NCERT Solutions Class 8 Maths Chapter 9 Mensuration Exercise 9.2

In Chapter 9 of Class 8 Maths, Mensuration Exercise 9.2 focuses on finding the surface area of cylinders, which is an important concept for solving real-world problems involving round objects like pipes, cans, and tubes. This NCERT Solutions Class 8 Maths Chapter 9 exercise helps students understand how curved and flat surfaces are measured together in such solids.

These NCERT Solutions provide step-by-step answers based on the official NCERT textbook, making it easier for students to grasp the concept and improve their problem-solving skills. With these solutions, students can practice better and feel confident in their understanding of measurement topics.

1.0Download NCERT Solutions Class 8 Maths Chapter 9 Mensuration Exercise 9.2: Free PDF

Download the free NCERT Solutions Class 8 Maths Chapter 9 Mensuration Exercise 9.2 PDF with clear, step-by-step answers to help you understand and solve each question easily.

2.0Key Concepts in Exercise 9.2 of Class 8 Maths Chapter 9

Exercise 9.2 of Chapter 9 – Mensuration focuses on calculating the surface area of a cube, cuboid, and cylinder, helping students understand how to apply geometry in real-world 3D shapes.

Key Concepts Covered in Exercise 9.2:

• Surface Area of a Cube
  A cube has six equal square faces. The formula for total surface area is:
  $\text{Surface Area of Cube}=6a^2$
  where aa is the length of a side.
• Surface Area of a Cuboid
  A cuboid has rectangular faces. The total surface area is calculated using:
  Surface Area of Cuboid=2(lb+bh+hl)
  where ll, bb, and hh are length, breadth, and height.
• Surface Area of a Cylinder
  A cylinder has two circular ends and one curved surface. Its total surface area is:
  Surface Area of Cylinder=2πr(h+r)
• where r is the radius and h is the height.
• Curved Surface Area (CSA)
  Students also learn to find just the side surface (excluding the top and bottom):
  CSA of Cylinder=2πrh
• Understanding Units
  Emphasis is placed on proper use of units like cm², m², etc., and converting when necessary.
• Real-Life Applications
  Students solve problems related to painting walls, wrapping gifts, and other practical uses of surface area.

3.0NCERT Class 8 Maths Chapter 9: Other Exercises

4.0NCERT Class 8 Maths Chapter 9 Exercise 9.2: Detailed Solutions

1. There are two cuboidal boxes as shown in the figure below. Which box requires the lesser amount of material to make?
   
   60 cm
   
   50 cm (b) Sol. Total surface area of first box $=2(\ell \mathrm{b}+\mathrm{bh}+\ell \mathrm{h})$ $=2(60\times 40+40\times 50+60\times 50){\mathrm{cm}}^2$ $=200(24+20+30){\mathrm{cm}}^2$ $=200\times 74{\mathrm{cm}}^2=14800{\mathrm{cm}}^2$ Total surface area of second box $=6(\text{ Edge }{)}^2=6\times 50\times 50{\mathrm{cm}}^2$ $=15000{\mathrm{cm}}^2$ Since the total surface area of first box is less than that of the second, therefore the first box i.e., (a) requires the least amount of material to make.
2. A suitcase of measures $80\mathrm{cm}\times 48\mathrm{cm}\times$ 24 cm is to be covered with a tarpaulin cloth. How many metres of tarpaulin of width 96 cm is required to cover 100 such suitcases? Sol. Total surface area of suitcase $=2[(80)(48)+(48)(24)+(24)(80)]$ $=2[3840+1152+1920]=13824{\mathrm{cm}}^2$ Total surface area of 100 suitcases $=(13824\times 100){\mathrm{cm}}^2=1382400{\mathrm{cm}}^2$ Required tarpaulin $=$ Length $\times$ Breadth $1382400{\mathrm{cm}}^2=$ Length $\times 96\mathrm{cm}$ Length $=\left(\frac{1382400}{96}\right)\mathrm{cm}=14400\mathrm{cm}$ $=144\mathrm{m}$ Thus, 144 m of tarpaulin is required to cover 100 suitcases.
3. Find the side of a cube whose surface area is $600{\mathrm{cm}}^2$. Sol. Let a be the side of the cube having surface area $600{\mathrm{cm}}^2$. $\therefore 6a^2=600\Rightarrow a^2=100\Rightarrow a=10$ Hence, the side of the cube $=10\mathrm{cm}$.
4. Rukhsar painted the outside of the cabinet of measure $1\mathrm{m}\times 2\mathrm{m}\times 1.5\mathrm{m}$. How much surface area did she cover if she painted all except the bottom of the cabinet.
   
