Question

Area of triangle in terms of inradius and circumradius

Answer 1

Let △ABC△ABC be a triangle whose sides are of lengths a,b,ca,b,c.


Then the area AA of △ABC△ABC is given by:

A=rsA=rs

where:

rr is the inradius of △ABC△ABCs=a+b+c2s=a+b+c2 is the semiperimeter of △ABC△ABC.


Proof



Let II be the incenter of △ABC△ABC.

Let rr be the inradius of △ABC△ABC.

The total area of △ABC△ABC is equal to the sum of the areas of the triangle formed by the vertices of △ABC△ABC and its incenter:

A=Area(△AIB)+Area(△BIC)+Area(△CIA)A=Area⁡(△AIB)+Area⁡(△BIC)+Area⁡(△CI
Let ABAB, BCBC and CACA be the bases of △AIB,△BIC,△CIA△AIB,△BIC,△CIA respectively.

The lengths of ABAB, BCBC and CACA respectively are c,a,bc,a,b.

The altitude of each of these triangles is rr.

Thus from Area of Triangle in Terms of Side and Altitude:

Area(△AIB)Area⁡(△AIB)==cr2cr2Area(△BIC)Area⁡(△BIC)==ar2ar2Area(△CIA)Area⁡(△CIA)==br2br2

Thus:

A=ra+b+c2A=ra+b+c2


That is:

A=rsA=rs

where s=a+b+c2s=a+b+c2 is the semiperimeter of △ABC△ABC.

Theorem

Let △ABC△ABC be a triangle whose sides are of lengths a,b,ca,b,c.


Then the area AA of △ABC△ABC is given by:

A=abc4RA=abc4R

where RR is the circumradius of △ABC△ABC.


Proof



Let OO be the circumcenter of △ABC△ABC.

Let AA be the area of △ABC△ABC.

Let a perpendicular be dropped from CC to ABAB at EE.

Let h:=CEh:=CE.

Then:

AA==ch2ch2 Area of Triangle in Terms of Side and Altitude(1):(1):⟹  ⟹  hh==2Ac2Ac


Let a diameter CDCD of the circumcircle be passed through OO.

By definition of circumradius, CD=2RCD=2R.

By Thales' Theorem, ∠CAD∠CAD is a right angle.

By Angles on Equal Arcs are Equal, ∠ADC=∠ABC∠ADC=∠ABC.

It follows from Sum of Angles of Triangle equals Two Right Angles that ∠ACD=∠ECB∠ACD=∠ECB.

Thus by Equiangular Triangles are Similar △DAC△DAC and △BEC△BEC are similar.


So:

CACDCACD==CECBCECB △DAC△DAC and △BEC△BEC are similar⟹  ⟹  b2Rb2R==haha==2Aac2Aac substituting for hh from (1)(1) above⟹  ⟹  AA==abc4Rabc4R

■◼

Answer 2

Area A = r × s, where r is the in radius and 's' is the semi perimeter.

The semi perimeter, s =2/3a

In-radius, 'r' for any triangle = A/s

∴ for an equilateral triangle its in-radius, 'r' = A/s = a2√3