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Q. 33

Expert-verifiedFound in: Page 541

Book edition
6th

Author(s)
Sullivan

Pages
1200 pages

ISBN
9780321795465

Area of a Segment Find the area of the segment (shaded in blue in the figure) of a circle whose radius is 8 feet, formed by a central angle of $70\xb0$.

[Hint: Subtract the area of the triangle from the area of the sector to obtain the area of the segment.]

The area of the segment is 9.03 square feet.

The given figure is:

Area of the sector is:

$\begin{array}{rcl}{A}_{1}& =& \frac{\theta}{360\xb0}\times \pi {r}^{2}\\ & =& \frac{70\xb0}{360\xb0}\times \pi \times (8{)}^{2}\\ & =& 39.0953...\\ & \approx & 39.10\end{array}$

The area of a triangle is:

$\begin{array}{rcl}{A}_{2}& =& \frac{1}{2}ab\mathrm{sin}C\\ & =& \frac{1}{2}\left(8\right)\left(8\right)\mathrm{sin}(70\xb0)\\ & =& 30.0701...\\ & \approx & 30.07\end{array}$

Subtract the area of the triangle from the area of the sector to obtain the area of the segment.

$\begin{array}{rcl}A& =& {A}_{1}-{A}_{2}\\ & =& 39.10-30.07\\ & =& 9.03\end{array}$

The area of the segment is 9.03 square feet.

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