Question

$P$ and $Q$ are points on the sides $CA$ and $CB$, respectively, of $△ABC$ right angled at $C.A{Q}^2+B{P}^2$ equals

• A. $B{C}^2+P{Q}^2$
• B. $A{B}^2+P{C}^2$
• C. $A{B}^2+P{Q}^2$
• D. $B{C}^2+A{C}^2$

Answer 1

Answer: (C)

$A{B}^2+P{Q}^2$

In rt. angled $△ACQ,$
$\Rightarrow$ $A{C}^2+C{Q}^2=A{Q}^2$ ....(i)
In rt. angled $△PCB$
$P{C}^2+C{B}^2=P{B}^2$ ....(ii)
Adding eqn (i) and (ii)
$A{C}^2+C{Q}^2+P{C}^2+C{B}^2=A{Q}^2+P{B}^2$
$\Rightarrow$ $(A{C}^2+C{B}^2)+(C{Q}^2+P{C}^2)=A{Q}^2+P{B}^2$
$\Rightarrow$ $A{B}^2+P{Q}^2=A{Q}^2+P{B}^2$ [$\left(rt\angle d△ABC\right)\left(rt\angle d△PQC\right)$ (Pythagoras' Theorem)]