The figure shows a rectangle $K G S T$ and a triangle $\triangle K G R$. The angles $K \widehat{R} T$ and $R \widehat{G} S$ are equal. If $T R=6$ and $R S=2$, what is the area of the triangle $\triangle K G R ?$
(a) 12
(b) 16
(c) $8 \sqrt{2}$
(d) $8 \sqrt{3}$
(e) 14

![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-012.jpg?height=257&width=457&top_left_y=1813&top_left_x=1371)
The correct option is (d).

The triangles $\triangle T K R$ and $\triangle G R S$ are proportional because they are right triangles with an equal acute angle. Therefore, we have $\frac{R S}{T K}=\frac{G S}{T R}$. Since $G S=T K$, it follows that $(T K)^{2}=R S \times T R=2 \times 6=12$, or $T K=2 \sqrt{3}$. Also,

$$
K G=T R+R S=6+2=8
$$

Thus, the area of the triangle $\triangle K G R$ is

$$
\frac{1}{2} \underbrace{K G}_{\text {base }} \times \underbrace{T K}_{\text {height }}=\frac{1}{2}[8 \times 2 \sqrt{3}]=8 \sqrt{3}
$$