go between the two. We obtain them by measuring. Data such as weight, time, and distance are continuous. After the data is classified and organized, then the data may be summarized using the measures of central tendency and spread. The data has been collected, classified, organised and summarised needs to be represented and interpreted graphically by different types of graphical representations such as bar graphs, pie charts, line graphs and histograms.
295 When deciding what kind of graph or chart best illustrates the data, you must think about the purpose of the graph or chart and what you want to present, then decide the variables to include and whether they should be expressed as frequencies, percentages, or categories. It should also be considered what type of data you are working with. Categorical data are grouped into non-overlapping categories (such as grade, race, and yes or no responses). Bar graphs, line graphs, and pie charts are useful for displaying categorical data. Continuous data are measured on a scale or continuum (such as weight or test scores). Histograms are useful for displaying continuous data. Bar Graph
A bar graph is composed of discrete bars that represent different categories of data. The length or height of the bar is equal to the quantity within that category of data. Bar graphs are appropriate for data that are non-numerical and discrete for at least one variable, i.e they are grouped into separate categories. There are no dependent or independent variables. In a bar graph, • Data is collected for discontinuous, non-numerical categories (e.g. place, colour, and species) so the bars do not touch. • Bar graphs are best used to compare values across categories.
Ex(1)
Favorite Sport 9 8
Number people
7 6 5 4 3 2 1 0
Football
Basketball
Volleyball
Tennis
Swimming
According to the bar graph given, answer the following questions a) How many people said volleyball was their favorite sport? b) Did more people like basketball or swimming? c) Which sport did exactly 7 people say was their favorite? d) What is the difference in the number of people who liked tennis and number who liked basketball? e) What is the combined number of people who liked football and tennis? f)
Which sport was liked by the largest number of people?
g) Which sport was liked by the fewest number of people?
296 Sales of different icecream flavors
Ex(2) Number of Icecreams sold
120 100 80
Shop A
60
Shop B
40 20 0
Chocolate
Vanilla
Strawberry
Caramel
Lemon
Flavors
According to the bar graph given, answe the following questions
a) How many vanilla ice cream were sold by Shop A?
b) Which flavor of ice cream do people like most in shop A?
c) Which flavor of ice cream do people like the most in shop B? Strawberry or vanilla?
d) Which flavor(s) sales are better in shop B than in shop A?
e) Find the total numbers of strawberry ice cream sold by shop A and shop B.
f)
Which shop sold 30 caramel ice cream?
g) Which shop earns more money by selling ice cream?
297 Pie Chart
Pie charts are best to use to compare parts of a whole. They do not show changes over time. A pie chart is divided into sectors and each sector represents a particular proportion of the data. In a pie chart, • The data for one variable are discontinuos. • The data for the dependent variable are usually in the form of counts, proportions or percentages. • They are not suitable for data sets with a large number of categories.
Ex(3) Favorite Holiday Activity walking
cycling 60°
30°
swimming
sunbathing
According to the pie chart given, answer the following questions
a) Which activity was liked by the largest number of people?
b) Which activity was liked by the fewest number of people?
5 people said that walking is their favourite holiday activity,
c) How many people took part in the survey?
d) What is the difference in the number of people who liked sunbathing and the number who liked cycling?
298 Line Graph
A line graph is a way of visually representing related data where individual items of data are plotted and joined by a line. They are useful for presenting data over time to compare changes to a variable over a set period. In a line graph, the variable represented is continuous; i.e. it is something that can be measured.
Ex(4)
Pages Read per Day for One Week 80 70
Number of Pages Read
60 50
= Tom = Jerry
40 30 20
Sat.
Fri.
Thurs.
Wed.
Tues.
Mon.
0
Sun.
10
Day of the Week
According to the graph given, answer the following questions a) How many pages did Tom read on Sunday? b) How many pages did Jerry read on thursday? c) On which day did Tom read the fewest pages? d) On which day did Jerry read the fewest pages? e) How many more pages did Tom read than Jerry on monday? f)
On which days did Tom read exactly five pages more than Jerry?
g) Which of them read the most pages during the week? h) Which of them was more consistent in doing the assigned reading?
299 Histograms are plots of continuos data and are often used to represent frequency distributions, where the y–axis shows the number of times a particular measurement or value was obtained. For this reason, they are often called frequency histograms. • The data are numerical and continuous (e.g. height or weight) so the bars touch. • The x–axis usually records the class interval. • The y–axis usually records the number of individuals in each class interval (frequency).
