Question

In a triangle ABC show that a+b+c=abc÷2Rr

Answer 1

Given:

a, b, c are the sides of a triangle.

R is the circumradius and r is the inradius.

To Show:

a+b+c=abc÷2Rr

Solution:

Let a, b, c be the sides and A, B and C be the angles opposite to these sides.

Let  s be the semi perimeter = ( a + b + c )/2

For a triangle, inradius r is the radius of the circle inscribed in it.

Therefore,

• r = Area of triangle/s = A/s
• Area of triangle = bcsinA/2
• r = (bcsinA)/2s

If R is the radius of a circle circumscribing the triangle,

• R = a/2sinA = b/2sinB = c/2sinC
• Let R = a/2sinA = >
• sinA = a/2R

Substituting sinA in equation of r ,

• r = bc a /2R .2s
• r = abc / 4R .s
• s = abc/4Rr
• (a + b + c ) /2  = abc/4Rr
• a + b + c = abc/2Rr

Thus showed that a + b + c = abc/2Rr.