Question

How do I find the area of the shaded region?

Answer 1

Answer:

Part a) The area of the shaded region is $A=25(\frac{2\pi}{3}-\sqrt{3})\ cm^2$

Part b) The area of the shaded region is $A=18(3\pi-2\sqrt{2})\ cm^2$

Step-by-step explanation:

Part a) we know that

step 1

Find the area of the sector

we know that

The area of a circle subtends a central angle of 2π radians

so

using proportion

Find the area of the sector by a central angle of π/3 radians

$\frac{\pi r^2}{2\pi}= \frac{x}{\pi/3} \\\\x=\frac{\pi r^2}{6}$

we have

$r=10\ cm$

substitute

$x=\frac{\pi (10)^2}{6}\\\\x=\frac{50\pi}{3}\ cm^2$

step 2

Find the area of triangle

The area of triangle is equal to

$A=\frac{1}{2} (10^2)sin(\frac{\pi}{3})$

Remember that

$\frac{\pi}{3}=60^o$

so

$A=50(\frac{\sqrt{3}}{2})=25\sqrt{3}\ cm^2$

step 3

Find the area of the shaded region

The area of the shaded region is equal to the area of the sector minus the area of isosceles triangle

so

$A=(\frac{50\pi}{3}-25\sqrt{3})\ cm^2$

Simplify

$A=25(\frac{2\pi}{3}-\sqrt{3})\ cm^2$

Part b) we know that

step 1

Find the area of the sector

we know that

The area of a circle subtends a central angle of 360 degrees

so

using proportion

Find the area of the sector by a central angle of 135 degrees

$\frac{\pi r^2}{360^o}= \frac{x}{135^o} \\\\x=0.375\pi r^2$

we have

$r=12\ cm$

substitute

$x=0.375\pi (12)^2\\\\x=54\pi\ cm^2$

step 2

Find the area of triangle

The area of triangle is equal to

$A=\frac{1}{2} (12^2)sin(135^o)$

$A=72(\frac{\sqrt{2}}{2})=36\sqrt{2}\ cm^2$

step 3

Find the area of the shaded region

The area of the shaded region is equal to the area of the sector minus the area of isosceles triangle

so

$A=(54\pi-36\sqrt{2})\ cm^2$

Simplify

$A=18(3\pi-2\sqrt{2})\ cm^2$