In right-angled, isosceles triangle $F G H, F H=\sqrt{8}$. Arc $F H$ is part of the circumference of a circle with centre $G$ and radius $G H$, as shown. The area of the shaded region is
(A) $\pi-2$
(B) $4 \pi-2$
(C) $4 \pi-\frac{1}{2} \sqrt{8}$
(D) $4 \pi-4$
(E) $\pi-\sqrt{8}$

![](https://cdn.mathpix.com/cropped/2024_04_20_6027bc27089ed4fc493cg-045.jpg?height=333&width=352&top_left_y=1102&top_left_x=1320)
In $\triangle F G H, F G=G H=x$ since they are both radii of the same circle.

By the Pythagorean Theorem, $F H^{2}=F G^{2}+G H^{2}=x^{2}+x^{2}$, or $F H^{2}=2 x^{2}$, and so $(\sqrt{8})^{2}=2 x^{2}$ or $2 x^{2}=8$ and $x^{2}=4$, so then $x=2$ (since $x>0$ ).

$F G, G H$ and $\operatorname{arc} F H$ form a sector of a circle with centre $G$ and radius $G H$.

Since $\angle F G H=90^{\circ}$, which is $\frac{1}{4}$ of $360^{\circ}$, then the area of this sector is one quarter of the area of the circle with centre $G$ and radius $G H=F G=2$.

The shaded area is equal to the area of sector $F G H$ minus the area of $\triangle F G H$.

The area of sector $F G H$ is $\frac{1}{4} \pi(2)^{2}$ or $\frac{1}{4} \pi(4)$ or $\pi$.

The area of $\triangle F G H$ is $\frac{F G \times G H}{2}$ or $\frac{2 \times 2}{2}$, so 2 .

Therefore, the area of the shaded region is $\pi-2$.

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