# Exercise E13.2 Surface Areas and Volumes NCERT Solutions Class 9

## Question 1

The curved surface area of a right circular cylinder of height \(14\,\rm{ cm}\) is \(88\,\rm{cm^2}\) . Find the diameter of the base of the cylinder.

### Solution

**Video Solution**

**Reasoning:**

Curved surface area \(\begin{align} = 2\pi rh \end{align}\)

**What is Known?**

Curved surface area of the cylinder and its height.

**What is Unknown?**

Diameter of the base of the cylinder.

**Steps:**

Curved surface area of a cylinder

\(\begin{align} = 2\pi rh = 88\,\rm{cm^2}\,\left( \rm{given} \right) \end{align}\)

Height of the cylinder \(=14\,\rm{ cm}\)

\[\begin{align}2 \times \frac{{22}}{7} \times r \times 14 &= 88\\r = \frac{{88 \times 7}}{{2 \times 22 \times 14}} &= 1\,\rm{cm}\\2 \times \rm{radius} &= \rm{diameter}\\2 \times 1 &= 2\,\rm{cm} \end{align}\]

The diameter of the base of the cylinder \(=\) \(2\, \rm{cm}.\)

## Question 2

It is required to make a closed cylindrical tank of height \(1 \,\rm{m}\) and base diameter \(140 \,\rm{cm}\) from a metal sheet. How many square metres of the sheet are required for the same?

### Solution

**Video Solution**

**Reasoning:**

The curved surface area of a right circular cylinder of base \(r\) and height \(h\) is and its total surface area is \(\begin{align}2\pi r(r + h) \end{align}\)

**What is the known?**

Height and diameter of the cylindrical tank.

**What is the unknown?**

Square metres of the sheet required for making the cylindrical tank.

**Steps:**

Height of the tank \((h)\) = \(1 \,\rm{m}\) = \(100 \,\rm{cm}\)

Diameter \(= 2r =\) \(140 \,\rm{cm}\)

\(\begin{align}r = \frac{{140}}{2} = 70\,\,\rm{cm} \end{align}\)

Total surface area of the closed cylindrical tank

\[\begin{align} &= 2\pi r\,(r + h)\\&= 2 \times \frac{{22}}{7} \times 70(70 + 100)\\ &= 74800\,\,\rm{c{m^2}} \end{align}\]

Since,

\[\begin{align}100 cm &= \rm{1 m}\\\rm{100^2}c{m^2} &= \rm1\,\,{m^2}\\\rm74800\,\,c{m^2} &= \frac{{748000}}{{100 \times 100}}\\ & = \rm7.48\,\,{m^2} \end{align}\]

\(\begin{align}7.48\,{\rm{m^2}} \end{align}\) of the sheet is required for making the cylindrical tank.

## Question 3

A metal pipe is \(77\,\rm{ cm}\) long. The inner diameter of a cross section is \(4\,\rm{ cm},\) the outer diameter being \(4.4\,\rm{ cm}\) . Find it

(i) Inner curved surface area

(ii) Outer curved surface area

(iii) Total surface area.

### Solution

**Video Solution**

**Reasoning:**

The curved surface area of a right circular cylinder of base radius \(r\) and height is \(\rm 2 \pi rh\) . And its total surface area is \(\begin{align}2\pi r(r + h) \end{align}\).

**What is the known?**

Length of the metal pipe, the inner and outer diameter of the cross sector of the metal pipe.

(i) Inner curved surface area

**Steps:**

\[\begin{align}{\rm{height}}(h) &= 77\,\,\rm{cm}\\{\rm{diameter} }&= 2r = 4\,\,\rm{cm}\\r &= 2\,\,\rm{cm} \end{align}\]

Inner curved surface area

\[\begin{align}&= 2\pi rh\\ &= 2 \times \frac{{22}}{7} \times 2 \times 77\\ &= 968\,\,\rm{c{m^2} } \end{align}\]

(ii) Outer curved surface area

**Steps:**

\[\begin{align}{\rm{{height}}} (h) &= 77\,\,\rm{cm}\\\rm{diameter} &= 2R = 4.4\,\,\rm{cm}\\R &= 2.2\,\,\rm{cm} \end{align}\]

Inner curved surface area

\[\begin{align}&= 2\pi Rh\\ &= 2 \times \frac{{22}}{7} \times 2.2 \times 77\\ &= 1064.8\,\,\rm{c{m^2}} \end{align}\]

(iii) Total surface area

**Steps:**

Total surface area of pipe = Curved Surface Area of inner surface + Curved Surface Area of outer surface + Area of both the circular ends of the pipe

