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Mathematics 22 Online
OpenStudy (anonymous):

What is the area of the shaded region in the figure below?

OpenStudy (anonymous):

OpenStudy (anonymous):

a. 32π −16 b. 16π − 16 c. 32π − 32 d. 16π − 32

OpenStudy (anonymous):

@ganeshie8

OpenStudy (anonymous):

@.Sam.

sam (.sam.):

Find the area of the sector then subtract out the triangle

OpenStudy (anonymous):

how do you find the area of the sector?

sam (.sam.):

\[\huge A_{shaded}=A_{sector}-A_{\triangle}\]

sam (.sam.):

\[\large A_{sector}=\frac{1}{2}r^2 \theta\] Where \(\theta\) is in radians

OpenStudy (anonymous):

would 8 be r?

sam (.sam.):

Yes

OpenStudy (anonymous):

okay so 32?

sam (.sam.):

You forgot \(\theta\)

sam (.sam.):

\[\theta=\frac{90}{180} \times \pi\]

sam (.sam.):

The 90 comes from the given angle

OpenStudy (anonymous):

1.57?

sam (.sam.):

Yes then multiply this to 32

sam (.sam.):

Because \[\large A_{sector}=\frac{1}{2}r^2 \theta\]

OpenStudy (anonymous):

50.24

sam (.sam.):

Yes

OpenStudy (anonymous):

okay so now i subtract 50.24- 90?

sam (.sam.):

why do you subtract by 90?

OpenStudy (anonymous):

oh the area of the triangle

sam (.sam.):

You sure the area is 90 for triangle?

sam (.sam.):

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