Find the area of an isosceles triangle each of

Question:

Find the area of an isosceles triangle each of whose equal sides measures 13 cm and whose base measures 20 cm.

Solution:

We have :

$a=13 \mathrm{~cm}$ and $b=20 \mathrm{~cm}$

$\therefore$ Area of isosceles triangle $=\frac{b}{4} \sqrt{4 a^{2}-b^{2}}$

$=\frac{20}{4} \times \sqrt{4(13)^{2}-20^{2}}$

$=5 \times \sqrt{676-400}$

$=5 \times \sqrt{276}$

$=5 \times 16.6$

$=83.06 \mathrm{~cm}^{2}$

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