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OpenStudy (jamesreed30):
Find the area of a circle circumscribed about an equilateral triangle whose side is 18 inches long
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OpenStudy (eliesaab):
Let x=18 be the side, then the height of the triangle is
\[h=\frac{x \sqrt 3}{2}
\]
OpenStudy (eliesaab):
The area is \[
\frac {x h }2
\]
OpenStudy (eliesaab):
Replace and you get your answer
OpenStudy (eliesaab):
\[
Area= \frac { x^2 \sqrt 3}4= \frac{18^2 \sqrt 3}4=81 \sqrt 3
\]
OpenStudy (jamesreed30):
So divide 81 and 3 and that's my answer?
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OpenStudy (jamesreed30):
Also my answer has the PI symbol at the end so would that change by answer?
OpenStudy (jamesreed30):
@eliesaab
OpenStudy (eliesaab):
Oh Sorry, I found the are of the triangle. You need the one for the circle. Wait a minute
OpenStudy (jamesreed30):
okay
OpenStudy (eliesaab):
The radius of the circle is one third of the height h. So
\[
Circle Area = \pi r^2= \pi (1/3 h)^2=\frac {h^2}9 \pi=\frac{\pi}9 h^2
\]
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OpenStudy (eliesaab):
Now
\[
h=\frac{x \sqrt 3}2\\
CircleArea=\frac \pi 9 \left( \frac{x \sqrt 3}2\right)^2
\]
Replace x by 18 and you are done
OpenStudy (jamesreed30):
So 6 PI?
OpenStudy (eliesaab):
No
OpenStudy (eliesaab):
\[
CircleArea=\frac \pi 9 \frac { 3 x^2}4=\frac \pi {12} x^2=\frac \pi {12} 18^2=
\frac \pi{12}324=27 \pi\]
OpenStudy (jamesreed30):
Thank you
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OpenStudy (eliesaab):
Let me double check my computations
OpenStudy (eliesaab):
The answer is \[27 \pi
\]
OpenStudy (jamesreed30):
Alright
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