Question

$ABCD$ is a rectangle with sides $36cm$ and $90cm$. P is a point on $BC$ which is one of the longer sides such that $PA=2PD$. The length of $PB$ is

• A. $80cm$
• B. $76cm$
• C. $72cm$
• D. $64cm$

Answer 1

The correct option is C $72cm$

$ABCD$ is a rectangle.

So, $\Delta ABP$ & $\Delta DPC$ are right ones with PA & PD as hypotenuses .

$PA=2PD$ i.e

${PA}^2={\left(2PD\right)}^2=4{PD}^2$

Let $PB=x$, i.e $PC=90-x$.

So, applying Pythagoras theorem,

${PA}^2=x^2+{36}^2$

$\Rightarrow 4{PD}^2=x^2+{36}^2$ ........(i)

${PD}^2={(90-x)}^2+{36}^2$

$\Rightarrow 4{PD}^2={4(90-x)}^2+4\times {36}^2$ ......(ii)

Comparing (i) & (ii),

$x^2+{36}^2={4(90-x)}^2+4\times {36}^2$

$\Rightarrow 37584-720x+36288=0$

$\Rightarrow (x-168)(x-72)=0$

$\Rightarrow x=(168,72)cm.$

We reject $x=PB=168$ as PB is the part of $BC=90cm.$

$\therefore x=PB=72cm.$