1985 AJHSME Problems/Problem 12

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

$\text{(A)}\ 24\text{ cm}^2 \qquad \text{(B)}\ 36\text{ cm}^2 \qquad \text{(C)}\ 48\text{ cm}^2 \qquad \text{(D)}\ 64\text{ cm}^2 \qquad \text{(E)}\ 144\text{ cm}^2$

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 (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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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