Question

in an eqiulateral triangle PQR,S is a point on side QR such that QS =1/3 QR prove that 9 PS²=7 PQ²​

Answer 1

Answer:

$pt \:be \: the \: perpendicular \: from \: p \: on \: qr \:$

$hence \: qt \: = \frac{1}{2} qr$

$st = qr( \frac{1}{2 } - \frac{1}{3} ) = \frac{1}{6} qr$

${ps}^{2} = {pt}^{2} + {st}^{2}$

$also \: pt = qt \tan(60)$

$pt \: = \sqrt{3qt}$

${ps}^{2} = 3 {qt}^{2} + \frac{ {qr}^{2} }{36}$

${ps}^{2} = \frac{3 {qr}^{2} }{4} + \frac{ {qr}^{2} }{36}$

${ps}^{2} = \frac{28 {qr}^{2} }{36} = \frac{7 {qr}^{2} }{9}$

$qr \: = \: pq$

${9 \: ps}^{2} = {7 \: pq}^{2}$