Answer

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**Hint:**

First draw a suitable diagram of the rectangle. Then, take a pint on BC. Take both right angled triangles and apply Pythagoras theorems. Compare both expressions together as $PA= 2 PD$. Solve the quadratic equation, thus obtained. Suitable root of it will be the result.

**Complete step by step solution:**

Draw the rectangle ABCD as below.

Sides AB and DC are equal and 36 cm.

Also, sided AC and AD are 90 cm.

Let us assume that P is appointed somewhere on side BC.

It is given that ,

PA = 2 PD …(1)

Now let us assume that length of BP is x cm,

So, the PC will be (90-x) cm.

In right angled triangle ABP, we apply the Pythagoras theorem, and hence we get,

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

Substituting the values in above equation, we have

$P{A^2} = {36^2} + {x^2}$ …(2)

Similarly in right angled triangle CPD, we apply the Pythagoras theorem, and hence we get,

$P{D^2} = P{C^2} + D{C^2}$

Substituting the values in above equation, we have

$P{D^2} = {36^2} + {(90 - x)^2}$ …(3)

Doing square on both sides of equation (1), we get

$P{A^2} = 4P{D^2}$

Now, from equations (2) and (3) in the above equation, we have

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

Further simplification, we get

$

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

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

\Rightarrow {x^2} = 3 \times {36^2} + 4 \times (8100 + {x^2} - 180x) \\

\Rightarrow {x^2} = 3888 + 32400 + 4{x^2} - 720x \\

\Rightarrow 3{x^2} - 720x + 36288 = 0 \\

\Rightarrow {x^2} - 240x + 12096 = 0 \\

$

Thus we have a quadratic equation. Two factors of 12096 are 168 and 72. Thus doing factorization of above equation we get,

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

Thus values of $x = 168$ and $72$.

Since $168 > 90$, which is not possible for point P.

Thus a suitable value of $x = 72$.

$\therefore $ The length of PB will be 72 cm.

**So, the correct option is C.**

**Note:**

This question is a direct application of Pythagoras theorem. Also, solving the quadratic equation is an important part of such problems. In this way, we can solve a geometrical problem with the help of algebraic computations.

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