The Original Area Mazes, Volume 2: 100 More Addictive Puzzles to Solve with Simple Math—and Clever Logic! 161519522X, 9781615195220

Table of contents :
Brain Training with Area Mazes
How to Solve Area Mazes
Puzzles
Level 1
Level 2
Level 3
Level 4
Level 5
Solutions

Citation preview

Contents Cover Brain Training with Area Mazes How to Solve Area Mazes Puzzles Level 1 Level 2 Level 3 Level 4 Level 5 Solutions

THE ORIGINAL AREA MAZES—VOLUME TWO: 100 More Addictive Puzzles to Solve with Simple Math—and Clever Logic! Copyright © 2012 by Gakken Publishing, 2018 by Gakken Plus Translation © 2018 by Paper Crane Agency & Paper Crane Editions Puzzles copyright © 2012, 2018 by Naoki Inaba Originally published in Japan as Logical Shikou Training Puzzle Menseki Meiro (ロジカル思考トレーニ ングパズル ⾯積迷路) by Gakken Publishing Co., Ltd, Tokyo in 2012. First published in North America by The Experiment, LLC, in 2018. English translation rights arranged with Gakken Plus Co., Ltd., through Paper Crane Agency. All rights reserved. Except for brief passages quoted in newspaper, magazine, radio, television, or online reviews, no portion of this book may be reproduced, distributed, or transmitted in any form or by any means, electronic or mechanical, including photocopying, recording, or information storage or retrieval system, without the prior written permission of the publisher. The Experiment, LLC 220 East 23rd Street, Suite 600 New York, NY 10010-4658 theexperimentpublishing.com Many of the designations used by manufacturers and sellers to distinguish their products are claimed as trademarks. Where those designations appear in this book and The Experiment was aware of a trademark claim, the designations have been capitalized. The Experiment’s books are available at special discounts when purchased in bulk for premiums and sales promotions as well as for fund-raising or educational use. For details, contact us at [email protected]. ISBN 978-1-61519-522-0 Ebook ISBN 978-1-61519-526-8 Cover design by Sarah Smith, based on a design by Lidija Tomas Text design by Sarah Schneider Manufactured in the United States of America First printing October 2018 10 9 8 7 6 5 4 3 2 1

BRAIN TRAINING WITH AREA MAZES

As long as I keep my brain active, does it matter how? What comes to mind when you hear the word “training”? Most likely a “workout” in which you use machines and barbells to strengthen your muscles. But today, more and more of us are “brain training”— stimulating the brain so that its function will not decline. It’s a daily step you can take for your health—just like brushing your teeth. Everything you do uses your brain somehow, so you might think it is enough to just focus your mind on any task. However, some activities are better than others.

Puzzles with math and logic in mind El Camino is a science and math coaching school for students from first grade through high school. To prepare our elementary school students for the Mathematical Olympiads, we give them puzzles that strengthen their logical and creative thinking. Saying “let’s solve a puzzle!” appeals to the children much more than “let’s do a math problem.” The students tackle the puzzles enthusiastically—having fun, developing their abilities, and deepening their interest in mathematics all at once. We invented “area mazes” (menseki meiro) especially for our third-grade students. Area mazes can seem impossible to solve without using fractions and decimals. However, our third-grade students haven’t learned those techniques yet, so of course you can solve them using only whole numbers! The challenge is to work out how. It takes more than calculation: It takes logic, spatial reasoning, and wits. Getting the answer without doing any complicated math is what makes area mazes fun!

Isn’t it too late for my brain to grow?

Let me return to the original topic. When our students’ parents ask me how to train their own brains, I recommend the puzzles we use at El Camino. You might say, “What? But those are for kids!” Yes, that is true—and that’s exactly my point. The young mind is a flexible “blank slate,” whereas the adult brain becomes inflexible. That’s why many puzzle books designed for adults fall short: The puzzles tend to follow a pattern. Once your brain gets used to it, you will focus on spotting and applying the pattern—rather than thinking creatively. To use our metaphor of strength training, doing that kind of puzzle is something like rehab. Repeating an action over and over again will keep your brain active and maintain its function to some extent. However, it cannot rejuvenate your brain: It may put off aging, but it won’t make you younger. Area mazes are different: They are designed to develop flexible thinking—youthful thinking, if you will. They cannot be solved by repeating a process. You often need a stroke of inspiration to solve them! As you work through these puzzles, you will feel your brain “waking up.” I hope you will enjoy area mazes. There is a delightful sense of achievement that comes with saying, “I got it!” —RYAN MURAKAMI, director of El Camino

HOW TO SOLVE AREA MAZES

Using the given lengths and areas, find the value of . Remember, the formula for the area of a rectangle is height × width. If your calculation creates a fraction or decimal, STOP and look for another way. Area mazes can be solved using whole numbers only!*

Example One Find length . . . 45 ÷ 5 = 9 in. Find length . . . This is the same as , so 9 in. Find length . . . 72 ÷ 9 = 8 in. Find length . . . This is the same as , so 8 in. Length is 32 ÷ 8 = 4 in. Solution 4 in. Note that the figures are not drawn to scale. You can’t solve by “eyeballing”— you have to prove it with math! Even after you have solved a problem, you can revisit it to look for a more elegant solution.

Example Two Find length . . . 42 ÷ 6 = 7 in. Find length . . . 13 − 7 = 6 in. Find length . . . 24 ÷ 6 = 4 in. Find length . . . (4 + 7) − 6 = 5 in. Find length . . . 20 ÷ 5 = 4 in. Find length . . . 28 ÷ 7 = 4 in. Length is 4 + 4 = 8 in. Solution: 8 in.

Example Three First, consider the two rectangles on the left together. Find the area with stripes. . . 5 × 10 = 50 in.2 Find area . . . 50 − 36 = 14 in.2 Next, we will look at the two rectangles on the bottom. Area is the same height as area , and exactly twice as wide. So, area must be exactly double area . Area is 14 × 2 = 28 in.2 Solution: 28 in.2 *However, do not assume that every length or area in the puzzle must be a whole number.

PUZZLES

The puzzles get more challenging as you go. This key will guide you!

