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surface area
- Thread starter mathe25
- Start date
Hello, mathe25!
Since \(\displaystyle PQ = P\!A = AQ = r\), triangle \(\displaystyle P\!AQ\) is equilateral.
. . Hence: \(\displaystyle \angle P = \angle Q = \angle A = 60^o\)
In the upper half of the shaded region, we have a sector and a segment.
. . The area of the sector is: .\(\displaystyle \frac{1}{6}\pi r^2\)
. . The area of the segment is:. \(\displaystyle \frac{1}{6}\pi r^2 - \frac{\sqrt{3}}{4}r^2\)
Hence, its area is: .\(\displaystyle \frac{1}{3}\pi r^2 - \frac{\sqrt{3}}{4}r^2\)
Therefore, the shaded region has area:
.. \(\displaystyle 2\times\left(\frac{1}{3}\pi r^2 - \frac{\sqrt{3}}{4}r^2\right) \;=\; \frac{1}{6}\left(4\pi - 3\sqrt{3}\right)r^2 \)
I have two circles and I have only the radius \(\displaystyle r.\)
I must find S, the area of the shaded region.
Code:* * * A * * * * *o* * * *:/:::* * * *:/:::::* * :/::::::: * */::::::::* * * P o:-:-:-:-:o Q * * *:::::::::* - - - - * :::::::::: * *:::::::* * * *:::::* * * *:* * * * * * * *
Since \(\displaystyle PQ = P\!A = AQ = r\), triangle \(\displaystyle P\!AQ\) is equilateral.
. . Hence: \(\displaystyle \angle P = \angle Q = \angle A = 60^o\)
In the upper half of the shaded region, we have a sector and a segment.
. . The area of the sector is: .\(\displaystyle \frac{1}{6}\pi r^2\)
. . The area of the segment is:. \(\displaystyle \frac{1}{6}\pi r^2 - \frac{\sqrt{3}}{4}r^2\)
Hence, its area is: .\(\displaystyle \frac{1}{3}\pi r^2 - \frac{\sqrt{3}}{4}r^2\)
Therefore, the shaded region has area:
.. \(\displaystyle 2\times\left(\frac{1}{3}\pi r^2 - \frac{\sqrt{3}}{4}r^2\right) \;=\; \frac{1}{6}\left(4\pi - 3\sqrt{3}\right)r^2 \)