Question

In a right triangle $ABC$ right angled at $C,$ $P$ and $Q$ are points on the sides $CA$ and $AB$ respectively, which divide these sides in the ratio $2:1.$ Then, which of the following statements is true$?$

• A. $9A{Q}^2=9B{C}^2+4A{C}^2$
• B. $9A{Q}^2=9A{P}^2+4B{C}^2$
• C. $9A{Q}^2=9B{C}^2+4P{Q}^2$
• D. $9A{Q}^2=9A{B}^2-4B{P}^2$

Answer 1

The correct option is B $9A{Q}^2=9A{P}^2+4B{C}^2$

Given:

$\Delta ABC$ is right angled at $C$.

$P$ and $Q$ are the points on $AC$ and $AB$ respectively such that

$AP:PC=2:1=AQ:QB$

i.e $\frac{AQ}{AB}=\frac{2}{2+1}$

$\Rightarrow AB=\frac{3}{2}\times AQ$

$\frac{AP}{AC}=\frac{2}{2+1}$

$\Rightarrow AC=\frac{3}{2}\times AP$ .......... $\left(i\right)$

Now, $\Delta ABC$ is right angled at $C$

So, applying Pythagoras' theorem,

${AB}^2={AC}^2+{BC}^2$

Using $\left(i\right),$ we get

${(\frac{3}{2}\times AQ)}^2={(\frac{3}{2}\times AP)}^2+{\left(BC\right)}^2$

$\Rightarrow \frac{9}{4}\times {AQ}^2=\frac{9}{4}\times {AP}^2+{BC}^2$

$\Rightarrow 9{AQ}^2=9A{P}^2+4{BC}^2$