A sector of a circle with radius 8 cm has a central angle of 45°. What is the area of the corresponding segment?

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Accepted Answer

Solution:
Radius of the circle:
$$ r = 8 	ext{ cm} $$
Central angle:
$$ 	heta = 45^\circ $$
Area of the corresponding segment:
$$ 	ext{Area of segment} = 	ext{Area of sector} - 	ext{Area of triangle} $$
First, find the area of the sector:
$$ 	ext{Area of sector} = \frac{	heta}{360^\circ} 	imes \pi r^2 $$
$$ = \frac{45}{360} 	imes \pi 	imes 8^2 $$
$$ = \frac{1}{8} 	imes \pi 	imes 64 $$
$$ = 8\pi $$
Now, find the area of the triangle formed by the two radii:
$$ 	ext{Area of triangle} = \frac{1}{2}r^2\sin	heta $$
$$ = \frac{1}{2} 	imes 8^2 	imes \sin 45^\circ $$
$$ = \frac{1}{2} 	imes 64 	imes \frac{1}{\sqrt{2}} $$
$$ = 32 	imes \frac{1}{\sqrt{2}} $$
$$ = 16\sqrt{2} $$
Now, subtract the area of the triangle from the area of the sector:
$$ 	ext{Area of segment} = 8\pi - 16\sqrt{2} $$
Therefore, the area of the corresponding segment is:
$$ \boxed{(8\pi - 16\sqrt{2})\ 	ext{cm}^2} $$
Final Answer: **$(8\pi - 16\sqrt{2})\ 	ext{cm}^2$**

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