The incircle of a triangle ABC touches AB at point P and has radius equal to 21.If AP = 23 and PB = 27, then find the perimeter of the triangle ABC.

Engineering

Mathematics

Basic Rules of Properties of Triangle

Question

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r = 21

c = 50

$\tan \frac{B}{2} = \frac{21}{27} = \frac{7}{9}$

$\tan \frac{A}{2} = \frac{21}{23}$

$\tan \frac{C}{2} = \tan (90^{\circ} - \frac{A + B}{2}) = \cot (\frac{A}{2} + \frac{B}{2}) = \frac{1}{\tan (\frac{A}{2} + \frac{B}{2})}$

$\tan \frac{C}{2} = \frac{1 - \tan \frac{A}{2} \tan \frac{B}{2}}{\tan \frac{A}{2} + \tan \frac{B}{2}} = \frac{1 - \frac{21}{23} \cdot \frac{7}{9}}{\frac{21}{23} + \frac{7}{9}} = \frac{70}{350} = \frac{1}{5}$

using r = (s – c)tan $\frac{C}{2}$ ; 21 = (s – 50) · $\frac{1}{5}$

s – 50 = 105 ⇒ s = 155 ⇒ 2s = 345.

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