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What is the area of a sector with radius 10" and measure of arc equal to 45°? help me please
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see circle area is pir^2 which is pix100....when theta = 90 its quarter pi25 is the area but 45 is half so the area must be 25/2*pi
Sorry, writing the final answer: \[45°=\frac{\pi}{4}rad\] \[A=\frac{\theta \times r^2}{2}\] With theta in radians \[A=\frac{\pi \times 100}{2 \times 4}=\frac{25\times \pi}{2}\] in square inches
\[\frac{x}{{\pi}r^2} = \frac{45}{360}\] \[\frac{x}{{\pi}10^2} = \frac{45}{360}\]
Cross Multiply, solve for x
\[360x = 4500\pi \] \[x = \frac{4500{\pi}} {360} \]
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x = 39.2 units squared
or x = 25pi/2 units squared
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