PUZZLE 1

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PUZZLE 2

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PUZZLE 3

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PUZZLE 4

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PUZZLE 5

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PUZZLE 6

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PUZZLE 7

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PUZZLE 8

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PUZZLE 9

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PUZZLE 10

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PUZZLE 11

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PUZZLE 12

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PUZZLE 13

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PUZZLE 14

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PUZZLE 15

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PUZZLE 16

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PUZZLE 17

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PUZZLE 18

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PUZZLE 19

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PUZZLE 20

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PUZZLE 21

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PUZZLE 22

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PUZZLE 23

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PUZZLE 24

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PUZZLE 25

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PUZZLE 26

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PUZZLE 27

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PUZZLE 28

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PUZZLE 29

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PUZZLE 30

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PUZZLE 31

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PUZZLE 32

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PUZZLE 33

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PUZZLE 34

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PUZZLE 35

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PUZZLE 36

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PUZZLE 37

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PUZZLE 38

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PUZZLE 39

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PUZZLE 40

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PUZZLE 41

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PUZZLE 42

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PUZZLE 43

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PUZZLE 44

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PUZZLE 45

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PUZZLE 46

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PUZZLE 47

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PUZZLE 48

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PUZZLE 49

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PUZZLE 50

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PUZZLE 51

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PUZZLE 52

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PUZZLE 53

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PUZZLE 54

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PUZZLE 55

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PUZZLE 56

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PUZZLE 57

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PUZZLE 58

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PUZZLE 59

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PUZZLE 60

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PUZZLE 61

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PUZZLE 62

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PUZZLE 63

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PUZZLE 64

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PUZZLE 65

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PUZZLE 66

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PUZZLE 67

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PUZZLE 68

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PUZZLE 69

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PUZZLE 70

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PUZZLE 71

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PUZZLE 72

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PUZZLE 73

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PUZZLE 74

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PUZZLE 75

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PUZZLE 76

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PUZZLE 77

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PUZZLE 78

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PUZZLE 79

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PUZZLE 80

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PUZZLE 81

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PUZZLE 82

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PUZZLE 83

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PUZZLE 84

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PUZZLE 85

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PUZZLE 86

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PUZZLE 87

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PUZZLE 88

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PUZZLE 89

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PUZZLE 90

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PUZZLE 91

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PUZZLE 92

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PUZZLE 93

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PUZZLE 94

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PUZZLE 95

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PUZZLE 96

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PUZZLE 97

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PUZZLE 98

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PUZZLE 99

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PUZZLE 100

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SOLUTIONS

Puzzle 1 Solution: 16in.2 Find length . . . 4 + 3 = 7in. Find length . . . 42 ÷ 7 = 6in. Find length . . . 10 − 6 = 4 in. Area is 4 × 4 = 16 in.2

Puzzle 2 Solution: 42 in.2

Find length . . . 10 − 6 = 4 in. Find length . . . 28 ÷ 4 = 7 in. Area is 7 × 6 = 42 in.2

Puzzle 3 Solution: 54 in.2 Find length . . . 56 ÷ 7 = 8 in. Find length . . . 8 − 4 = 4 in. Find length . . . 5 + 4 = 9 in. Area is 6 × 9 = 54 in.2

Puzzle 4 Solution: 30 in.2 Find length . . . 9 − 5 = 4 in. Find length . . . 24 ÷ 4 = 6 in. Area is 6 × 5 = 30 in.2

Puzzle 5 Solution: 6 in. Add the two areas on the bottom . . . 13 + 17 = 30 in.2

Find length . . . 30 ÷ 10 = 3 in. Find length . . . 10 − 3 = 7 in. Length is 42 ÷ 7 = 6 in.

Puzzle 6 Solution: 6 in. Find length . . . 9 ÷ 3 = 3 in. Find length . . . 3 + 4 = 7 in. Find length . . . 28 ÷ 7 = 4 in. Find length . . . 4 + 5 = 9 in. Find length . . . 54 ÷ 9 = 6 in. Find length . . . 6 + 6 = 12 in. Length is 72 ÷ 12 = 6 in.

Puzzle 7 Solution: 20 in.2 Find length . . . 45 ÷ 5 = 9 in. Find length . . . 13 − 9 = 4 in. Find length . . . 40 ÷ 4 = 10 in. Find length . . . 16 − 10 = 6 in. Find length . . . 48 ÷ 6 = 8 in. Find length . . . 8 − 4 = 4 in. Find length . . . 10 − 5 = 5 in. Area is 4 × 5 = 20 in.2

Puzzle 8 Solution: 10 in. Find length . . . 20 ÷ 5 = 4 in. Find length . . . 9 − 4 = 5 in. Find length . . . 30 ÷ 5 = 6 in. Find length . . . 14 − 6 = 8 in. Find length . . . 40 ÷ 8 = 5 in. Find length . . . 10 − 5 = 5 in. Length is 50 ÷ 5 = 10 in.

Puzzle 9 Solution: 4 in. Find length . . . 3 + 4 = 7 in. Find length . . . 35 ÷ 7 = 5 in. Find length . . . 5 + 5 = 10 in. Find length . . . 40 ÷ 10 = 4 in. Find length . . . 4 + 4 = 8 in. Find length . . . 48 ÷ 8 = 6 in. Find length . . . 6 + 3 = 9 in. Length is 36 ÷ 9 = 4 in.

Puzzle 10 Solution: 7 in. Find length . . . 6 + 4 = 10 in. Find length . . . 10 − 5 = 5 in. Find length . . . 3 + 5 = 8 in. Find length . . . 8 − 5 = 3 in. Length is 3 + 4 = 7 in.

Puzzle 11 Solution: 3 in. Find length . . . 12 ÷ 3 = 4 in. Find length . . . 20 ÷ 4 = 5 in. Find length . . . 5 + 3 = 8 in. Find length . . . 32 ÷ 8 = 4 in. Find length . . . 4 + 4 = 8 in. Find length . . . 40 ÷ 8 = 5 in. Find length . . . 3 + 5 = 8 in. Length is 24 ÷ 8 = 3 in.

Puzzle 12 Solution: 6 in. Find length . . . 18 ÷ 6 = 3 in. Find length . . . 3 + 4 = 7 in. Find length . . . 35 ÷ 7 = 5 in. Find length . . . 9 − 5 = 4 in. Find length . . . 52 ÷ 4 = 13 in. Find length . . . 13 + 4 = 17 in. Length is 102 ÷ 17 = 6 in.

