# equilateral triangle inscribed in a circle..

• Feb 28th 2010, 08:02 AM
snigdha
equilateral triangle inscribed in a circle..
In the given fig., ABC is an equilateral triangle inscribed in a circle of radius 4 cm. Find the area of the shaded region.
• Feb 28th 2010, 12:16 PM
masters
Quote:

Originally Posted by snigdha
In the given fig., ABC is an equilateral triangle inscribed in a circle of radius 4 cm. Find the area of the shaded region.

Hi snigdha,

The area of the circle is $A_c=\pi r^2=16 \pi$

We need to find the area of the triangle, subtract that area from the area of the circle and then divide it by 3.

Find the area of the equilateral triangle:

Draw the radius from O to B.
Draw the perpendicular bisector of BC through O intersecting BC at X.
Solve the right triangle we just made.

Hypotenuse = 4
Since it's a 30-60-90 right triangle, angle OBX = 30 degrees and angle BOX = 60 degrees.

With a little trig or 30-60-90 rules, we determine that OX = 2, and BX = $2\sqrt{3}$.
Since triangle OBC is isosceles, BX = CX.
The base of the equilateral triangle ABC is $4\sqrt{3}$.

All we need now is the height of the equilateral triangle.

The area of the inscribed equilateral triangle is $A_t=\frac{1}{2}bh=\frac{1}{2}(4\sqrt{3})(6)=12\sqr t{3}$.

Now you have both pieces I talked about in the beginning.
You may finish up now.
Turn the lights out when you leave.
• Feb 28th 2010, 01:33 PM
Plato
You can also look at this: circular segement.
• Feb 28th 2010, 02:40 PM
masters
Quote:

Originally Posted by Plato
You can also look at this: circular segement.

I thought about doing it that way, then I thought again.
• Feb 28th 2010, 10:52 PM
snigdha
Well.....i found masters' solution better and way simpler....!
thanks a lot!!