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Triangle ABC is isosceles with AB = BC.  If AC = 20 and the inradius is 5, then find the perimeter of triangle ABC.

 Jun 7, 2024
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inradius = area  / semi-perimeter

 

Let AB = BC  = x

 

semi-perimeter  = [ 20 + 2x ] / 2 =  10 + x

 

area =  sqrt [ (10 + x) (10 + x -20) (10 +x -x)^2]  =  sqrt [ (10+x) (x -10) (100) ]

 

5 =   sqrt [ (10+x) (x -10) (100) ] / [ 10 +x]

 

5 [10 + x] = sqrt [ 100 ( x^2 - 100) ] 

 

5 [ 10 + x] = 10 sqrt [ (x^2 - 100) ]

 

[10 + x] = 2 sqrt [x^2 -100]        square both sides

 

x^2 + 20x + 100  =  4x^2 - 400

 

3x^2 - 20x - 500 =  0

 

x = 50/3

 

Perimeter  =  2 (50/3) + 20  =  160 / 3

 

 

cool cool cool

 Jun 7, 2024

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