Question

prove the area of isosceles triangle

Answer 1

Solution :

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Derivation of Area of isosceles triangle ;

Let ABC be an isosceles triangle with equal sides a and unequal side b.

Area of ΔADC = 1 / 2 × base × height

From ΔADC, we have ;

AC² = AD² + DC²

$\sf ⇒a {}^{2} = h {}^{2} + (\frac{b}{2} ) {}^{2} \\ \\ ⇒ \sf h {}^{2} = a {}^{2} - \frac{b {}^{2} }{4} \\ \\ \sf ⇒h {}^{2} = \frac{4a {}^{2} - b {}^{2} }{4} \\ \\ \sf ⇒h = \frac{ \sqrt{4a {}^{2} - b {}^{2} }}{2}$

$\sf area \: of \: \Delta = \frac{1}{2} \times b \times h \\ \\ \sf area \: of \: \Delta = \frac{1}{2} \times b \times \frac{ \sqrt{4a {}^{2} - b {}^{2}}}{2} \\ \\ \sf area \: of \: \Delta = \frac{b}{4} \sqrt{4a {}^{2} - b {}^{2} }$

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