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What is the area of the shaded region in the figure below?
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a. 32π −16 b. 16π − 16 c. 32π − 32 d. 16π − 32
@ganeshie8
@.Sam.
Find the area of the sector then subtract out the triangle
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how do you find the area of the sector?
\[\huge A_{shaded}=A_{sector}-A_{\triangle}\]
\[\large A_{sector}=\frac{1}{2}r^2 \theta\] Where \(\theta\) is in radians
would 8 be r?
Yes
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okay so 32?
You forgot \(\theta\)
\[\theta=\frac{90}{180} \times \pi\]
The 90 comes from the given angle
1.57?
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Yes then multiply this to 32
Because \[\large A_{sector}=\frac{1}{2}r^2 \theta\]
50.24
Yes
okay so now i subtract 50.24- 90?
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why do you subtract by 90?
oh the area of the triangle
You sure the area is 90 for triangle?
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