Puzzle 13 Solution: 5 in. Find length . . . 11 − 6 = 5 in. Find length . . . 35 ÷ 5 = 7 in. Find length . . . 4 + 7 = 11 in. Find length . . . 11 − 5 = 6 in. Length is 30 ÷ 6 = 5 in.

Puzzle 14 Solution: 8 in.

Find length . . . 49 ÷ 7 = 7 in. Find length . . . 7 − 2 = 5 in. Find length . . . 5 + 3 = 8 in. Find length . . . 56 ÷ 8 = 7 in. Find length . . . 7 − 4 = 3 in. Find length . . . 5 + 3 = 8 in. Find length . . . 64 ÷ 8 = 8 in. Find length . . . 8 − 6 = 2 in. Find length . . . 2 + 7 = 9 in. Length is 72 ÷ 9 = 8 in.

Puzzle 15

Solution: 64 in.2 Find length . . . 80 ÷ 5 = 16 in. Find length . . . 16 − 3 − 5 = 8 in. Find length . . . 60 ÷ 4 = 15 in. Find length . . . 15 − 3 − 8 = 4 in. Find length . . . 70 ÷ 5 = 14 in. Find length . . . 14 − 4 − 3 = 7 in. Find length . . . 4 + 5 + 7 = 16 in. Area is 4 × 16 = 64 in.2

Puzzle 16 Solution: 9 in. Find length . . . 42 ÷ 7 = 6 in. Find length . . . 48 ÷ 6 = 8 in.

Find length . . . 40 ÷ 8 = 5 in. Find length . . . 45 ÷ 5 = 9 in. Lengths and are the same, so length is 9 in.

Puzzle 17 Solution: 6 in. Find length . . . 78 ÷ 6 = 13 in. Find length . . . 13 − 3 − 5 = 5 in. Find length . . . 2 + 5 + 7 = 14 in. Length is 84 ÷ 14 = 6 in.

Puzzle 18 Solution: 64 in.2 Find length . . . 49 ÷ 7 = 7 in. Find length . . . 10 − 7 = 3 in. Find length . . . 36 ÷ 6 = 6 in. Find length . . . (5 + 3) − 6 = 2 in. Find length . . . 11 − 2 = 9 in. Find length . . . 54 ÷ 9 = 6 in. Find length . . . 35 ÷ 5 = 7 in. Find length . . . 9 − 7 = 2 in. Area is (5 + 3) × (6 + 2) = 64 in.2

Puzzle 19 Solution: 12 in.2 Find length . . . 25 ÷ 5 = 5 in. Find length . . . 18 ÷ 3 = 6 in. Find length . . . (5 + 4) − 6 = 3 in. Find length . . . 15 ÷ 3 = 5 in. Find length . . . 20 ÷ 4 = 5 in. Find length . . . 21 ÷ 3 = 7 in. Find length . . . (5 + 5) − 7 = 3 in. Area is 3 × 4 = 12 in.2

Puzzle 20 Solution: 25 in.2 Find length . . . 28 ÷ 7 = 4 in. Find length . . . 20 ÷ 4 = 5 in. Find length . . . 30 ÷ 5 = 6 in. Find length . . . 60 ÷ (6 + 4) = 6 in. Find length . . . 44 ÷ (5 + 6) = 4 in. Find length . . . 12 ÷ 4 = 3 in. Find length . . . 36 ÷ (7 − 3) = 9 in. Find length . . . 9 − 4 = 5 in. Area is 5 × 5 = 25 in.2

Puzzle 21 Solution: 25 in.2 Add the bottom areas . . . 25 + 35 = 60 in.2 Find length . . . 60 ÷ 6 = 10 in. Find the total area of the top two rectangles . . . 10 × 7 = 70 in.2 Area is 70 − 45 = 25 in.2

Puzzle 22 Solution: 19 in.2 Find the total area of the two rectangles on the right . . . 9 × 5 = 45 in.2 Area is 45 − 26 = 19 in.2

Puzzle 23 Solution: 5 in. The five rectangles on top all have the same area and height, so they must also have the same width. Find length . . . 3 × 5 = 15 in. The three rectangles on the bottom all have the same area and height, so they must also have the same width. Length is 15 ÷ 3 = 5 in.

Puzzle 24 Solution: 8 in. Add the top two areas . . . 28 + 35 = 63 in.2 Find length . . . 63 ÷ 9 = 7 in. Find length . . . 35 ÷ 7 = 5 in. Add the middle two areas . . . 21 + 49 = 70 in.2 Find their combined width . . . 5 + 5 = 10 in. Find length . . . 70 ÷ 10 = 7 in. Find length . . . 49 ÷ 7 = 7 in. Add the bottom two areas . . . 42 + 56 = 98 in.2 Find their combined width . . . 7 + 7 = 14 in. Find length . . . 98 ÷ 14 = 7 in. Length is 56 ÷ 7 = 8 in.

Puzzle 25 Solution: 21 in.2 The two 19 in.2 areas are the same height, so they must also be the same width: Length equals . The two 20 in.2 areas are the same width, so they must also be the same height: Length equals . The two areas with stripes are the same height and width, so they must be equal. Area is 21 in.2

Puzzle 26 Solution: 21 in.2 The two rectangles on the right are the same width, and 84 in.2 is triple 28 in.2 So, length must be triple length . The two rectangles on the left are the same width, so 63 in.2 must be triple area . Area is 63 ÷ 3 = 21 in.2

Puzzle 27 Solution: 30 in.2 Create areas and .

Find area . . . 6 × 5 = 30 in.2 Find area . . . 47 − 30 = 17 in.2 Area is equal to the bottom right area and has the same width, so it must be the same height: Length equals 6 in. Area is 6 × 5 = 30 in.2

Puzzle 28 Solution: 12 in. All four rectangles are the same width, and 84 in.2 is double 42 in.2 So, length is 9 × 2 = 18 in. Because 84 in.2 is triple 28 in.2, length is 18 ÷ 3 = 6 in. Because 56 in.2 is double 28 in.2, length is 6 × 2 = 12 in.

Puzzle 29 Solution: 12 in. Find the total area without stripes . . . 11 × 5 = 55 in.2 Find area . . . 55 − 22 = 33 in.2 Area is the same height and width as area , so it must also be 33 in.2

Find the total area with stripes . . . 27 + 33 = 60 in.2 Length is 60 ÷ 5 = 12 in.

