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VARVARA [1.3K]
3 years ago
9

How do I find the area of the shaded region?

Mathematics
1 answer:
vovangra [49]3 years ago
4 0

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

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