Question

Triangle $ABC$ is right angled at $A$. The points $P$ and $Q$ are on the hypotenuse $BC$ such that $BP=PQ=QC$.
If $AP=3$ and $AQ=4$, then the length $BC$ is equal to

• A. $\sqrt{27}$
• B. $\sqrt{36}$
• C. $\sqrt{45}$
• D. $\sqrt{54}$

Answer 1

Answer: (C)

$\sqrt{45}$

$BP=PQ=QC=x\left(let\right)$

$AP=3;AQ=4$

$In△AQB$

$APismedianbyApolloniusThm$

${AB}^2+{AQ}^2=2({AP}^2+{PQ}^2)$

$⟹{AB}^2+16=2(9+x^2)$

$⟹{AB}^2=2x^2+2-\left(1\right)$

$Similarly,in△APC$

$AQismedian$

$So,$

${AC}^2+{AP}^2=2({AQ}^2+{QC}^2)$

$\therefore {AC}^2+9=2(16+x^2)$

$⟹{AC}^2=2x^2+23-\left(2\right)$

$\left(1\right)+\left(2\right)$

${AB}^2+{AC}^2=4x^2+25$

$⟹{BC}^2=4x^2+25[In△ABC,usingpythagorasthm]$

$⟹{\left(3x\right)}^2=4x^2+25$

$⟹9x^2=4x^2+25$

$⟹x=\sqrt{5}$

BC=3x=$\sqrt{45}$