Segments formed by an inscribed equilateral triangle

**norifurippu** · May 29th 2008, 07:16 AM

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.

**masters** · May 29th 2008, 08:06 AM

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, $A=\frac{1}{2}(24\sqrt3)(4)=48\sqrt3$

**norifurippu** · May 30th 2008, 07:28 PM

Thanks a lot.

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