70 60 Frequency
Histogram
50 40 30 20 10 0
5
10 15
20 25
30
35 40
45 50 55 60
65 70 75
Age (in years)
The histogram below shows the ages of people attending the showing of a new movie.
a) How many people attended the movie?
b) What would you estimate the median age of the group be?
c) What kind of movie might this be?
300 REVIEW EXERCISES 1. Calories Burned Per Hour
1.000
2. 60%
800
58%
700
56%
600
54%
500
52% 50%
400
48%
300
44%
Handball
Walking
Tennis
Bicycling
Swimming
Skiing
46% Running
200
42% 40%
Type of Exercise
Use the bar graph to answer these questions. a) How many calories would you burn playing handball for one hour? b) Approximately how many calories would you burn bicycling? c) How many more calories would you lose running rather than playing tennis? 1 d) How many calories would you burn on a 2 2 hour skiing trip? e) Which two activities are almost the same in terms of the amount of calories they burn? f)
Male/Female Attendance at Colleges
900
Approximately how many calories would you burn after one hour of running and one hour of swimming?
g) Which two activities would have to be done for one hour each, to equal one hour of skiing? h) Would you burn more calories on a 3–hour walk or a 1–hour run?
UCLA
NYU = male
Yale
Harvard = female
According to the bar graph, answer the following questions a) What percentage of students at Yale is male?
What percentage is female?
b) Which college has the greatest disparity between the percentage of male and female students? c) Using the graph as a representative of college attendance, are more males or more females attending these colleges?
301 3.
Elements as a Percentage of the Earth’s Crust
28% Sillicon
4.
Percentage of Body Weight
Water 62% 9% Other 3.5% Calcium
8%
47% Oxygen
4.5% İron
in
um Al um
This circle graph illustrates which elements are most abundant in the earth’s crust.
a) Which is the abundant element in the earth’s crust?
b) Which two elements make up three – fourth’s of the earth’s crust?
c) Which two elements together are equal to the amount of aluminum in the earth's crust?
d) Where would carbon, hydrogen, and sodium be included?
Protein 17%
Oth tro er 3% ge n 3%
Ni Fat 15%
Use the circle graph to answer these questions. a) Which component makes up the highest percentage of body weight? b) Which single component on the graph has the lowest percentage of body weight? c) What percentage of body weight do fat and protein make up together? d) Where do you think calcium, sodium, and iron are included in the graph? e) How much greater is the percentage of water than fat?
302 5.
Minutes of Guitar Practice Each Week for Six Weeks
6. This double line graph shows the average heights of boys and girls by age from 7 through 16. Study the graph and answer the questions below. Average Heights of Children 175
75 60
= Carlos
170
= Santana
165 160 Height in cm
Number of Practice Minutes
90
45 30 15
150 145 140 135
6 th
5 th
4 th
3 rd
2 nd
130
1 st
0
155
125 120
7
8
9
10
11
12
13
14
15
16
Week
According to the line graph, answer the following questions
a) At which two ages do boys and girls average the same heights?
a) How many minutes did Carlos practice the first week? b) At which two ages are girls on average taller than b) How many minutes did Santana practice the fifth week?
boys?
c) At what age do boys average 10 cm taller than girls? c) Which student practiced more in the sixth week? d) At what three ages do boys and girls grow at a about d) How many minutes did Carlos practice for the entire six weeks?
the same amount before the girls catch up to boys?
e) At what age do boys catch up and pass girls? e) How many minutes did Santana practice for the entire six weeks?
303
No. Of Personnel
7.
20 18 16 14 12 10 8 6 4 2 0
4
8
12
16
20
24
28
32
36
Percent Body Fat
Description
Men
Essential Fat
2 - 5%
Athletes
6 - 13%
Fitness
14 - 17%
Average
18 - 24%
Obese
25% +
The histogram shows the percent body fat of 80 male personnel working in a office. According to the histogram and the table from the American Council on Exercise, answer the following questions
a) Approximately how many of the personnel can be considered as fit or average?
b) Approximately how many of the personnel can be considered as fit or obese?
c) Estimate the description for the median percent body fat.
d) If you were a dietitian, do you think that this office is a good place for you to consult people on how to loose weight?
