Question

In the given figure below the in-circle is tangent to triangle ABC at points D,E ,F. 2root 14, 4root 3 root 42 are the lenghts of the common tangents to the semi circles of diameter ( BD, CD), (CE, AE) and (AF, BF) respectively

Answer 1

Given : in-circle is tangent to triangle ABC

To Find : Radius of incircle

Solution:

Let say radius of  yellow semicircle = a

green semicircle = b

purple semi circle = c

Sides of triangle = 2a + 2c  ,  2b + 2c  ,  2a + 2b

Draw tangent between semicircle

Tangent  lengths = ( a + c)/2 , ( b + c)/2 , (a + b)/2

√42 = 2 x  ( a + c)/2 => a  + c = √42

2√14 = 2 x  (b + c)/2 => b  + c = 2√14

4√3 = 2 x  (a + b)/2 => a  + c = 4√3

Sides of triangles are = 2√42  , 4√14 , 8√3

s = ( 2√42 + 4√14 + 8√3)/2  = √42 + 2√14 + 4√3

Area of triangle = r.s  = r. (√42 + 2√14 + 4√3)

Area of triangle using heron

√ (√42 + 2√14 + 4√3) (-√42+2√14 + 4√3)(√42 - 2√14 + 4√3)(√42 + 2√14 - 4√3)

(√42 + 2√14 + 4√3) (-√42+2√14 + 4√3) = ( 2√14 + 4√3)² - 42 = 62 + 16√42

(√42 - 2√14 + 4√3)(√42 + 2√14 - 4√3) = 42 -  ( 2√14 - 4√3)²  =  16√42 - 62

= √6,908

r. (√42 + 2√14 + 4√3) = √6,908

=> r = √6,908/ (√42 + 2√14 + 4√3)

=> r = 3.978

 Radius of incircle = 3.978

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