Segments formed by an inscribed equilateral triangle

• May 29th 2008, 07:16 AM
norifurippu
Segments formed by an inscribed equilateral triangle
Hello. This is a problem from a book I'm studying on, I can't understand the solution.

"Find each segment formed by an inscribed equilateral triangle if the radius of the circle is 8"

Solution
R=8. Since $s=R\sqrt{3}=8\sqrt{3}$, the area of $\vartriangle ABC$ is $\tfrac{1}{4}s^{2}\sqrt{3}=48\sqrt{3}$.
Also, area of circle O = $\pi R^{2}=64\pi$.
Hence area of segment BDC = $\tfrac{1}{3}\left( 64\pi -48\sqrt{3} \right)$

These are my questions:
1.- Why is $s=R\sqrt{3}=8\sqrt{3}$?

2.And why is the area of $\vartriangle ABC$ equal to $\tfrac{1}{4}s^{2}\sqrt{3}=48\sqrt{3}$? I know the area of a triangle is equal to $\tfrac{1}{2}bh$, how do you get the height?
Thanks.
• May 29th 2008, 08:06 AM
masters
Inscribed equilateral triangle:

Draw apothem from center perpendicular to a side.

Draw radius from center to vertex of same side.

Triangle formed is 30-60-90.

If hypotenuse(radius) is 8, then side opposite the 30 deg. angle is 4 (apothem).

Side opposite 60 degree angle is $4\sqrt3$

Thus, the length of a side is $2(4\sqrt3)=8\sqrt3$

Perimeter = $3(8\sqrt3)=24\sqrt3$

$Area=\frac{1}{2}Pa$, where P=perimeter, a=apothem

Finally, $A=\frac{1}{2}(24\sqrt3)(4)=48\sqrt3$
• May 30th 2008, 07:28 PM
norifurippu
Thanks a lot.