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
Answers
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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Answer:
3.978 is a answer to the question above which is given