Puzzle 30 Solution: 57 in.2 Because 51 in.2 is triple 17 in.2, length must be triple length . So, area must be triple 7 in.2 . . . 7 × 3 = 21 in.2 This is half of 42 in.2 So, length must be half of length . Area must be double 9 in.2 . . . 9 × 2 = 18 in.2 Because 54 in.2 is triple 18 in.2, length must be triple length . Area is 19 × 3 = 57 in.2

Puzzle 31 Solution: 9 in. Find the bottom area with stripes . . . 5 × 6 = 30 in.2 Find area . . . 30 − 9 − 14 = 7 in.2 Find the total area with height 7 in. and width . . . 20 + 15 + 7 = 42 in.2 Find length . . . 42 ÷ 7 = 6 in. Find the total area with height 8 in. and width . . . 8 × 6 = 48 in.2 Find area . . . 48 − 13 − 20 = 15 in.2 Find the top area with stripes . . . 25 + 14 + 15 = 54 in.2 Length is 54 ÷ 6 = 9 in.

Puzzle 32 Solution: 10 in. Find the total area . . . 27 + 54 = 81 in.2 This is triple 27 in.2 So, length is 15 ÷ 3 = 5 in. Length is 15 − 5 = 10 in.

Puzzle 33 Solution: 10 in.

Create areas and . Find area . . . 5 × 3 = 15 in.2 Find area . . . 5 × 4 = 20 in.2 Find the area with stripes . . . 64 − 15 − 20 = 29 in.2 This is half of 58 in.2 So, 5 in. is half of length . Length is 5 × 2 = 10 in.

Puzzle 34 Solution: 10 in. Find length . . . 30 ÷ 5 = 6 in. Find length . . . 42 ÷ 6 = 7 in. Find length . . . 56 ÷ 7 = 8 in. Find area . . . 5 × 8 = 40 in.2 Find area . . . 72 − 40 = 32 in.2 This is equal to the bottom right area, so length is 5 × 2 = 10 in.

Puzzle 35 Solution: 9 in. Find length . . . 16 ÷ 4 = 4 in. Find length . . . 48 ÷ 4 = 12 in. Because 34 in.2 is double 17 in.2, length must be double length . Find length . . . 12 ÷ 2 = 6 in. Find length . . . 18 ÷ 6 = 3 in. Find length . . . 54 ÷ 3 = 18 in. Because 38 in.2 is double 19 in.2, length must be double length . Length is 18 ÷ 2 = 9 in.

Puzzle 36 Solution: 17 in.2 Find area . . . 9 × 4 = 36 in.2 Find area . . . 70 − 36 = 34 in.2 Because 36 in.2 is double 18 in.2, area must be double area . Area is 34 ÷ 2 = 17 in.2

Puzzle 37

Solution: 12 in. Find area . . . 5 × 6 = 30 in.2 Find area . . . 35 − 30 = 5 in.2 Find the area with stripes . . . 5 + 21 = 26 in.2 This is half of 52 in.2 So, 6 in. must be half of length . Length is 6 × 2 = 12 in.

Puzzle 38 Solution: 13 in.2 Because 9 in. is triple 3 in., the total area with stripes must be 14 × 3 = 42 in.2 Add the bottom two areas . . . 27 + 15 = 42 in.2 This is equal to the area with stripes, so length must equal length . Area is 27 − 14 = 13 in.2

Puzzle 39 Solution: 7 in. The two 10 in.2 areas are the same height, so length must equal length . Create areas and . Area must equal 39 in.2 Find area . . . 53 − 39 = 14 in.2 Find length . . . 14 ÷ 2 = 7 in. Length also equals 7 in. Find the total area with stripes . . . 39 + 10 = 49 in.2 Length is 49 ÷ 7 = 7 in.

Puzzle 40 Solution: 12 in.2 Create area of 20 in.2 This is equal to the leftmost area, so length must equal 7 in. Find area . . . 32 − 20 = 12 in.2 Find length . . . 9 − 7 = 2 in. Find length . . . 12 ÷ 2 = 6 in. Find the total area of the three leftmost rectangles . . . 6 × 7 = 42 in.2 Area is 42 − 20 − 10 = 12 in.2

Puzzle 41 Solution: 14 in.

Add the two areas on the right . . . 13 + 10 = 23 in.2 Add the two areas on the left . . . 33 + 13 = 46 in.2 This is double 23 in.2 So, length is 7 × 2 = 14 in.

Puzzle 42 Solution: 5 in. Find length . . . 21 ÷ 7 = 3 in. Find length . . . 8 − 3 = 5 in. Create area . Find the total area with stripes . . . 5 × 9 = 45 in.2 Find area . . . 45 − 22 = 23 in.2 This is equal to the top right area, so length must equal length . Length is 5 in.

Puzzle 43 Solution: 75 in.2 Find length . . . 30 ÷ 5 = 6 in. Find length . . . 15 − 6 − 6 = 3 in. Find the area with stripes . . . 13 × 6 = 78 in.2 Find area . . . 78 − 28 = 50 in.2 Area is the same width as area and twice as tall, so it must be twice as big. Find area . . . 50 ÷ 2 = 25 in.2 Area is 50 + 25 = 75 in.2

Puzzle 44 Solution: 3 in. Find the total area with stripes . . . 13 + 15 = 28 in.2 This is double 14 in.2 So, length must be double length . Area must be double the area to its right . . . 15 × 2 = 30 in.2 Find area . . . 58 − 30 = 28 in.2 This is equal to the area with stripes, so length must equal 3 in.

Puzzle 45 Solution: 18 in.2 Find length . . . 18 ÷ 6 = 3 in. Add the top left and right areas . . . 20 + 15 = 35 in.2 Add the bottom left and right areas . . . 13 + 22 = 35 in.2 The combined width of the bottom left and right areas is length − 3 in. This combined width of the top left and right areas is also length − 3 in. So, the height of the top row must equal the height of the bottom row: Length equals 6 in. Area is 6 × 3 = 18 in.2

Puzzle 46 Solution: 8 in. Create area of 29 in.2

This is equal to the rightmost area, so length must equal 9 in. Find length . . . 13 − 9 = 4 in. Find area . . . 58 − 29 = 29 in.2 This is equal to area , so areas and are identical. Length is 4 × 2 = 8 in.

