Chapter 12, Problem 32CR

To determine

To find: The radius of the drainpipe when the water in it is 18 cm deep and the width of the surface of water is 48 cm.

Answer to Problem 32CR

The radius of the drainpipe is 25 cm.

Explanation of Solution

Given Information:

Depth of the water = 18 cm

Width of the surface of water = 48 cm

Formula used:

By Pythagoras theorem,

  ${\text{Hypotenuse}}^2={\text{Base}}^2+{\text{Altitude}}^2$

Calculation:

Let $x$ be the radius of the pipe.

Draw radius $OP$ and then draw a line $\perp$ the surface of the water and extend it to meet the circle.

We know that, a perpendicular from the centre of a circle to the chord bisect it.

So, $PR=\frac{48}{2}=24\text{cm}$

Now, $OR\perp PR$

  

Here, $\Delta ORP$ forms a right triangle with $\angle ORP={90}^{\text{o}}$

By Pythagoras theorem,

  ${\text{Hypotenuse}}^2={\text{Base}}^2+{\text{Altitude}}^2$

  $\begin{array}{l}\therefore O{P}^2=P{R}^2+O{R}^2\\ \therefore x^2={24}^2+{\left(x-18\right)}^2\\ \therefore x^2=576+x^2-36x+324\\ \therefore 36x=900\\ \therefore x=25\text{cm}\end{array}$

Hence, radius of the pipe is 25 cm.