# Ex.13.3 Q4 Surface Areas and Volumes Solution - NCERT Maths Class 9

## Question

A conical tent is \(10\,\rm m\) high and the radius of its base is \(24\,\rm m.\) Find

**i.** Slant height of the tent.

**ii.** Cost of the canvas required to make the tent, if the cost of \( 1 \, \rm {m^2} \) canvas is \(\rm Rs. \,70.\)

## Text Solution

**Reasoning:**

Curved surface area of the cone of base radius \(r\) and slant height \(l\) is\(\begin{align}\pi rl \end{align}\). Where, \(\begin{align}l = \sqrt {{r^2} + {h^2}} \end{align}\) using the **Pythagoras Theorem.** And cost of canvas required will be the product of area and cost per meter square of canvas.

**What is known?**

Height of the cone and its base radius.

**What is unknown?**

**i.** Slant height of the tent.

**Steps:**

Slant height \(l = \sqrt {{r^2} + {h^2}} \)

Radius \((r) = 24\rm\, m\)

Height \((h) = 10\rm \, m\)

\[\begin{align}l &= \sqrt {{r^2} + {h^2}} \\ l &= \sqrt {{{(24)}^2} + {{(10)}^2}} \\ & = \sqrt {576 + 100} \\ &= \sqrt {676} \end{align}\]

Slant height of the conical tent \(= 26\, \rm m\)

**ii.** Cost of canvas required to make if \(1 \rm{m^2} \) canvas is \(\rm Rs\, 70.\)

Canvas required to make the tent is equal to the curved surface area of the cone.

Curved surface area of the cone \( = \pi rl \)

Radius \((r) = 24\rm\, m\)

Slant height \(\begin{align}(l) = 26\,\, \rm m \end{align}\)

\[\begin{align}CSA = \frac{{22}}{7} \times 24 \times 26\,\, \rm {m^2} \end{align}\]

\(\begin{align}\therefore \end{align}\) Cost of \(\begin{align}\frac{{22}}{7} \times 24 \times 26\,\rm {m^2} \end{align}\) canvas \(\begin{align} = \frac{{22}}{7} \times 24 \times 26 \times 70 = \rm Rs\,\,137280 \end{align}\)

Cost of the canvas required to make the tent \(= \rm Rs. 137280\)

**Answer:**

(i) Slant height of the tent is \(26\,\rm m.\)

(ii) The cost of the canvas is \(\rm Rs. 137280\)