Total surface area

\[\begin{align} &= \,\,2\pi Rh + 2\pi rh + 2\pi ({R^2} - {r^2}]\\ &= 1064.8 \!+\! 968 \!+\! 2\! \times \!\frac{{22}}{7} \!\times\! [{(2.2)^2}\! - \!{(2)^2}]\\ &= 1064.8 + 968 + 5.28\\ &= 2038.08\,\,\,\rm{c{m^2}} \end{align}\]

Inner curved surface area \(\begin{align} = 968\,\,\rm{c{m^2}} \end{align}\)

Outer curved surface area \(\begin{align} = 1064.8\,\,\rm{c{m^2}} \end{align}\)

Total surface area \(\begin{align} = 2038.08\,\,\rm{c{m^2}} \end{align}\)

## Question 4

The diameter of a roller is \(84\,\rm{ cm}\) and its length is \(120\,\rm{ cm}.\). It takes \(500\) complete revolutions to move once over to level a playground. Find the area of the playground in \(\begin{align}\rm{{m^2}} \end{align}\).

### Solution

**Video Solution**

**Reasoning:**

The roller is cylindrical in shape and hence it is considered as a right circular cylinder.

The curved surface area of a right circular cylinder of base radius \(r\) and height \(h\) is

\(\begin{align}2\pi rh \end{align}\)

**What is the known?**

Radius and height of the cylinder.

**What is the unknown?**

Area of the playground in\(\begin{align}\rm{{m^2}}. \end{align}\)

**Steps:**

Curved surface area of the cylinder = \(\begin{align}2\pi rh \end{align}\)

\(\begin{align}{\rm{diameter}} ={ 2r} &= {84\,\,\rm{cm}}\\r &= {42\,\,\rm{cm}}\\h &={ 120\,\,\,\rm{cm}}\\{} \end{align}\)

Area of the playground leveled in taking \(1\) complete revolution = The curved surface area of the cylinder having \( r =\) \(42\,\rm{ cm}\) and \( h =\) \(120\,\rm{ cm}.\)

\[\begin{align} &= 2\pi rh\\ &= 2 \times \frac{{22}}{7} \times 42 \times 120\\ &= 31680\,\,\rm{c{m^2}} \end{align}\]

Area of the playground = Area levelled by the cylinder in 500 revolutions.

\[\begin{align} &= 500 \times 31680\\ &= 15840000\,\rm{c{m^2}}\\\ \rm\end{align}\]

Since,

\(\begin{align}&100\,\rm{cm} = 1\,m\\&{100^2}\rm{c{m^2}} = 1\,\rm{{m^2}}\end{align}\)

\[\begin{align} &= \frac{{15840000}}{{100 \times 100}}\\ &= 1584\,\rm{{m^2}} \end{align}\]

Area of the play ground \(=1584 \rm{m^{2}}\)

## Question 5

A cylindrical pillar is \(50\,\rm{ cm}\) in diameter and \(3.5\,\rm{ m}\) in height. Find the cost of painting the curved surface of the pillar at the rate of \(\rm{Rs}\, 12.50\) per \(\begin{align}{m^2} \end{align}\).

### Solution

**Video Solution**

**Reasoning:**

The curved surface area of a right circular cylinder of base radius \(r\) and height \(h\) is \(\begin{align}2\pi rh \end{align}\). So, the cost of painting curved surface area will be the product of curved surface are and cost of painting per meter square.

**What is the known?**

Height and diameter of the pillar.

**What is the unknown?**

Cost of painting the curved surface of the pillar.

**Steps:**

Diameter = \(2r\) = \(50\,\rm{ cm}\)

\(r\) = \(25\,\rm{ cm}\) = \(0.25\,\rm{ cm}\)

Height = \(h\) = \(3.5\,\rm{ m}\)

Curved surface area

\[\begin{align}&= {2\pi rh }\\ &= 2 \times \frac{{22}}{7} \times 0.25 \times 3.5\\ &= 5.5\,\,\rm{{m^2}} \end{align}\]

Cost of painting the curved surface area per

\(\begin{align}\rm{{m^2}} \end{align}\)= \(\rm{Rs}\, 12.50.\)

Cost of painting

\(\begin{align}\,5.5\,\,\rm{{m^2}} = 12.50 \times 5.5\,\, = Rs\,\,68.75 \end{align}\)

Cost of painting the curved surface of the pillar is \(\rm{Rs}\, 68.75\)

## Question 6

Curved surface area of a right circular cylinder is \(4.4\) \(\begin{align}\rm{{m^2}}. \end{align}\). If the radius of the base of the cylinder is \(0.7 \,\rm{m},\) find its height.

### Solution

**Video Solution**

**Reasoning:**

The curved surface area of a right circular cylinder of the base radius \(r\) and height \(h\) is \(\begin{align}2\pi rh. \end{align}\)

**What is the known?**

Curve surface area and radius of the base.