304 6.2.2. Scatter Plot The statistical study of the relationship between variables is called regression. A scatter plot is helpful in understanding the form, direction and strength of the relationship between two variables. The direction and strength of the linear relationship between the two variables is called correlation. Scatter Plot
Ex(5)
Scatter plot (scatter diagram) is a type of mathematical diagram using cartesian coordinates to display values for two variables for a set of data. It is used to investigate the possible relationship between two variables that both relate to the same event. For the point A(x,y) plotted in the scatter diagram, A(x,y) dependent variable (vertical axis)
independent variable (horizontal axis)
Imagine drawing a straight line or curve through the data so that it ‘fits’ as well as possible. The more the points cluster closely around the imaginary line of best fit, the stronger the correlation that exists between the two variables. In a collection of ordered pairs (x, y), if y tends to increase as x increases, the collection has a positive correlation.
y
y
x
strong positive correlation
weak positive correlation
x
If y tends to decrease as x increases, the collection has a negative correlation.
y
y
strong negative correlation
x
weak negative correlation
x
305 • If it is hard to see where you would draw a line, and if the points show no significant clustering, there is probably no correlation.
y
no correlation
x
If there is a strong linear relationship between two variables (positive or negative), a line of best fit, or a line that best fits the data, can be used to make predictions. This is also called a trend line.
Ex(1)
The table shows the results of an experiment in a chemistry lab. x
10
20 30
40 50
60
70
y
17
21 25
28 33
40
49
Draw the scatter diagram and explain the relation between x and y. y
x
306 Ex(2)
The following table gives the test results for 10 children A
B
C
D E
F
G
math mark (x)
1
8
15
18 23
28
33 39
45 45
english mark (y)
3
14 8
20 19
17
36 26
14 29
child
H
I
J
a) Draw the scatter diagram y
x
b) y = 0,43x + 7,57 represents the equation of English marks from the Math marks. What mark in English would you expect with a mark of 20 in Math?
6.2.3. Box and Whisker Plot Box and whisker plot is a convenient way of graphically depicting groups of numerical data through their quartiles. It is used to display differences between populations without making any assumptions of the underlying statistical distribution. Interquartile Range
Q3 - Q1 Range max - min
OUTLIER Less than 3/2 times of IQR
MEDIAN MAXIMUM OUTLIER 50% of data is Greatest value, greater than this excluding outliers More than 3/2 times of IQR value; middle of dataset UPPER QUARTILE LOWER QUARTILE 25% of data greater 25% of data less than than this value this value
MINIMUM Least value, excluding outliers
Outliers are the points lying more than 1, 5 times the IQR above Q3 or Iying less than 1,5 times the IQR below Q1. Mark any outlier with ‘•’ or ‘*’ .
307 Ex(1)
Find the following for the information given in the box and whisker plots a)
b) 0 10 20 30 40 50 60 70 80 90 100 1
lovest value
:
lovest value
:
highest value :
highest value :
median
median
:
:
lower quartile :
lower quartile :
upper quartile :
upper quartile :
range
:
range
:
IQR
:
IQR
:
c)
Ex(2)
0 10 20 30 40 50 60 70 80 90 100 2
d) 0 10 20 30 40 50 60 70 80 90 100 3
0 10 20 30 40 50 60 70 80 90 100 4
lovest value
lovest value
:
:
highest value :
highest value :
median
median
:
:
lower quartile :
lower quartile :
upper quartile :
upper quartile :
range
:
range
:
IQR
:
IQR
:
A box and whisker plot has been drawn to show the heights of a special plant, in cm
0
5
10
15
a) What is the i) tallest height :
ii) shortest height :
iii) 75th percentile :
iv) median height :
v) range :
vi) IQR :
20
25
308 b) Give an example for any outlier Steps for creating a box and whisker plot : 1. Order your data from least to greatest 2. Find the median of your data 3. Find the quartiles of your data (the median of the upper and lower half) 4. Find the extremes of your data (the least and greatest values) 5. Plot the median, quartiles, and extremes below a number line 6. Draw the box and the whiskers
Ex(3)
Make the box and whisker plot of the following data 44, 28, 51, 62, 60, 43, 51, 66, 32, 57, 54
Ex(4)
A class of students have played a computer game wich tested how quickly they reached to a visual instruction to press a particular key. The box and whisker plot shows their reaction times in seconds. girls boys 1
2 3
4
5
6 7
8 9 10 11 12 13 14 15 16 17 18 19 20
a) Explain the performance of the boys and girls in this game.
b) Find the IQR of the reaction times of the boys and girls.