Puzzle 47 Solution: 9 in.2 Create area and find the total area with stripes . . . 5 × 10 = 50 in.2 Find area . . . 50 − 18 = 32 in.2 This is double 16 in.2 So, length must be double length . Area is 18 ÷ 2 = 9 in.2

Puzzle 48 Solution: 5 in. Create areas and . Find area . . . 5 × 4 = 20 in.2 Find area . . . 49 − 20 = 29 in.2 This is equal to the area below it, so length must equal 5 in. Create areas and . Find area . . . 5 × 6 = 30 in.2 Find area . . . 57 − 30 = 27 in.2 This is equal to the area below it, so length must equal 5 in.

Puzzle 49 Solution: 50 in.2 Create areas and and find the total area with stripes . . . 9 × 14 = 126 in.2 Find the combined area of and . . . 126 − 50 = 76 in.2 Add the top and bottom areas on the left . . . 57 + 19 = 76 in.2 This equals the combined area of and . So, length must equal 9 in. Area is 50 in.2

Puzzle 50 Solution: 75 in.2 Create area and find the total area with stripes . . . 6 × 14 = 84 in.2 Find area . . . 84 − 70 = 14 in.2 Area is the same height as area and half as wide, so it must be half as big . . . 14 ÷ 2 = 7 in.2 Find area . . . 36 − 14 − 7 = 15 in.2 Because 70 in.2 is five times 14 in.2, area must be five times area . Area is 15 × 5 = 75 in.2．

Puzzle 51 Solution: 10 in. Create area . Because 27 in.2 is triple 9 in.2, area must be triple the area to its left . . . 10 × 3 = 30 in.2 Length is 30 ÷ 3 = 10 in.

Puzzle 52 Solution: 3 in. Find length . . . 6 − 4 = 2 in. Create area . Area is the same height as the 14 in.2 area and half as wide, so it must be half as big . . . 14 ÷ 2 = 7 in.2 Find the total area with stripes . . . 7 + 14 + 29 = 50 in.2 This is double 25 in.2 So, 6 in. must be double length . Length is 6 ÷ 2 = 3 in.

Puzzle 53 Solution: 5 in. Add the two bottom areas on the left . . . 16 + 15 = 31 in.2 This is equal to the bottom area on the right, so length must equal length . Find the total area with stripes . . . 17 + 16 = 33 in.2 Add the two areas on the right . . . 35 + 31 = 66 in.2 This is double 33 in.2 So, 10 in. must be double length . Length is 10 ÷ 2 = 5 in.

Puzzle 54 Solution: 5 in. Create areas , , and . Find area . . . 7 × 6 = 42 in.2 This is double 21 in.2 So, area must be double 11 in.2 Find area . . . 11 × 2 = 22 in.2 Find area . . . 99 − 42 − 22 = 35 in.2 Length is 35 ÷ 7 = 5 in.

Puzzle 55 Solution: 7 in. Find the total area with stripes . . . 14 + 28 = 42 in.2 Find length . . . 42 ÷ 6 = 7 in. Because 14 in.2 is half of 28 in.2, length must be half of length . Because 15 in.2 is half of 30 in.2, length must also be half of length . Length is equal to length . So, length must equal length . Length is 7 in.

Puzzle 56 Solution: 7 in.2

Find the total area on the left . . . 5 × 10 = 50 in.2 Find area . . . 50 − 14 = 36 in.2 Create area . Find the total area with stripes . . . 5 × 9 = 45 in.2 Find area . . . 45 − 36 = 9 in.2 Because 36 in.2 is four times 9 in.2, 28 in.2 must be four times area . Area is 28 ÷ 4 = 7 in.2

Puzzle 57 Solution: 20 in.2 Create areas and . Find area . . . 3 × 4 = 12 in.2 Find area . . . 30 − 12 = 18 in.2 Create areas , , and .

Find area . . . 3 × 6 = 18 in.2 This is equal to area , so area must equal area . Find area . . . 5 × 6 = 30 in.2 Find area . . . 50 − 30 = 20 in.2 Area is 20 in.2

Puzzle 58 Solution: 21 in.2 Create area of 17 in.2 This is equal to the area on top. So, length must equal 8 in. Find area . . . 55 − 17 = 38 in.2 Find length . . . 16 − 8 = 8 in. The area with stripes is the same height and width as area , so it must also be 38 in.2 Area is 38 − 17 = 21 in.2

Puzzle 59 Solution: 8 in. Create areas and . Find area . . . 4 × 3 = 12 in.2 Find area . . . 27 − 12 = 15 in.2 This is half of 30 in.2 So, 4 in. must be half of length . Find length . . . 4 × 2 = 8 in. Find length . . . 8 − 2 = 6 in. Create areas and .

Find area . . . 3 × 6 = 18 in.2 Find area . . . 31 − 18 = 13 in.2 This is half of 26 in.2 So, 6 in. must be half of length . Find length . . . 6 × 2 = 12 in. Length is 12 − 4 = 8 in.

Puzzle 60 Solution: 39 in.2 Length is 36 ÷ 6 = 6 in. Find length . . . 6 − 3 = 3 in. Find length . . . 3 + 4 − 5 = 2 in. The width of area is length + 2 in. This equals the width of the 26 in.2 area. Find length . . . 42 ÷ 6 = 7 in.

Find length . . . 7 − 4 = 3 in. The width of the 26 in.2 area is length + 3 in. This equals the width of the area with stripes. Area is the same width and height as the area with stripes, so it must also be 39 in.2

Puzzle 61 Solution: 35 in.2 Create areas and . Find the total area with stripes . . . 5 × 7 = 35 in.2 Find area . . . 35 − 17 = 18 in.2 Find the combined area of and . . . 5 × 8 = 40 in.2 Find area . . . 40 − 18 = 22 in.2 This is equal to the bottom right area, so length must equal 5 in. Area is 5 × 7 = 35 in.2

Puzzle 62 Solution: 12 in. Create areas and . Because 14 in.2 is half of 28 in.2, area must be half of 30 in.2 Find area . . . 30 ÷ 2 = 15 in.2 Find area . . . 24 − 15 = 9 in.2 Find the total area with stripes . . . 30 + 15 = 45 in.2 This is five times area , so 10 in. must be five times length . Find length . . . 10 ÷ 5 = 2 in. Length is 10 + 2 = 12 in.

