An equilateral triangle with a side of length 6√3 cm is inscribed in a circle. Find the radius of the circle.
![](https://www.edugain.com/egdraw/draw.php?num=2&sx=200&sy=200&x0=100&y0=100&A1=shape:ellipse;cx:0;cy:0;r1:150;r2:150&A2=shape:polygon;points:0,75,-65,-37,65,-37;textc:[-5.20]A,[-15.0]B,[10.0]C )
Answer:
6 cm
- ΔABC is inscribed in a circle. Let O be the center of the circle.
- Now, connect all vertices of the triangle to O, and draw a perpendicular from O meet the side BC of the triangle at point D.
- We know, all the angles of an equilateral triangle measure 60°.
So, angle ACB = ABC = CBA = 60°
OB and OC are bisectors of ∠B and ∠C respectively,
∠OBD = 30° - Since, triangle ODB is a right- angled triangle.
We have,
= cos 30° =BD OB √3 2
OB =BD √3 2
=3√3 √3 2
= 6 cm - Therefore, the radius of the circle is 6 cm.