Chapter 10.9, Problem 16PSC

a)

To determine

To find : Distance from the corner point to the can (PA).

Answer to Problem 16PSC

PA = $9.94$ cm

Explanation of Solution

Given Information :

We know that, Diameter of the circle = $48$ cm

Construction :

Let us consider O as the centre of the circle. PA is extended and it passes through the centre O.

  

Formula Used :

Radius = $\frac{\text{diameter}}{\text{2}}$

Calculation :

  $\text{OB}=\frac{1}{2}\times 48=24\text{ cm}$

  $\angle \text{BPQ}=90°$ [Rectangular corner]

  $\angle \text{OBP}=\angle \text{OQP}=90°$ [Radius $\perp$ to tangent at that point]

OBPQ is a square.

So $\Delta \text{OBP}$ is a right angle triangle.

  ${(\text{OB})}^2+{(\text{BP})}^2={(\text{OP})}^2$ (By Pythagoras Theorem)

  ${(\text{OP})}^2={(\text{24})}^2+{(\text{24})}^2$

OP = $24\sqrt{2}=33.94\text{ cm}$

Now, OA + AP = OP

  $\Rightarrow 24+\text{AP}=24\sqrt{2}$ [OA = radius = $24$ cm]

  $\begin{array}{l}\Rightarrow \text{PA}=24(\sqrt{2}-1)\\ \Rightarrow \text{PA}=9.94\text{ cm}\end{array}$

b)

To determine

To find : PB

Answer to Problem 16PSC

  $24$ cm = PB

Explanation of Solution

Given Information :

We know that, Diameter of the circle = $48$ cm

Formula Used:

Sides of square are equal in length.

Calculation :

As OBPQ is a square.

So, OB = PB

PB = $24$ cm [OB = radius = $24$ cm]