Question

An equilateral triangle is circumscribed by a circle of radius is 6cm. Find its side

Answer 1

Lets call the triangle as triangle ABC. Let the circle have centre O. Now, draw OB and OC. And draw the perpendicular from O to the chord BC. Let it intersect BC at M. 

Now consider triangles OBM and OCM. We have: 
OB = OC = radius 
OM = OM = common side 
angle(OMB) = angle(OMC) = 90deg (by construction) 

So the two triangles are congruent. Hence we have: 
i. angle BOM = angle COM 
ii. BM = CM 

But: angle (BOM) + angle (COM) = angle (BOC) 
=> 2*angle(BOM) = angle(BOC) 

Now: angle (BOC) = 2*angle (BAC) [angle subtended at the centre is twice that subtended ant any point on the circle] 

ABC is an equilateral triangle. So angle (BAC) = 60 deg. So we have: 
angle (BOC) = 2*angle (BAC) 
2*angle(BOM) = 2*angle(BAC) 
=> angle(BOM) = angle(BAC) = 60 deg 

Now consider the triangle BOM. It is a right angled triangle. Then we can use trigonometric relations to get: 
BM/OB = sin(angle(BOM)) = sin(60 deg) 
BM/OB = {sqrt(3)}/2 
BM = OB*{sqrt(3)}/2 

From ii. we have BM = CM. But BM + CM = BC 
=> BC = 2*BM = {sqrt(3)}*(OB) = 6*sqrt(3) 

Thus BC = AC = AB = 6*sqrt(3)