Puzzle 63 Solution: 54 in.2 Find the total area with stripes . . . 7 × 9 = 63 in.2 Find area . . . 63 − 34 = 29 in.2 Create areas , , , and . Find the combined area of and . . . 7 × 7 = 49 in.2 Find area . . . 49 − 29 = 20 in.2 Find the combined area of and . . . 7 × 8 = 56 in.2 Find area . . . 56 − 20 = 36 in.2 Find area . . . 30 − 20 = 10 in.2 This is half of area , so area must be half of area . Find area . . . 36 ÷ 2 = 18 in.2 Area is 36 + 18 = 54 in.2

Puzzle 64 Solution: 2 in. Create area . Find length . . . 9 − 6 = 3 in. This is half of 6 in., so 19 in.2 must be half of area . Find area . . . 19 × 2 = 38 in.2 Create areas and . Because 6 in. is triple 2 in., area must be triple 13 in.2 Find area . . . 13 × 3 = 39 in.2 Find area . . . 89 − 38 − 39 = 12 in.2 Length is 12 ÷ 6 = 2 in.

Puzzle 65 Solution: 10 in. Add the dotted lines to create two overlapping areas with height 10 in. The left area is 9 in. wide, and the right one is 11 in. wide. Find their difference in width . . . 11 − 9 = 2 in. Find their difference in area . . . 10 × 2 = 20 in.2 The two overlapping areas share the region with stripes, so the regions they do not share must have a difference of 20 in.2

Find area . . . 51 − 20 = 31 in.2 Area has the same width as the given area of 31 in.2, so they are identical. Length must be 10 in.

Puzzle 66 Solution: 11 in. Create areas and . Find area . . . 4 × 3 = 12 in.2 Find area . . . 22 − 12 = 10 in.2 Create areas , , and . Find area . . . 4 × 5 = 20 in.2 This is double area , so area must be double area .． Find area . . . 3 × 5 = 15 in.2 Find area . . . 57 − 20 − 15 = 22 in.2 Area is 22 ÷ 2 = 11 in.2

Puzzle 67 Solution: 8 in. Create areas and . Find area . . . (4 × 11) − 22 = 22 in.2 Find area . . . (4 × 13) − 22 = 30 in.2 This is half of 60 in.2 So, 4 in. must be half of length . Length is 4 × 2 = 8 in.

Puzzle 68

Solution: 13 in.2 Find area . . . (7 × 4) − 18 = 10 in.2 Create area and find it . . . 5 × 4 = 20 in.2 This is double 10 in.2 So, area must be double area . Find area . . . (5 × 9) − 19 = 26 in.2 Area is 26 ÷ 2 = 13 in.2

Puzzle 69 Solution: 2 in. Create areas and . Find area . . . (4 × 9) − 21 = 15 in.2 Find area . . . 36 − 15 = 21 in.2 This is equal to the area to its left. So, the total area with stripes must be double area .

Find the area with stripes . . . 15 × 2 = 30 in.2 This is double 15 in.2 So, 4 in. must be double length . Length is 4 ÷ 2 = 2 in.

Puzzle 70 Solution: 3 in.2 Find area . . . 8 × 6 = 48 in.2 Find area . . . 72 − 48 = 24 in.2 This is half of area , so area must be half of 64 in.2 Find area . . . 64 ÷ 2 = 32 in.2 Find area . . . 48 − 32 = 16 in.2 This is half of area , so area must be half of area . Find area . . . 24 ÷ 2 = 12 in.2

Find area . . . 18 − 12 = 6 in.2 This is half of area , so area must be half of area . Find area . . . 16 ÷ 2 = 8 in.2 Find area . . . 12 − 8 = 4 in.2 This is half of area , so area must be half of area . Area is 6 ÷ 2 = 3 in.2

Puzzle 71 Solution: 4 in. Create area . Find the total area with stripes . . . 8 × 11 = 88 in.2 Find area . . . 88 − 28 = 60 in.2 This is double 30 in.2 So, 8 in. must be double length .

Length is 8 ÷ 2 = 4 in.

Puzzle 72 Solution: 10 in. Create areas and . Find area . . . (7 × 8) − 28 − 15 = 13 in.2 Add area and the area to its right . . . 13 + 15 = 28 in.2 This equals the given area of 28 in.2 So, area must equal 22 in.2 Add areas and . . . 22 + 13 = 35 in.2 Find the total area with stripes . . . 35 × 2 = 70 in.2 Length is 70 ÷ 7 = 10 in.

Puzzle 73 Solution: 39 in.2 Add the top two areas with width . . . 7 + 8 = 15 in.2 This equals the given area of 15 in.2 So, length must equal length . Therefore, area must be 7 in.2 Create and find area . . . 26 ÷ 2 = 13 in.2 Find the total area with stripes . . . 8 + 13 = 21 in.2 This is triple 7 in.2 So, area must be triple area . Find area . . . (3 × 9) − 7 − 7 = 13 in.2 Area is 13 × 3 = 39 in.2

Puzzle 74 Solution: 7 in. Find area . . . (7 × 6) − 14 = 28 in.2 This is double 14 in.2 So, area must be double area . Find area . . . 62 − 28 = 34 in.2 Find area . . . 34 ÷ 2 = 17 in.2 Find area . . . 51 − 17 = 34 in.2 Area is equal to area and has the same width, so it must also have the same height. The combined height of areas and is 7 in., so the combined height of areas and must also be 7 in.

Length is 7 in.