**What is the unknown?**

Height of the cylinder.

**Steps:**

Curved surface area \(\begin{align}2\pi rh \end{align}\)\(=\) \(4.4\)\(\begin{align}\rm{{m^2}} \end{align}\)

\[\begin{align}\rm radius(r) &= 0.7m\\\rm height(h) &= ? \end{align}\]

\[\begin{align}2\pi rh &= 4.4\\2 \times \frac{{22}}{7} \times 0.7 \times h &= 4.4\\h &= \frac{{4.4 \times 7}}{{2 \times 22 \times 0.7}}\\\,\,\,\,\, &= 1\,\,\rm{m} \end{align}\]

The height of the right circular cylinder is \(\begin{align} = 1\,\rm{m} \end{align}\).

## Question 7

The inner diameter of a circular well is \(3.5\, \rm{m}\,.\) It is \(10 \,\rm{m}\) deep. Find

- Its inner curved surface area
- The cost of plastering this curved surface at the rate of \(40\) per\(\begin{align}\rm{{m^2}} \end{align}\)

### Solution

**Video Solution**

**Reasoning:**

The curved surface area of a right circular cylinder of base radius and height h is \(\begin{align}2\pi rh \end{align}\)

**What is the known?**

- The inner diameter and the depth of the well.
- Cost of plastering the curved surface area per\(\begin{align}\rm{{m^2}} \end{align}\)

1.Its inner curved surface area

**What is the unknown?**

The inner curved surface area.

**Steps:**

Diameter \(= 2r = 3.5 \rm{m}\)

\(\begin{align} r &= \frac{{3.5}}{2}\,\,\rm{m}\\height = h &= 10\,\,\rm{m}\end{align}\)

The Curved surface area

\[\begin{align}&= 2\pi rh\\& = 2 \times \frac{{22}}{7} \times \frac{{3.5}}{2} \times 10\\& = 110\,\,{m^2} \end{align}\]

2. Cost of plastering the curved surface area per \(\begin{align}\rm{{m^2}} \end{align}\).

**What is the unknown?**

Cost of plastering the curved surface area.

**Steps:**

Cost of plastering \(\begin{align}\,1\,\rm{{m^2}} = Rs\,\,40 \end{align}\)

Cost of plastering

\(\begin{align}110\,\,\rm{{m^2}} = 110 \times 40 = Rs\,\,4400 \end{align}\)

Its inner curved surface \(\begin{align} = 110\,\,\rm{{m^2}} \end{align}\)

Cost of plastering the curved surface

\(= \rm{Rs}\, 4400\)

## Question 8

In a hot water heating system, there is a cylindrical pipe of length \(28 \,\rm{m}\) and diameter \(5 \,\rm{cm}\). Find the total radiating surface in the system.

### Solution

**Video Solution**

**Reasoning:**

The curved surface area of a right circular cylinder of base radius \(r\) and height \(h \) is\(\begin{align}2\pi rh \end{align}\).

**What is the known?**

The length and diameter of the cylindrical pipe of a water heater.

**What is the unknown?**

Radiating surface area in the system.

**Steps:**

Diameter \(= 2r = 5\,\rm{cm}\)

\(\begin{align}r = \frac{5}{2}\,cm &= \frac{5}{{2 \times 100}}\,\,m\end{align}\)

height \(= h = 28 \,m\)

Curved surface area

\[\begin{align}&= 2\pi rh\\ &= 2 \times \frac{{22}}{7} \times 28 \times \frac{5}{{2 \times 100}}\\ &=\frac{{22}}{5} \\&= 4.4\,\,\rm{{m^2}} \end{align}\]

The total radiating surface \(\begin{align} = \frac{{22}}{5} = 4.4\,\,\rm{{m^2}} \end{align}\)

## Question 9

Find

- The lateral or curved surface area of a closed cylindrical petrol storage tank that is \(4.2 \,\rm{m}\) in diameter and \(4.5 \,\rm{m}\) high.
- How much steel was actually used, if \(1/12\) of the steel actually used was wasted in making the tank?

### Solution

**Video Solution**

**Reasoning:**

The curved surface area of a right circular cylinder of base radius \(r\) and height \(h\) is \(\begin{align}2\pi rh \end{align}\)

and its total surface area is\(\begin{align}2\pi r(r + h) \end{align}\).

