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1990 AJHSME Problems/Problem 15

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Problem

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

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);

[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} \qqu...

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 (Problems • 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
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