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1985 AJHSME Problems/Problem 12

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Problem

A square and a triangle have equal perimeters. The lengths of the three sides of the triangle are 6.2 \text{ cm}, 8.3 \text{ cm} and 9.5 \text{ cm}. The area of the square is

\text{(A)}\ 24\text{ cm}^2 \qquad \text{(B)}\ 36\text{ cm}^2 \qquad \text{(C)}\ 48\text{ cm}^2 \qquad \text{(D)}\ 64\text{ cm...

Solution

We are given the three side lengths of the triangle, so we can compute the perimeter of the triangle to be 6.2+8.3+9.5=24. The square has the same perimeter as the triangle, so its side length is \frac{24}{4}=6. Finally, the area of the square is 6^2=36, which is choice \boxed{\text{B}}

See Also

1985 AJHSME (ProblemsResources)
Preceded by
Problem 11
Followed by
Problem 13
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