# Exercise E13.3 Surface Areas and Volumes NCERT Solutions Class 9

## Question 1

Diameter of the base of a cone is \(10.5\,\rm cm\) and its slant height is \(10\,\rm cm.\) Find its curved surface area.

### Solution

**Video Solution**

**Reasoning:**

Curved surface area of a right circular cone of base radius *r* and slant height *l* is \(\begin{align}\pi rl. \end{align}\)

**What is known?**

Diameter of the base and slant height.

**What is unknown?**

Curved surface area.

**Steps:**

Diameter \(= 2r = 10.5{\rm\, {cm}}\)

\(\begin{align}\therefore r = \frac{{10.5}}{2}cm \end{align}\)

Slant height \(\begin{align}l = 10\,\,\,cm \end{align}\)

Curved surface area\(\begin{align} = \pi rl \end{align}\)

\[\begin{align} &= \frac{{22}}{7} \times \frac{{10.5}}{2} \times 10\\ &= 165\,\,c{m^2} \end{align}\]

**Answer:**

Curved surface area of the cone \(\begin{align} = 165\,\,c{m^2} \end{align}\)

## Question 2

Find the total surface area of a cone, if its slant height is \(21\,\rm m\) and diameter of its base is \(24 \,\rm m.\)

### Solution

**Video Solution**

**Reasoning:**

The total surface area of the cone is the sum of the curved surface area and area of the base which is a circle.

**What is known?**

Diameter of the box and slant height of the cone.

**What is unknown?**

Total surface area of the cone.

**Steps:**

Total surface area of the cone is

\(= \pi rl + \pi {r^2} = \pi r(l + r) \)

Where \(l\) is slant height and radius \(r.\)

Diameter \(= 2r = 24 \rm\,m\)

\(r = 12{\rm m}\\ \\ l = 21 \rm m\)

**T S A **\(= \pi r(l + r) \)

\(\begin{align} &= \frac{{22}}{7} \times 12 \times (12 + 21)\\ &= \frac{{22}}{7} \times 12 \times 33 \\ &= \frac{{8712}}{7} = 1244.57\,\,{\rm{m}}^2 \end{align}\)

**Answer:**

Total surface area of the cone \(\begin{align} = 1244.57\,{\rm{m}}^2 \end{align}\)

## Question 3

Curved surface area of a cone is \(\begin{align}308\,\,c{m^2} \end{align}\) and its slant height is \(14\rm\, cm.\) Find

**(i)** Radius of the base

**(ii)** Total surface area of the cone.

### Solution

**Video Solution**

**Reasoning:**

The total surface area of the cone is the sum of the curved surface area and its base area which is a circle.

**What is known?**

Curved surface area of the cone and its slant height.

**What is unknown?**

**(i)** Radius of the base.

**Steps:**

Curved surface area \(=\) \(\begin{align}\,\,\pi rl = 308\,\,{\rm{cm}}^2 \end{align}\)

Slant height (\(l\)) \(= 14\,\rm cm\)

\[\begin{align}\pi rl &= 308 \\\frac{{22}}{7} \times r \times 14 &= 308 \\r &= \frac{{308}}{{14}} \times \frac{7}{{22}} = 7\,\,\rm cm\end{align}\]

**(ii)** Total surface area of the cone.

**Steps:**

Total surface area \(\begin{align}\, = \pi r\,(l + r) \end{align}\)

Radius \(= 7 \rm\,cm\)

Slant height (\(l\)) \(= 14\,\rm cm\)

\[\begin{align}\rm TSA &= \pi r(r + l) \\ &= \frac{{22}}{7} \times 7 \times (7 + 14) \\ &= 22 \times 21 \\ & = 462\,\,\rm c{m^2} \end{align}\]

**Answer:**

Radius of the cone \(= 7\,\rm cm\)

Total surface area of the cone \( = 462\, \rm cm^2 \)

## Question 4

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 \(\rm Rs. \,70.\)

### Solution

**Video 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 \(\rm Rs. 137280\)

## Question 5

What length of tarpaulin \(3\,\rm m\) wide will be required to make conical tent of height \(8\,\rm m\) and base radius \(6\,\rm m\)? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately \(20\rm\, cm\) (Use \(\pi = 3.14\))

### Solution

**Video Solution**

**Reasoning:**

Curved surface area of a right circular cone of base radius r and slant height is \(\pi r l \) and \(\begin{align}l = \sqrt {{r^2} + {h^2}} \end{align}\)

**What is known?**

Height and radius of the cone width of the tarpaulin and its margin.

**What is unknown?**

Length of the tarpaulin.

**Steps:**

Radius \((r) = 6\rm\, m\)

Height \((h) = 8 \rm\,m\)

Slant height

\[\begin{align}l = \sqrt {{r^2} + {h^2}} \\\,\,\, = \sqrt {{6^2} + {8^2}} \\\,\,\, = \sqrt {36 + 64} \\\,\,\, = \sqrt {100} \\\,\,\, = 10\,\, \rm m \end{align}\]

Curved surface area\(\begin{align}\, = \pi rl \end{align}\)

\[\begin{align} &= \pi \times 6 \times 10\\ &= 3.14 \times 6 \times 10\\& = 188.4\,\, \rm {m^2} \end{align}\]

Curved surface area \(=\) Area of the tarpaulin

Tarpaulin area \(=\) length \(\times\) width

Width of the tarpaulin \(= 3 \rm\, m\)

\[\begin{align} \text{length $\times$ width } & = 188.4\\ \text{length} \times 3 & = 188.4\\ \text{length} & = \frac{{188.4}}{3}\\ &= 62.8\,\, \rm m \end{align}\]

Extra length of the material

\(\rm= 20\, cm = 0.2\, m\)

Actual length required

\(\rm = 62.8 +0.2 = 63\, m\)

**Answer:**

Length of the tarpaulin required \(= 63\rm\, m\)

## Question 6

The slant height and base diameter of a conical tomb are \(25\rm\, m\) and \(14\rm\, m\) respectively. Find the cost of white washing its curved surface at the rate of \( \rm\,Rs \,210\) per \(100\) \(\rm\,{m^2} \).

