Find the area of an isosceles triangle whose one side is $10$ cm greater then its equal side and its perimeter is $100$ cm ( Take $\sqrt{5} = 2.23$ )

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Solution

Given $Δ A B E$ is isosceles with $A B = A C$ and $B C = A B + 10$
Perimeter of $Δ A B C = 100 c m$ & $[\sqrt{5} = 2.23]$
Perimeter $= A B + B C + A C$
$100 = 2 A B + A B + 10$
$100 = 3 A B + 10$
$90 = 3 A B$
$[A B = 30 c m]$
now $B C = 30 + 10 = 40 c m$
and $B O = \frac{B C}{2} = 20 c m$
using Pythaoras is $Δ A O B$
$A B^{2} = A O^{2} + O B^{2}$
$(30)^{2} = (A O)^{2} + (20^{2}$
$(A O)^{2} = 900 - 400 = 500$
$[A O = 10 \sqrt{5} c m]$
area of $Δ A B C = \frac{1}{2} \times B C \times A O$
$= \frac{1}{2} \times 40 \times 10 \sqrt{5}$
$= 200 \sqrt{5} = 200 \times 2.23$
area of $Δ A B C = 446 c m^{2}$

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