Question

$P$ and $Q$ are the mid points of the sides $CA$ and $CB$ respectively of a triangle $ABC$, right angled at $C$. Then, find the value of $4(A{Q}^2+B{P}^2)$.

• A. $5A{B}^2$.
• B. $3A{B}^2$.
• C. $2A{B}^2$.
• D. $4A{B}^2$.

Answer 1

The correct option is A $5A{B}^2$.

In $rt.\angle d\Delta ACQ,A{Q}^2=Q{C}^2+A{C}^2$
In $rt.\angle d\Delta BCP,B{P}^2=B{C}^2+P{C}^2$
$\therefore A{Q}^2+B{P}^2=(Q{C}^2+A{C}^2)+(B{C}^2+P{C}^2)$
$=(A{C}^2+B{C}^2)+(Q{C}^2+P{C}^2)$
$=A{B}^2+P{Q}^2=A{B}^2+{\left(\begin{array}{c}\frac{1}{2}AB\end{array}\right)}^2$
[ $(\because PQ=\frac{1}{2}AB)$ Line joining the mid points of two sides of a $\Delta$ is always half the third side]
$=A{B}^2+\frac{1}{4}A{B}^2=\frac{5}{4}A{B}^2$
$\Rightarrow 4(A{Q}^2+B{P}^2)=5A{B}^2$.