Question

 the area of an isosceles triangle also, prove it.

Answer 1

Please diagram. Base is BC = a,  Two Sides  b = c ie., AB =  AC.  and  Angle B = angle C  : Isosceles triangle.

The perpendicular bisector AD bisects BC, because AB=AC. AD is perpendicular.

$AD^2 + BD^2 = AB^2, \ \ \ Pythagoras\ law \\ AD^2 = b^2 - (\frac{BC}{2})^2 = \frac{1}{4} (4b^2 - a^2) \\ AD = \frac{1}{2} \sqrt{4b^2-a^2} \\ \\ Area\ of\ ISOSCELES\ Triangle\ =\ \frac{1}{2} base\ *\ altitude\ \\ = \frac{1}{2} * a * \frac{1}{2} \sqrt{4b^2 - a^2} \\ \\ = \frac{1}{4} a\ \sqrt{4b^2 - a^2}$

Another formula:

$Sin\ B = \frac{Height}{b},\ \ \ Height = b\ Sin\ B \\ \\ Area\ =\ \frac{1}{2}\ a\ b\ Sin\ B$

Answer 2

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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