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, 8.3 and 9.5. 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 $\boxed{\text{B}}$

Video Solution

https://youtu.be/C1_dFnM-G00

~savannahsolver

1985 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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1985 AJHSME Problems/Problem 12

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Problem

Solution

Video Solution

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