Question

QT and RS are medians of a triangle PQR right angled at P prove that a (QT square + RS square ) = 5 QR square​

Answer 1

The question should be

QT and RS are medians of a triangle PQR right angled at P prove that $4(QT^2+ RS^2) = 5 QR^2$​

$\textsf{Given:}$

$\textsf{QT and RS are medians}$

$\mathsf{PT=\frac{PR}{2}\;\;\&\;\;PS=\frac{PQ}{2}}$

$\textsf{In \triangle PQT, Using pythagoras theorem}$

$\mathsf{QT^2=PT^2+PQ^2}$.....(1)

$\textsf{In \triangle PSR, Using pythagoras theorem}$

$\mathsf{RS^2=PS^2+PR^2}$......(2)

$\textsf{Consider}$

$\mathsf{QT^2+ RS^2}$​

$\mathsf{=(PT^2+PQ^2)+(PS^2+PR^2)}$​

$\mathsf{=\frac{PR^2}{4}+PQ^2+\frac{PQ^2}{4}+PR^2}$​

$\mathsf{=(\frac{PR^2}{4}+PR^2)+(\frac{PQ^2}{4}+PQ^2)}$​

$\mathsf{=\frac{5\;PR^2}{4}+\frac{5\;PQ^2}{4}}$​

$\mathsf{=\frac{5}{4}[PR^2+PQ^2]}$​            $(\because\;\mathsf{QR^2=PR^2+PQ^2})$

$\mathsf{=\frac{5}{4}QR^2}$​

$\implies\mathsf{QT^2+ RS^2=\frac{5}{4}QR^2}$​

$\implies\boxed{\mathsf{4(QT^2+ RS^2)=5\,QR^2}}$​

Answer 2

Answer:

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