### Solution

**Video Solution**

**Reasoning:**

Curved surface area of a right circular cone of \(r\) base radius and slant height is \(\pi rl \).

**What is known?**

Slant height and base diameter rate of white washing \(\rm Rs\, 210\) per \(100\)\(\begin{align}\rm\,{m^2} \end{align}\).

**What is unknown?**

Cost of white washing its curved surface area.

**Steps:**

Slant Height \((l) = 25\rm\, m\)

\[\begin{align}r = \frac{{14}}{2} = 7\rm\,m \end{align}\]

Curved surface area \(\begin{align} = \pi rl \end{align}\)

\[\begin{align} &= \frac{{22}}{7} \times 7 \times 25\\&=550\rm\,\,{m^2} \end{align}\]

Cost of white washing per \(\begin{align}100\rm\,\,{m^2} \end{align}\) \(= 210\)

\[\begin{align}{\rm{For}}\,\,550\rm\,\,{m^2} &= \frac{{210 \times 550}}{{100}}\\ &= {\rm{ Rs }}\,1155 \end{align}\]

**Answer:**

Cost of white washing the conical tomb is \(\rm Rs.\, 1155.\)

## Question 7

A joker’s cap is in the form of a right circular cone of base radius \(7\rm\, cm\) and height \(24\,\rm cm.\) Find the area of the sheet required to make \(10\) such caps.

### Solution

**Video Solution**

**Reasoning:**

Curved surface area of a right circular cone of base radius \(r\) and slant height \(l\) is \( = \pi rl \) and \(l = \sqrt {{r^2} + {h^2}} \)

**What is known?**

**(i)** Caps box radius and height.

**(ii)** Number of caps.

**What is unknown?**

Area of the sheet required to \(10\) caps.

**Steps:**

Base Radius \((2r) = 7\rm\,cm\)

Height \(h = 24\,\rm cm\)

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

\[\begin{align} &= \sqrt {{7^2} + 24} \\& = \sqrt {49 + 576} \\ &= \sqrt {625} \\ &= 25\rm\,\,cm \end{align}\]

Curved surface area of cone \(=\) Surface area of the sheet required for \(1\) cap\(\begin{align}\,\pi rl. \end{align}\)

\[\begin{align} &= \frac{{22}}{7} \times 7 \times 25\\ &= 550\rm\,\,{m^2} \end{align}\]

Curved surface area for \(10\) caps \(\begin{align} = 550 \times 10 = 5500\rm\,\,{m^2} \end{align}\)

**Answer:**

The area of the sheet required to make \(10\) such caps is \(\begin{align}5500\rm\,\,{m^2}. \end{align}\)

## Question 8

A bus stop is barricaded from the remaining part of the road, by using \(50\) hollow cones made of recycled cardboard. Each cone has a base diameter of \(40\rm\, cm\) and height \(1\rm\, m.\) If the outer side of each of the cones is to be painted and the cost of painting is \(\rm Rs\, 12\) per \(\rm m^{2}\) . What will be the cost of painting all these cones?

(Use \(\pi = 3.14\) and take \(\sqrt{1.04}= 1.02\))

### Solution

**Video Solution**

**Reasoning:**

Curved surface area of a right circular cone of base radius \(r\) and slant height l is \( = \pi rl \) and \(l = \sqrt {{r^2} + {h^2}} \)

**What is known?**

Base diameter and height of the cone number of cones and cost per \(\rm\,{m^2}. \)

**What is unknown?**

Cost of painting the \(50\) cones.

**Steps:**

\(\begin{align} \text{Base diameter} &=40 {\rm{cm}}\\2r & =40 {\rm{cm}} \\ r &= \frac{{40}}{2} \\ &= 20\,\,{\rm{cm}}\\ & = 0.2 \, {\rm{m}}\; \begin{bmatrix} \because 100\,{\rm{cm}} \\ = 1\,{\rm{m}}\end{bmatrix} \end{align}\)

Height \( h = 1\rm\, m\)

\[\begin{align}l &= \sqrt {{r^2} + {h^2}} \\ &= \sqrt {{{(0.2)}^2} + {{(1)}^2}} \\ &= \sqrt {1.04} \\ &= 1.02\rm \,m \end{align}\]

Curved surface are \(=\) \(\begin{align}\pi rl \end{align}\)

\[\begin{align} &= \pi \times 0.2 \times 1.02\\ &= 3.14 \times 0.2 \times 1.02\\ &= 0.64056\,\,\rm\,{m^2} \end{align}\]

Curved surface area of \(50\) cones

\[\begin{align} &= 0.64056 \times 50\\ &= 32.028\,\,\rm{m^2} \end{align}\]

Cost of painting per \(\begin{align}\rm{m^2} \end{align}\) \(=\rm Rs\, 12\)

Cost of painting \(32.028\rm\,{m^2}\)

\[\begin{align} &= \rm Rs\,\,32.028 \times 12 \\ &= \rm Rs\,\,\,\,384.34\,\,\rm\,{m^2} \end{align}\]

**Answer:**

Cost of painting all the cones

\(=\rm Rs\, 384.34\)

- Live one on one classroom and doubt clearing
- Practice worksheets in and after class for conceptual clarity
- Personalized curriculum to keep up with school