**What is the known?**

The diameter and height of the storage tank.

i. The lateral or curved surface area

**Steps:**

\(\begin{align} \rm Diameter\text{ }=2r &=4.2\,\rm{m} \\ r&=\frac{4.2}{2} \\ \,\,\,\,&=2.1\,\,\rm{m} \\ \rm height=h&=4.5\,\,\rm{m} \end{align}\)

Curved surface area

\[\begin{align}&=\,2\pi rh \\ & =2\times \frac{22}{7}\times 2.1\times 4.5 \\ & =59.4\,\,\rm{{{m}^{2}}} \\ \end{align}\]

ii. Steel actually used

**Steps:**

Total surface area

\[\begin{align} &=2\pi r(r+h) \\ & =2\times \frac{22}{7}\times 2.1(4.5+2.1) \\ &=2\times \frac{22}{7}\times 2.1\times 6.6 \\ &=87.12\,\,\rm{{{m}^{2}}} \\ \end{align}\]

This is the steel used\(\begin{align} = 87.12\rm{{m^2}} \end{align}\), \(\begin{align}\frac{1}{{12}} \end{align}\) of this is wasted.

The area of the steel which has gone into the task\(\begin{align} = 1 - \frac{1}{{12}} = \frac{{11}}{{12}} \end{align}\) of the steel used to make the tank \(= 87.12\)

This means actual area of steel used

\(\begin{align} = \frac{{12}}{{11}} \times 87.12 = 95.04\,\,\rm{{m^2}} \end{align}\)

Curve surface area \(\begin{align} = 59.4\,\rm{{m^2}} \end{align}\)

Steel actually used \(\begin{align} = 95.04\,\rm{{m^2}} \end{align}\)

## Question 10

In the below figure, you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of \(20 \,\rm{}cm\) and height of \(30 \,\rm{}cm.\) A margin of \(2.5 \,\rm{}cm.\) is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade.

### Solution

**Video Solution**

**Reasoning:**

The curved surface area of a right circular cylinder of base radius r and height h is and its total surface area \(\begin{align}2\pi r(r + h) \end{align}\).The amount of cloth required to cover the cylinder will be equal to surface are of cylinder. (Here height will will equal to the height of cylinder plus \(2.5 \,\rm{}cm.\) margin on both sides).

**What is the known?**

The height and diameter of the frame of lampshade. And the margin of \(2.5 \,\rm{}cm.\) for folding.

**What is the unknown?**

Cloth required for covering the lamp shade.

**Steps:**

Diameter \(= 2r = 20\,\rm{}cm\)

\(r = 10 \rm{}\,cm\)

Height of the lamp shade \(= 30\,\rm{} cm + 2.5 \,\rm{}cm\) for both side folding

Total height of the cloth \(= 35 \,\rm{}cm\)

Cloth required is the curved surface area

\[\begin{align}&=2\pi rh\\ &= 2 \times \frac{{22}}{7} \times 10 \times 35\\ &= 2200\,\,\rm{c{m^2}} \end{align}\]

Cloth required \(\begin{align} = 2200\,\,\rm{{cm^2}} \end{align}\)

## Question 11

The students of a Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius \(3 \,\rm{}cm\) and height \(10.5 \,\rm{}cm.\) The Vidyalaya was to supply the competitors with cardboard. If there were \(35\) competitors, how much cardboard was required to be bought for the competition?

### Solution

**Video Solution**

**Reasoning:**

The curved surface area of a right circular cylinder of base radius \(r\) and height \(h\) is \(\begin{align}2\pi rh. \end{align}\)

And total surface area =\(\begin{align}2\pi r(r + h) \end{align}\). So, total amount of cardboard required will be the product of surface area of pen holders and total number of participating students.

**What is the known?**

Radius and height of the card board.

**What is the unknown?**

Card board required to be bought for \(35\) competitors.

**Steps:**

radius(r) \(\begin{align}= 3\,\,\rm{}cm \end{align}\)

height(h) \(\begin{align} = 10.5\,\,\rm{}cm \end{align}\)

Card board required for \(1\) competitor = Total surface area of the cylinder

Since the pen holder is open at the top.

One area of \(\begin{align}\pi {r^2} \end{align}\) should be subtracted.

Cardboard required for \(1\) competitor

\[\begin{align} &= 2\pi rh + \pi {r^2}\\&= 2 \times \frac{{22}}{7} \times 3 \times 10.5 + \frac{{22}}{7} \times 3 \times 3\\ &= 198 + \frac{{198}}{7} = 198(1 + \frac{1}{7})\\ &= \frac{{198 \times 8}}{7}\,c{m^2} \end{align}\]

Cardboard required for \(35\) competitions:

\[\begin{align} = \frac{{198 \times 8}}{7} \times 35 = 7920\,\,\rm{}c{m^2} \end{align}\]

\(\begin{align}7920\,\,\rm{}c{m^2} \end{align}\) of cardboard was required to be bought for the competition.

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