2015 CEMC Gauss (Grade 8) Problems/Problem 16

Problem

There is a square whose perimeter is the same as the perimeter of the triangle shown.

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The area of that square is

$\textbf{(A)}\ 12.25 \text { cm}^{2} \qquad\textbf{(B)}\ 196 \text { cm}^{2} \qquad\textbf{(C)}\ 49 \text { cm}^{2} \qquad\textbf{(D)}\ 36 \text { cm}^{2} \qquad\textbf{(E)}\ 144 \text { cm}^{2}$

Solution 1

We can first find the perimeter of the triangle, and then use the equation for the perimeter of a square to find its side length.

We can see that the triangle would be a $6$ - $8$ - $10$ right triangle. Thus, the hypotenuse is $10$ .

Summing all of the side lengths to find the perimeter, we get $6 + 8 + 10 = 14 + 10 = 24 \text { cm}$ .

The perimeter of a square is just $4$ multiplied by its side length because all of its side lengths are equal. Let $s$ be the side length of the square. We then have: