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Answer 1

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Step-by-step explanation:

Answer 2

Answer:

your proof is as follows

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Step-by-step explanation:

$to \: prove \: that : \\ A(∆ABC) = \frac{1}{4} b .\sqrt{4a {}^{2} - b {}^{2} } \\ \\ for \: ∆ABC \\ AB = AC = a \\ BC = b \\ \\ thus \: then \\ s(∆ABC) = \frac{a + b + c}{2} \\ \\ s= \frac{2a + b}{2} \\ \\by \: herons \: formula \\ A(∆ABC) = \sqrt{s(s - a)(s - b)(s - c)} \\ \\ = \sqrt{( \frac{2a + b}{2}) ( \frac{ 2a + b}{2} - a)( \frac{2a + b}{2} - b)( \frac{2a + b}{2} - c} \\ \\ = \sqrt{( \frac{2a + b}{2})( \frac{2a - 2a + b}{2} )( \frac{2a + b - 2b}{2})( \frac{2a + b - 2a}{2} ) } \\ \\ = \sqrt{ (\frac{2a + b}{2} )( \frac{2a - b}{2}) \times \frac{b {}^{2} }{4} } \\ \\ = \sqrt{ (\frac{4a {}^{2} - b {}^{2} }{16}) .b {}^{2} } \\ \\ = \frac{b}{4} \sqrt{ 4a {}^{2} - b {}^{2} } \\ \\ = \frac{1}{4} \sqrt{4a {}^{2} - b {}^{2} } \\ \\ thus \: proved$