Find the area of an isosceles triangle whose base is $16 c m$ and length of each of the equal sides is $10 c m$ .

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Solution

Given
Base $= 16 c m$
Lengh of both equal side $= 10 c m$
We will use Herons's formula to find the area
$a = 16 c m$
$b = 10 c m$
$c = 10 c m$
Therefore,
Halfperimeter $= s = \frac{1}{2} \times (16 + 10 + 10)$
$= \frac{36}{2}$
$s = 18 c m$
using Heron's formula
$\sqrt{s (s - a) (s - b) (s - c)}$
$\begin{matrix} = \sqrt{18 (18 - 16) (18 - 10) (18 - 10)} = \sqrt{18 \times 2 \times 8 \times 8} = \sqrt{2 \times 9 \times 4 \times 2 \times 4 \times 2 \times 2} = 3 \times 2 \times 2 \times 2 \times 2 = 48 c m^{2} \end{matrix}$
hence,
The area of given triangle is $48 c m^{2}$

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