   Sol. Here $\ell =2\mathrm{m},\mathrm{b}=1\mathrm{m}$ and $\mathrm{h}=1.5\mathrm{m}$ Area to be painted $=2\mathrm{bh}+2\ell \mathrm{h}+\ell \mathrm{b}$ $=(2\times 1\times 1.5+2\times 2\times 1.5+2\times 1){\mathrm{m}}^2$ $=(3+6+2){\mathrm{m}}^2=11{\mathrm{m}}^2$
5. Daniel is painting the walls and ceiling of a cuboidal hall with length, breadth and height of $15\mathrm{m},10\mathrm{m}$ and 7 m respectively. From each can of paint $100{\mathrm{m}}^2$ of area is painted. How many cans of paint will she need to paint the room? Sol. Here $\ell =15\mathrm{m},\mathrm{b}=10\mathrm{m}$ and $\mathrm{h}=7\mathrm{m}$ Area to be painted $=2\mathrm{bh}+2\ell \mathrm{h}+\ell \mathrm{b}$ $=(2\times 10\times 7+2\times 15\times 7+15\times 10){\mathrm{m}}^2$ $=(140+210+150){\mathrm{m}}^2=500{\mathrm{m}}^2$ Since each can of paint covers $100{\mathrm{m}}^2$, therefore number of cans required $=\frac{500}{100}=5$.
6. Describe how the two figures given below are alike and how they are different. Which box has larger lateral surface area?
   
   Sol. Similarity between the figures is that both have the same heights. The difference between the two figures is that one is a cylinder and the other is a cube. Lateral surface area of cube $=4(7{)}^2$ $=196{\mathrm{cm}}^2$ Lateral surface area of cylinder $=2\times \frac{22}{7}\times \frac{7}{2}\times 7$ $=154{\mathrm{cm}}^2$ Hence, the cube has larger lateral surface area.
7. A closed cylindrical tank of radius 7 m and height 3 m is made from a sheet of metal. How much sheet of metal is required ? Sol. Here, $r=7\mathrm{m}$ and $\mathrm{h}=3\mathrm{m}$. Sheet of metal required to make a closed cylinder $$\begin{gathered} =\text{ Total surface area of the cylinder } \\ =\left(2\pi r\mathrm{rh}+2\pi {\mathrm{r}}^2\right)\text{ sq. units. } \\ =\left(2\times \frac{22}{7}\times 7\times 3+2\times \frac{22}{7}\times 7\times 7\right){\mathrm{m}}^2 \\ =(132+308){\mathrm{m}}^2=440{\mathrm{m}}^2 \end{gathered}$$
8. The lateral surface area of a hollow cylinder is $4224{\mathrm{cm}}^2$. It is cut along its height and formed a rectangular sheet of width 33 cm . Find the perimeter of rectangular sheet. Sol. A hollow cylinder is cut along its height to form a rectangular sheet. Area of cylinder = Area of rectangular sheet $4224{\mathrm{cm}}^2=33\mathrm{cm}\times$ Length Length $=\frac{4224{\mathrm{cm}}^2}{33\mathrm{cm}}=128\mathrm{cm}$ Thus, the length of the rectangular sheet is 128 cm . Perimeter of the rectangular sheet = 2 (Length + Width) $=[2(128+33)]\mathrm{cm}$ $=(2\times 161)\mathrm{cm}$ $=322\mathrm{cm}$
9. A road roller takes 750 complete revolutions to move once over to level a road. Find the area of the road if the diameter of a road roller is 84 cm and length is 1 m .
   
   Sol. In one revolution, the roller will cover an area equal to its lateral surface area. Thus, in 1 revolution, area of the road covered $=2\pi$ rh $=2\times \frac{22}{7}\times 42\mathrm{cm}\times 1\mathrm{m}$ $=2\times \frac{22}{7}\times \frac{42}{100}\mathrm{m}\times 1\mathrm{m}=\frac{264}{100}{\mathrm{m}}^2$ In 750 revolutions, area of the road covered $=\left(750\times \frac{264}{100}\right){\mathrm{m}}^2=1980{\mathrm{m}}^2$
10. A company packages its milk powder in cylindrical container whose base has a diameter of 14 cm and height 20 cm . Company places a label around the surface of the container (as shown in the figure). If the label is placed 2 cm from top and bottom, what is the surface area of the label.
    
    Sol. Since the company places a label around the surface of the cylindrical container of radius 7 cm and height 20 cm such that it is placed 2 cm from top and bottom. $\therefore$ We have to find the curved surface area of a cylinder of radius 7 cm and height (20-4) cm i.e., 16 cm . This curved surface area $=\left(2\times \frac{22}{7}\times 7\times 16\right){\mathrm{cm}}^2$ $=704{\mathrm{cm}}^2$

5.0Key Features and Benefits of Class 8 Maths Chapter 9 Exercise 9.2

• Surface Area of Cylinders: Students learn how to use simple formulas to get the curved and total surface area of cylinders in Exercise 9.2.
• Real-World Applications: Students gain problem-solving skills pertaining to everyday objects such as drums, pipes, and cans.
• Step-by-Step Practice: Students can gradually grasp each subject by answering well-structured questions in this practice.
• Enhances Formula Application: Students who practice frequently are better able to retain and confidently apply surface area formulas.
• Aligned with NCERT Syllabus: All questions and methods follow the latest NCERT textbook, making this perfect for exam preparation and daily revision.