Question

If an isosceles triangle has base b and equal sides a, then its area will be

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Answer 1

$\underline {\blue {Given :}}$

$ABC \:is \:an \: isosceles \: triangle.$

$AB = AC = a , \:and \: BC = b$

$\underline {\blue {Construction :}}$

$Draw \: AD \: perpendicular \:to \: BC$

$\underline {\blue {Solution:}}$

$In \: \triangle ABD , \\</p><p>\angle {ADB} = 90\degree ,\\BD = \frac{BC}{2} = \frac{b}{2} \: ( Altitude \:bisects \: base )$

$AB^{2} = BD^{2} + AD^{2} \\(By \: Pythagoras \: theorem )$

$\implies a^{2} = \big(\frac{b}{2}\big)^{2} + h^{2}$

$\implies h^{2} = a^{2} - \big(\frac{b}{2}\big)^{2}$

$\implies h^{2} = \frac{(4a^{2} - b^{2})}{4}$

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

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

$\red { Area \: of \:\triangle ABC} = \frac{1}{2} \times base \times corresponding \:height\\= \frac{1}{2} \times BC \times AD$

$= \frac{1}{2} \times b \times \frac{\sqrt{(4a^{2} - b^{2})}}{2}[From \:(1) ]$

$= \frac{b\sqrt{(4a^{2} - b^{2})}}{4}$

Therefore.,

$\red { Area \: of \:\triangle ABC}$

$\green { = \frac{b\sqrt{(4a^{2} - b^{2})}}{4}}$

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