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QPR is a right triangle right angled at Q If QS=SR then show that PR×PR=4PS×PS-3PQ×PQ? |
| Answer» Given: PQR is a right Triangle, right-angled at QAlso QS = SRTo prove:- {tex}P{R^2} = 4P{S^2} - 3P{Q^2}{/tex}Proof:- In right-angled triangle PQR right angled at Q.{tex}P{R^2} = P{Q^2} + Q{R^2}{/tex}[By Pythagoras theorem] ....(i)Also {tex}QS = {1 \\over 2}QR{/tex} [QS=SR] .....(ii)In right-angled triangle PQS, right angled at Q.PS2 = PQ2 + QS2{tex}\\Rightarrow {/tex}{tex}P{S^2} = P{Q^2} + {\\left( {\\frac{1}{2}QR} \\right)^2}{/tex} [From (ii)]{tex}\\Rightarrow {/tex} 4PS2 = 4PQ2 + QR2 .....(iii)From (i) and (iii), we getPR2 = PQ2 + 4PS2 - 4PQ2{tex}\\Rightarrow {/tex} PR2 = 4PS2 - 3PQ2 | |