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 $\displaystyle A_c=\pi r^2=16 \pi$
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 = $\displaystyle 2\sqrt{3}$.
Since triangle OBC is isosceles, BX = CX.
The base of the equilateral triangle ABC is $\displaystyle 4\sqrt{3}$.
All we need now is the height of the equilateral triangle.
We just add the radius 4 to OX and get 6.
The area of the inscribed equilateral triangle is $\displaystyle A_t=\frac{1}{2}bh=\frac{1}{2}(4\sqrt{3})(6)=12\sqr t{3}$.
Now you have both pieces I talked about in the beginning.
You may finish up now.
Turn the lights out when you leave.
You can also look at this: circular segement.