Puzzle 75 Solution: 16 in.2 Create area of 20 in.2 This is equal to the rightmost area, so length must equal 7 in. Find area . . . 32 − 20 = 12 in.2 Find length . . . 9 − 7 = 2 in. Find length . . . 12 ÷ 2 = 6 in. Find length . . . 11 − 6 = 5 in. Find length . . . 8 − 5 = 3 in. This is half of 6 in., so area must be half of 32 in.2 Area is 32 ÷ 2 = 16 in.2

Puzzle 76 Solution: 48 in.2 Find area . . . (7 × 8) − 26 − 10 = 20 in.2 This is double 10 in.2 So, area must be double the area with stripes. Find the combined area of and the area with stripes . . . 8 × 9 = 72 in.2 Because area is double the area with stripes, this combined area must be triple the area with stripes. Find the area with stripes . . . 72 ÷ 3 = 24 in.2 Area is 24 × 2 = 48 in.2

Puzzle 77 Solution: 16 in.2 Find length . . . 21 ÷ 3 = 7 in. Find length . . . 28 ÷ 4 = 7 in. Length equals length + , and length equals length + . Since equals , must equal . Create area of 26 in.2 Find area . . . 39 − 26 = 13 in.2 This is half of area , so 3 in. must be half of length . Find length . . . 3 × 2 = 6 in. Area is (6 × 7) − 26 = 16 in.2

Puzzle 78 Solution: 5 in. Create area of 33 in.2 This is equal to the top right area, so length + 3 in. must equal length + length . The two 16 in.2 areas are the same height, so they must also be the same width: Length equals length . So, length must equal 3 in. Find length . . . 6 − 3 = 3 in. Find area . . . 48 − 33 = 15 in.2 Length is 15 ÷ 3 = 5 in.

Puzzle 79 Solution: 37 in.2 Find area . . . (7 × 8) − 15 − 25 = 16 in.2 Find area . . . 30 − 16 = 14 in.2 This is half of 28 in.2 So, area + 15 in.2 must be half of the area with stripes. Find the area with stripes . . . (16 + 15) × 2 = 62 in.2 Area is 62 − 25 = 37 in.2

Puzzle 80 Solution: 5 in.

Find length . . . 6 + 7 + 8 + 9 = 30 in. This is five times 6 in., so 50 in.2 must be five times area . Find area . . . 50 ÷ 5 = 10 in.2 Find length . . . 7 + 8 = 15 in. This is half of length , so area must be half of 50 in.2 Find area . . . 50 ÷ 2 = 25 in.2 Find area . . . 50 − 10 − 25 = 15 in.2 Find the total area with stripes . . . 30 + 15 = 45 in.2 Length is 45 ÷ 9 = 5 in.

Puzzle 81 Solution: 22 in.2 Add the top two areas on the left . . . 34 + 12 = 46 in.2 Add the bottom two areas on the left . . . 12 + 11 = 23 in.2 This is half of 46 in.2 So, area must be half of the total area with stripes. Find the area with stripes . . . 11 × 4 = 44 in.2 Area is 44 ÷ 2 = 22 in.2

Puzzle 82 Solution: 7 in. Find area . . . (7 × 12) − 50 − 18 = 16 in.2 Because 18 in.2 is double 9 in.2, area must be double area . Find area . . . 16 ÷ 2 = 8 in.2 Find the total area with stripes . . . 8 + 9 = 17 in.2 The area with stripes is equal to the given area of 17 in.2 and has the same width, so it must also have the same height. Length is 7 in.

Puzzle 83 Solution: 13 in.2 Find area . . . (6 × 8) − 16 = 32 in.2 This is double 16 in.2 So, area must be double area .． Find area . . . 52 − 32 = 20 in.2 Find area . . . 20 ÷ 2 = 10 in.2 Find area . . . 26 − 10 = 16 in.2 This is equal to the top left area, so area must be equal to area , or 32 in.2 Find area . . . 40 − 32 = 8 in.2 Because 32 in.2 is four times 8 in.2, the 52 in.2 area must be four times area . Area is 52 ÷ 4 = 13 in.2

Puzzle 84 Solution: 4 in. Find area . . . 3 × 4 = 12 in.2 Find area . . . 25 − 12 = 13 in.2 Find the total area with stripes . . . 14 + 12 = 26 in.2 This is double area , so 30 in.2 must be double area . Find area . . . 30 ÷ 2 = 15 in.2 Find area . . . 28 − 15 = 13 in.2 This is equal to area , so length is 4 in.

Puzzle 85 Solution: 20 in.2 Find area . . . 4 × 5 = 20 in.2 Find area . . . 51 − 20 = 31 in.2 This is equal to the bottom left area, so area must equal 34 in.2 Find area . . . 65 − 34 = 31 in.2 This is equal to the bottom left area, so length must equal 5 in. Area is 4 × 5 = 20 in.2

Puzzle 86 Solution: 36 in.2 The top right area is double the top left area, so area + 25 in.2 must be double 20 in.2 Find area . . . (20 × 2) − 25 = 15 in.2 Find area . . . 27 − 15 = 12 in.2 Find the total area with stripes . . . 20 + 25 = 45 in.2 This is triple area , so area must be triple area . Area is 12 × 3 = 36 in.2

Puzzle 87 Solution: 7 in. Find area . . . (5 × 8) − 32 = 8 in.2 Find area . . . 17 − 8 = 9 in.2 The top left area is four times area , so area must be four times area .

Find area . . . 9 × 4 = 36 in.2 Find area . . . 68 − 36 = 32 in.2 This is equal to the top left area, so length must be equal to 5 in. So, area must be equal to area , or 8 in.2 Find area . . . 43 − 8 = 35 in.2 Length is 35 ÷ 5 = 7 in.

Puzzle 88 Solution: 12 in.2 The 36 in.2 area is double the 18 in.2 area and exactly twice as tall, so it must have the same width: Length equals length . Find length . . . 8 − 7 = 1 in. Find length . . . 4 − 3 = 1 in. The 12 in.2 area and area are both exactly 1 in. shorter than the area of 24 in.2 They also have the same width, so they must be identical. Area is 12 in.2

Puzzle 89 Solution: 4 in. Add the three given areas . . . 30 + 35 + 57 = 122 in.2 Add the dotted lines to create two overlapping areas. Find the one on the left . . . 6 × 13 = 78 in.2 Find the one on the right . . . 4 × 11 = 44 in.2 Add the two overlapping areas . . . 78 + 44 = 122 in.2 This is equal to the sum of the three given areas. So, the two larger areas must not actually overlap when drawn to scale. The two dotted lines are the same line. Length must be equal to the given length of 4 in.

Puzzle 90 Solution: 22 in.2 Find the difference in width between the four rectangles on the right and the four on the left . . . 13 − 12 = 1 in. The three center rectangles are in both groups, so the 36 in.2 area must be 1 in. wider than the 29 in.2 area. Find the difference between the two areas . . . 36 − 29 = 7 in.2 Find length . . . 7 ÷ 1 = 7 in. Find area . . . (13 × 7) − (6 × 7) − 36 = 13 in.2 Area is (5 × 7) − 13 = 22 in.2

Puzzle 91 Solution: 48 in.2 Find area . . . 6 × 7 = 42 in.2 This is double 21 in.2 So, area must be double 45 in.2 Find area . . . 45 × 2 = 90 in.2 This is triple 30 in.2 So, area must be triple 16 in.2 Area is 16 × 3 = 48 in.2

Puzzle 92 Solution: 10 in.

