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

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Chapter 13 Ex.13.2 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 \(=\) \(2\, \rm{cm}.\)

Chapter 13 Ex.13.2 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.

Chapter 13 Ex.13.2 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 \(\begin{align} = 2038.08\,\,\rm{c{m^2}} \end{align}\)

Chapter 13 Ex.13.2 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 \(=1584 \rm{m^{2}}\)

Chapter 13 Ex.13.2 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 \(\rm{Rs}\, 68.75\)

Chapter 13 Ex.13.2 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 \(\begin{align} = 1\,\rm{m} \end{align}\).

 

Chapter 13 Ex.13.2 Question 7

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

  1. Its inner curved surface area
  2. 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?

  1.  The inner diameter and the depth of the well.
  2.  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

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

Chapter 13 Ex.13.2 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 \(\begin{align} = \frac{{22}}{5} = 4.4\,\,\rm{{m^2}} \end{align}\)

Chapter 13 Ex.13.2 Question 9

Find

  1. 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.
  2. 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 \(\begin{align} = 95.04\,\rm{{m^2}} \end{align}\)

Chapter 13 Ex.13.2 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 \(\begin{align} = 2200\,\,\rm{{cm^2}} \end{align}\)

Chapter 13 Ex.13.2 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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