309 REVIEW EXERCISES 1. Each day, a factory recorded the number (x) of boxes it produces and the total production cost (y) dollars. The result of seven days are shown in the following table x
26
y
400 582 784 625 699 448 870
44 65
43 50
31
68
a) Draw the scatter diagram y
x
b) y = 10,6x + 132 represents the equation of the production cost from the number of boxes. Estimate the cost of producing 60 boxes.
2. Find the following for the intformation given in the box and whisker plots
0 10 20 30 40 50 60 70 80 90 100
0 10 20 30 40 50 60 70 80 90 100
lovest value
lovest value
:
:
highest value :
highest value :
median
median
:
:
lower quartile :
lower quartile :
upper quartile :
upper quartile :
range
:
range
:
IQR
:
IQR
:
0 10 20 30 40 50 60 70 80 90 100
0 10 20 30 40 50 60 70 80 90 100
lovest value
lovest value
:
:
highest value :
highest value :
median
median
:
:
lower quartile :
lower quartile :
upper quartile :
upper quartile :
range
:
range
:
IQR
:
IQR
:
310 3. The box and whisket plot of the data
5. The box and whisker plot shown below represents the marks received by 32 students
48, 48, 49, 49, 50, 52, 52, 53, 57, 58, 58, 61, 64, 64, 66
is shown below
A
B
C
D
1
2
3
4
5
6
7
8
9
10
a) Write down the value of the median mark.
E
b) Write down the value of the upper quartile. a) Write down the values of A, B, C, D and E. c) Estimate the number of students who received a mark greater than 6. b) Find the IQR.
6. According to the box-whisker plot given,
4. Make the box and whisker plot of the following data
10
15
20
25
30
35
a) Find the median
19, 27, 19, 24, 21, 20, 23, 29, 25, 26, 33
b) Find the IQR
c) Find the maximum and minimum values
d) What percentage of the data is less than 14?
311 7. PROBABILITY 7.1. Probability 7.1.1. Basic Definitions and Computing Probability Probability Sample Space Outcome Event
Probability is the calculation of the chance that something will happen. The set of all outcomes (results) of an experiment is called a sample space. Each element of the sample space is called an outcome or sample point. Any subset of a sample space is called an event.
S
: sample space
A
: event
n(S)
: number of possible outcomes
n(A)
: number of elements in the set A
P(A)
: The probability that the event A happens
0
: The probability of impossible event
1
: The probability of certain event n (A) P (A) = n (S)
Ex(1)
(A Õ S)
0 ≤ P(A) ≤ 1
What is the probability of getting a head when a coin is tossed? outcomes
: head and tail
sample space (S) : {head, tail} event (A)
: {head}
n(S) : 2 n(A) : 1 = P (A)
Ex(2)
n (A) 1 n= (S) 2
What is the probability of getting an even number on the top when a die is rolled? outcomes
: 1, 2, 3, 4, 5, 6
sample space (S) : {1, 2, 3, 4, 5, 6} event (A)
: {2, 4, 6}
n(S)
:6
n(A) :3 n (A 3 1 P= (A) n= (S) 6 = 2
312 Ex(3)
A die is rolled. Find the probability that the number obtained is a) a six b) less than 3 c) less than 7 d) greater than 6 e) a factor of 6
Ex(4)
From a set of cards numbered from 1 to 20, a card is drawn at random. Find the probability that the number is a) even b) prime c) divisible by 4 d) a multiple of 5
Combined Events
A
B x
y
z
t
n(A » B) = n(A) + n(B) – n(A « B) so; P(A » B) = P(A) + P(B) – P(A « B)
Ex(5)
A card is selected at random from an ordinary deck of 52 cards. Find the probability that the card is a) a heart b) a five and a heart c) a five or a heart d) a black and a six e) a black or a six
313 Ex(6)
A die is tossed. Find the probability that the number obtained is a) even and prime b) even or prime c) less than 5 and odd d) less than 5 or odd
Ex(7)
In a bag there are 12 discs numbered from 1 to 12. A disc is selected at random from the bag. Find the probability that the number on the selected disc is a) greater than 6 and even b) greater than 6 or even c) divisible by 2 and divisible by 3 d) divisible by 2 or divisible by 3
Mutually Exclusive Events
Two events are said to be mutually exclusive if they cannot occur at the same time. It is same as disjoint. A
B
n(A ∩ B) = 0 so, P(A » B) = P(A) + P(B) When a coin is flipped, getting a head and getting a tail is mutually exclusive. Either one of them is get.