Because 20 in.2 is double 10 in.2, length must be double 3 in. Find length . . . 3 × 2 = 6 in. Find the total area with stripes . . . (10 × 6) − 20 = 40 in.2 This is double 20 in.2 So, length must be double length . Find area . . . 40 − 26 = 14 in.2 Find area . . . 21 − 14 = 7 in.2 This is half of area , so length must be half of length and the same as length . Length + must equal length + , so length is 10 in.

Puzzle 93 Solution: 20 in.2 Lengths and are both equal to length − 4 in.

The 76 in.2 area is double the 38 in.2 area and has the same width, so it must be twice as tall: Length is double length . So, 34 in.2 must be double area . Find area . . . 34 ÷ 2 = 17 in.2 Find area . . . 30 − 17 = 13 in.2 Find the total area with stripes . . . 13 + 38 = 51 in.2 This is triple area , so the 60 in.2 area must be triple area . Area is 60 ÷ 3 = 20 in.2

Puzzle 94 Solution: 7 in. Find area . . . (6 × 12) − 57 = 15 in.2 Find the total area with stripes . . . (6 × 7) − 15 = 27 in.2 This is triple 9 in.2 So, 6 in. must be triple length .

Find length . . . 6 ÷ 3 = 2 in. Find area . . . (2 × 7) − 9 = 5 in.2 Find area . . . 38 − 15 − 5 = 18 in.2 This is equal to the given area of 18 in.2 So, area must be equal to area , or 15 in.2 Length is (57 − 15) ÷ 6 = 7 in.

Puzzle 95 Solution: 3 in. Find length . . . 15 − 3 = 12 in. This is triple 4 in., so area must be triple 14 in.2 Find area . . . 14 × 3 = 42 in.2 Find area . . . 74 − 42 = 32 in.2

Find length . . . 5 − 2 = 3 in. Length is four times 3 in., so area must be four times area . Find area . . . 32 ÷ 4 = 8 in.2 Find area . . . 19 − 8 = 11 in.2 Because 15 in. is five times 3 in., area must be five times area . Find area . . . 11 × 5 = 55 in.2 Find area . . . 82 − 55 = 27 in.2 Because 15 in. is triple 5 in., area must be triple area . Find area . . . 27 ÷ 3 = 9 in.2 Find area . . . 24 − 9 = 15 in.2 Length is 15 ÷ 5 = 3 in.

Puzzle 96 Solution: 42 in.2

Find length . . . 42 ÷ 6 = 7 in. Find length . . . 56 ÷ 8 = 7 in. This equals length . Because length + equals length + , length must equal length . Because 56 in.2 is double 28 in.2, length must be double length , which means length must be double length . So, 42 in.2 must be double area . Find area . . . 42 ÷ 2 = 21 in.2 Find area . . . 39 − 21 = 18 in.2 Find length . . . 18 ÷ 6 = 3 in. Length + 3 in. equals length + + 3 in., so length must equal length + . Find the total area with stripes . . . 28 + 56 = 84 in.2 The area with stripes is the same width as area and twice as tall, so it must be twice as big. Area is 84 ÷ 2 = 42 in.2

Puzzle 97 Solution: 16 in. Find area . . . (9 × 4) − 15 = 21 in.2 Find area . . . 35 − 21 = 14 in.2 The bottom area with stripes is the same height as area and twice as wide, so it must be twice as big . . . 21 × 2 = 42 in.2 Find area . . . 42 − 14 = 28 in.2 Find area . . . 46 − 28 = 18 in.2 Area is double area , so area must be double area . Find area . . . 18 ÷ 2 = 9 in.2 Find the top area with stripes . . . (8 × 6) − 18 − 9 = 21 in.2 This is half of 42 in.2 So, 8 in. must be half of length .

Length is 8 × 2 = 16 in.

Puzzle 98 Solution: 2 in. Find area . . . (3 × 8) − 17 = 7 in.2 Find area . . . 21 − 7 = 14 in.2 This is double area , so length must be double 3 in. Find length . . . 3 × 2 = 6 in. Find the total area with stripes . . . (6 × 11) − 26 − 14 = 26 in.2 This is equal to the bottom left area, so the height of the area with stripes must be equal to length . Find length . . . 6 ÷ 2 = 3 in. Find length . . . 13 − 3 = 10 in. Find the total area (not depicted) with height and width 10 in. . . . 10 × 10 = 100 in.2

This is four times 25 in.2 So, the total area with height and width 10 in. must be four times area . This means 15 in.2 must be triple area . Find area . . . 15 ÷ 3 = 5 in.2 Length is (5 + 15) ÷ 10 = 2 in.

Puzzle 99 Solution: 31 in.2 The given area of 18 in.2 is double the given area of 9 in.2 and twice as tall, so it must be exactly the same width. Therefore, length must equal length . Find the total area with stripes . . . 31 + 21 = 52 in.2 The area with stripes is four times the given area of 13 in.2 and also four times as wide, so it must be exactly the same height: Length equals length . Area is the same height and width as the given area of 31 in.2, so it must also be 31 in.2

Puzzle 100 Solution: 25 in.2 Area with stripes is the same width as the given area of 31 in.2 and twice as tall, so it must be twice as big . . . 31 × 2 = 62 in.2 Find area . . . 62 − 34 − 17 = 11 in.2 Find area with stripes . . . (8 × 5) + 17 + 11 = 68 in.2 This is equal to the given area of 68 in.2 and has the same width, so it must also have the same height. Area has the same height and width as the given area of 17 in.2 So, area must equal 17 in.2 The 51 in.2 area is triple area . So, area + 34 in.2 must be triple area + 17 in.2 Find area . . . (11 + 17) × 3 − 34 = 50 in.2 Area is the same width as area and half as tall, so it must be half as big. Area is 50 ÷ 2 = 25 in.2

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We hope you enjoyed The Original Area Mazes, Volume 2.

Don’t miss The Original Area Mazes, available now, along with Amazing Area Mazes, available September 3, 2019!