# Ex.13.3 Q3 Surface Areas and Volumes Solution - NCERT Maths Class 10

## Question

A \(20 \,\rm{m}\) deep well with diameter \(7 \,\rm{m}\) is dug and the earth from digging is evenly spread out to form a platform \(22\, \rm{m}\) by \(14 \,\rm{m}\). Find the height of the platform.

[Use \(\begin{align}\pi \;=\;\frac{22}{7}\end{align}\)]

## Text Solution

**What is known?**

Depth of the well is \(20\rm{m}\) and diameter is \(7\rm{m}\). Length of the platform is \(22\rm{m}\) and breadth is \(14\rm{m.}\)

**What is unknown?**

The height of the platform.

**Reasoning:**

Draw a figure to visualize the shapes better

The shape of the well will be cylindrical, and soil evenly spread out to form a platform will be in a cuboidal shape.

Therefore, volume of the soil dug from the well will be equal to the volume of soil evenly spread out to form a platform.

Volume of soil dug out from the well \(=\) Volume of soil used to make such platform

Hence, Volume of the cylindrical well \(=\) Volume of the cuboidal platform

We will find the volume of the cylinder and cuboid by using formulae;

Volume of the cylinder \( = \pi {r^2}h\)

where *\(r\)* and *\(h\)* are radius and height of the cylinder respectively

Volume of the cuboid \( = lbH\)

where *\(l, b\)* and *\(H\)* are length breadth and height of the cuboid respectively

**Steps:**

Depth of the cylindrical well,\(h = 20 \rm m\)

Radius of the cylindrical well,\(r = \frac{7}{2} \rm m\)

Length of the cuboidal platform,\(l = 22 \rm m\)

Breadth of the cuboidal platform,\(b = 14 \rm m\)

Let the height of the cuboidal platform \(= H\)

Volume of the cylindrical well \(=\) Volume of the cuboidal platform

\[\begin{align}\pi {r^2}h &= lbH\\H &= \frac{{\pi {r^2}h}}{{lb}}\\&= \frac{{ \begin{bmatrix} \frac{{22}}{7} \times \frac{7}{2} \rm m \times\\ \frac{7}{2} \rm m \times 20 \rm m \end{bmatrix} }}{{22 \rm m \times 14m}}\\&= \frac{5}{2} \rm m\\&= \,2.5 \rm m\end{align}\]

Therefore, the height of such platform will be \(2.5 \rm{m}\)