Question

A circle is inscribed in a triangle ABC. It touches AB
at D. AD = 5 cm, DB=3 cm and angle A=60°. Then
BC is
(1) 8 cm (2) 13 cm (3) 20 cm (4) 31 cm​

Answer 1

Given : A circle is inscribed in a triangle ABC. It touches AB at D.

AD = 5 cm, DB=3 cm and angle A=60°.  

To Find : BC  

(1) 8 cm (2) 13 cm (3) 20 cm (4) 31 cm​

Solution:

Let say r is in radius

Center of incircle meets at angle bisector of triangle

then tan(60°/2)   = r/AD

=> r = 5/√3

Let say circle touches AC at E and BC at F

AD = AE  = 5 cm ( equal tangent)

BD = BF  = 3 cm ( equal tangent)

CE = CF = x  cm   ( equal tangent)

Area of triangle = (1/2) AB * AC Sin∠A

= (1/2) * (5 + 3) (5 + x)  Sin(60°)

= 4(5 + x) √3 / 2

= 2(5 +x)√3

Area of triangle = (1/2) (AB + BC + AC)r

= (1/2)(5 + 3 +  3 + x + x + 5) 5/√3

= (8 + x)  5/√3

Equate area

(8 + x)  5/√3 = 2(5 +x)√3

=> (8 + x)  5 = 2(5 +x) 3

=> 40 + 5x = 30 + 6x

=> 10 = x

BC = BF + FC

= 3 + 10

= 13  cm

BC = 13 cm

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