There is a square whose perimeter is the same as the perimeter of the triangle shown. The area of that square is
(A) $12.25 \mathrm{~cm}^{2}$
(B) $196 \mathrm{~cm}^{2}$
(C) $49 \mathrm{~cm}^{2}$
(D) $36 \mathrm{~cm}^{2}$
(E) $144 \mathrm{~cm}^{2}$

![](https://cdn.mathpix.com/cropped/2024_04_20_6027bc27089ed4fc493cg-036.jpg?height=284&width=285&top_left_y=1907&top_left_x=1321)
First, we must determine the perimeter of the given triangle.

Let the unknown side length measure $x \mathrm{~cm}$.

Since the triangle is a right-angled triangle, then by the Pythagorean Theorem we get $x^{2}=8^{2}+6^{2}$ or $x^{2}=64+36=100$ and so $x=\sqrt{100}=10($ since $x>0)$.

Therefore the perimeter of the triangle is $10+8+6=24 \mathrm{~cm}$ and so the perimeter of the square is also $24 \mathrm{~cm}$.

Since the 4 sides of the square are equal in length, then each measures $\frac{24}{4}=6 \mathrm{~cm}$.

Thus, the area of the square is $6 \times 6=36 \mathrm{~cm}^{2}$.

ANswer: (D)