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 $10(5)=50\rightarrow \boxed{\text{E}}$.

See Also

1990 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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All AJHSME/AMC 8 Problems and Solutions

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