An isosceles triangle has perimeter $36 c m$ and each of its equal sides is $13 c m$ long . The area of the triangle is

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Solution

$\begin{matrix} W e h a v e p e r i m e t e r o f t r i a n g l e = 36 c m a n d, i t s t w o e q u a l s i d e s = 13 c m L e t t h i r d s i d e b e x c m T h e n p e r i m e t e r = 36 c m 13 + 13 + x = 36 ∴ x = 10 c m N o w S e m i - p e r i m e t e r S = \frac{p e r i m e t e r}{2} = \frac{36}{2} c m = 18 c m N o w A r e a o f t r i a n g l e = \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{18 (18 - 13) (18 - 13) (18 - 10)} c m^{2} = \sqrt{18 (5) (5) (8)} c m^{2} = \sqrt{2 \times 3 \times 3 \times 5 \times 5 \times 2 \times 2 \times 2} c m^{2} = 2 \times 3 \times 5 \times 2 c m^{2} = 60 c m^{2} . H e n c e, A r e a o f t r i n a g l e = 60 c m^{2} . \end{matrix}$

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