Question

solve it fast as u can.....​

Answer 1

$given \: that \: yz \: = b \: \\ and \: area \: of \: triangle \: = a \: \\ we \: know \: that \: area \: of \: triangle \: \\ = > \frac{1}{2} \times base \: \times height \: \\ so \: we \: can \: write \: \\ = > a = \frac{1}{2} \times xy \times b \\ = > xy = 2 \frac{a}{b} \\ the \: given \: triangle \: is \: right \: angled \: so \: we \: can \: find \: hypotenuse \: \\ {xz}^{2} = {xy}^{2} + {yz}^{2} \\ = > xz = \sqrt{ {b}^{2} + {(2 \frac{a}{b} )}^{2} } \\ = > xz = \frac{ \sqrt{ {b}^{4} + 4 {a}^{2} } }{b} = hypotenuse \\ now \: i \: we \: take \: hypotenuse \: as \: base \: we \: will \: get \: py \: as \: the \: height \: \\ so \: again \: euating \: the \: area \: \\ = > a = \frac{1}{2} \times py \times \frac{ \sqrt{ {b}^{4} + 4 {a}^{2} } }{b} \\ = > py \: = \frac{2ab}{\sqrt{ {b}^{4} + 4 {a}^{2} } }$

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