1985 AJHSME Problem 12

Problem 12

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

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

In-depth Solution by BoundlessBrain!

https://youtu.be/rIEgWFY_afM

Solution

The perimeter of the triangle is $6.2+8.3+9.5 = 24$ cm, so the side length of the square is $\frac{24}{4} = 6$ cm. The area of the square is $6^2 = \boxed{\text{(B)} 36}$ square centimeters.