1990 AJHSME Problems/Problem 15

Problem

The area of this figure is $100\text{ cm}^2$ . Its perimeter is

$[asy] draw((0,2)--(2,2)--(2,1)--(3,1)--(3,0)--(1,0)--(1,1)--(0,1)--cycle,linewidth(1)); draw((1,2)--(1,1)--(2,1)--(2,0),dashed); [/asy]$

[figure consists of four identical squares]

$\text{(A)}\ \text{20 cm} \qquad \text{(B)}\ \text{25 cm} \qquad \text{(C)}\ \text{30 cm} \qquad \text{(D)}\ \text{40 cm} \qquad \text{(E)}\ \text{50 cm}$

Solution

Since the area of the whole figure is $100$ , each square has an area of $25$ and the side length is $5$ .

There are $10$ sides of this length, so the perimeter is $10(5)=50\rightarrow \boxed{\text{E}}$ .

1990 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

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1990 AJHSME Problems/Problem 15

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