314 Ex(8)
In a bag there are 15 discs numbered from 1 to 15. A disc is selected at random from the bag. Find the probability that the number on the selected disc is a) an even number or a prime number greater than 3.
b) less than 4 or greater than 10.
Ex(9)
Given that the events A and B are mutually exclusive with P(A) = 3x, P(B) = 4x and P(A » B) = 1 – x. Find the value of x.
Complementary Events
The complementary of an event A (written A’), consists of all outcomes in the sample space which are not contained in A.
A A'
So, A and A' are mutually exclusive. P(A » A') = P(A) + P(A') = 1 P(A') = 1 – P(A)
Ex(10)
P(A) = 0,55 P(A » B) = 0,7 P(A « B) = 0,2
are given. Find P(B').
315 Ex(11)
Events A and B are such that P(A) = 0,3 ,P(B) = 0,6 and P(A » B) = 0,7 A
B q
r
s
t
The values q, r, s and t represent the probabilities a) Write down the value of t.
b) i) show that r = 0,2
ii) write down the values of q and s
c) Write down P(B’) and P(A’).
Ex(12)
A fisherman goes fishing at his favourite place. The information below shows four different types of fish; anchovy, sea bass, sea bream and sword fish. The fish are either undersized or normal. size / type of fish
anchovy
sea bass
sword fish
sea bream
TOTAL
undersized
18
12
3
9
42
normal
24
11
0
13
48
TOTAL
42
23
3
22
The fisherman catches a fish. Find the probability that he
a) catches a normal fish.
b) catches an undersized anchovy.
c) does not catch a normal sea bream.
d) catches a sea bream and an undersized fish.
e) catches a sea bass or an undersized fish.
316 REVIEW EXERCISES 1. A ball is randomly selected from a box containing 3 black, 4 red and 5 blue balls. Determine the probability of getting a) A red ball
3. In a class of 30 students, 19 listen to pop music, 17 listen to classical music and 15 listen to both of them
Find the probability that a randomly selected student listens to a) Both musics
b) At least one of the musics b) A green ball
c) Pop music but not classical music
c) A black or a blue ball. d) Neither of them.
2. A die is rolled once. Find the probability of a) Getting a 5
b) Not getting a 5
4. 40 boys were asked whether they gave their girl friend flower or chocolate for the last Valentine’s day. The results were; 28 gave chocolates, 11 gave flowers and 10 gave both chocolate and flowers. If one of the boys were chosen at random, find the probability that he has given his girlfriend a) Chocolates or flowers
b) Chocolates but not flowers c) Getting 3 or 4
c) Only flowers
d) Getting an even number or a number greater than 3.
d) Neither chocolates, nor flowers
317 5. The medical records for a hospital of 30 patients on Friday showed whether they had previously had measles or mumps. The records showed 24 had had measles, 12 had had both measles and mumps and 26 had had measles or mumps. If one patient from this hospital is selected randomly, find the probability that he/she has had
8. In a group of 45 people, 22 play tennis, 33 play football, 17 play both of them. A person is selected randomly from this group. Find the probability that the person plays a) Football b) Tenis
a) Mumps c) Neither football nor tenis
b) Mumps but not measles
d) Only football
e) Football or tenis c) Neither mumps nor measles
9.
Use the Venn diagram below to find
A
B a
d) Only measles
b
c
d
U a) P(A) b) P(B)
c) P(A ∩ B) 6. A die is rolled once. Find the probability of getting
d) P(A » B)
a) A multiple of 3 e) P(A’)
b) A prime number
f) P(B’)
c) A factor of 8 or less than 5
10. In a class of 40 students, 30 study Mathematics, 22 study Physics and each student study at least one of these subjects. If a student is randomly selected from this group, find the probability that he/she studies a) Mathematics but not Physics
b) Both Mathematics and Physics 7. In a bag there are 7 red, 5 blue and 3 white balls.
A ball is randomly chosen. Find the probability that it is white.
c) Only Physics
d) At most two subjects
ISBN 978-605-5485-51-1