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Find the area of the shaded region.

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Find the area of the shaded region.

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In the figure above, AC = BC = 8, angle C = 90°, and the circular arc has its center at point C. Find the area of the shaded region.

$$A.\ 8\pi-32$$
$$B.\ 16\pi-32$$
$$C.\ 16\pi-64$$
$$D.\ 32\pi-32$$
$$E.\ 32\pi-64$$

The OA is B.

In this PS question, I just need to find the area of the circular arc and then subtract the area of the triangle, right?

It will be,
$$A_{ARC}-A_{\triangle}=\frac{1}{2}r^2\theta-\frac{1}{2}b\cdot h=\frac{1}{2}8^2\frac{\pi}{2}-\frac{1}{2}8\cdot8=16\pi-32$$

Is there a strategic approach to this question? Can any experts help me, please?

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GMAT/MBA Expert

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AAPL wrote:


In the figure above, AC = BC = 8, angle C = 90°, and the circular arc has its center at point C. Find the area of the shaded region.

$$A.\ 8\pi-32$$
$$B.\ 16\pi-32$$
$$C.\ 16\pi-64$$
$$D.\ 32\pi-32$$
$$E.\ 32\pi-64$$

Area of shaded region = (area of sector) - (area of triangle)

Area of circle = π(radius)²
Area of triangle = (base)(height)/2

The sector ABC is 1/4 of a circle of radius 8
So, area of sector = (1/4)(π)(8²)
= 16π


area of triangle = (8)(8)/2
= 32


So, area of shaded region = (16π) - (32)

Answer: B

Cheers,
Brent

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Post
AAPL wrote:


In the figure above, AC = BC = 8, angle C = 90°, and the circular arc has its center at point C. Find the area of the shaded region.

$$A.\ 8\pi-32$$
$$B.\ 16\pi-32$$
$$C.\ 16\pi-64$$
$$D.\ 32\pi-32$$
$$E.\ 32\pi-64$$
We see that the radius of the quarter circle is 8, so the area of the quarter circle is:

1/4 x 8^2 x π = 16π

The area of the triangle is 8 x 8 x 1/2 = 32

Thus, the area of the shaded region is 16π - 32.

Answer: B

_________________
Scott Woodbury-Stewart Founder and CEO

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