NCERT Solutions for Class 8 Maths Chapter 11: Mensuration - Exercise 11.3

Exercise 11.3

1. There are two cuboidal boxes as shown in the adjoining figure. Which box requires the lesser amount of material to make.

Ans: When it is asked about the material required to make any object, we calculate the total surface area of the object. The formula of total surface area of the cuboid of length $l$, breadth $b$, and height $h$, is $T = 2\left( {lb + bh + hl} \right)$.

Let the box with dimensions 60 cm, 40 cm, and 50 cm be the box 1. Let the total surface area of box (a) be ${T_1}$.

$ \Rightarrow {T_1} = 2\left( {60 \times 40 + 40 \times 50 + 50 \times 60} \right) $

$ \Rightarrow {T_1} = 2\left( {2400 + 2000 + 3000} \right) $

$\Rightarrow {T_1} = 2 \times 7400 $

 $\Rightarrow {T_1} = 14800\,\,{\text{c}}{{\text{m}}^2} $ 

Let the box with dimensions 50 cm, 50 cm, and 50 cm be the box 2. Let the total surface area of box (b) be ${T_2}$. Observe that this cuboid has sides of equal length, thus, we can say it is a cube. Then, ${T_2} = 6{\left( {{\text{side}}} \right)^2}$.

$\Rightarrow {T_2} = 6 \times {\left( {{\text{50}}} \right)^2} $

$\Rightarrow {T_2} = 6 \times {\text{2500}} $

$ \Rightarrow {T_2} = 150{\text{00}}\,{\text{c}}{{\text{m}}^2} $ 

Hence, ${T_1} < {T_2}$. So, the cuboidal box will require a lesser amount of material because its surface area is lesser than the surface of the cube.

2. A suitcase with measures $80\,{\text{cm}} \times 48\,{\text{cm}} \times 24\,{\text{cm}}$ is to be covered with a tarpaulin cloth. How many metres of tarpaulin of 96 cm is required to cover 100 such suitcases?

Ans: The suitcase has a cuboidal shape. The formula of total surface area of the cuboid of length $l$, breadth $b$, and height $h$, is $T = 2\left( {lb + bh + hl} \right)$.

Let the total surface area of the suitcase be $T$.

$\Rightarrow {T_1} = 2\left( {80 \times 48 + 48 \times 24 + 24 \times 80} \right) $

$\Rightarrow {T_1} = 2\left( {3840 + 1152 + 1920} \right) $

 $\Rightarrow {T_1} = 2 \times 6912 $

 $\Rightarrow {T_1} = 13824\,\,{\text{c}}{{\text{m}}^2} $ 

The total area of 100 suitcases will be, 

$\Rightarrow 100 \times {T_1} = 100 \times 13824 $

$\Rightarrow 100 \times {T_1} = 1382400\,\,{\text{c}}{{\text{m}}^2} $ 

The tarpaulin will be rectangular in shape. Equate the area of tarpaulin with the total surface area of 100 suitcase to get the length of tarpaulin required.

Let the length of the tarpaulin be $l$.

$ \Rightarrow {\text{required tarpaulin}} = {\text{length}} \times {\text{breadth}} $

$ \Rightarrow 1382400 = l \times 96$

On dividing both sides by 96, we get,

$ \Rightarrow \dfrac{{1382400}}{{96}} = l $

$\Rightarrow 14400 = l $ 

The required length of the tarpaulin is 14400 cm.

Convert 14400 cm into meters. To convert 14400 cm into meters, divide by 100.

$ \Rightarrow \dfrac{{14400}}{{100}} = 144{\text{ m}}$

Hence, 144 meters of tarpaulin is required to cover 100 suitcases.

3. Find the side of a cube whose surface area is 600 square centimeter.

Ans: The surface area of the cube is $T = 6{a^2}$. 

For the given question, $T$ is 600.

$ \Rightarrow 600 = 6{a^2}$

Divide both sides by 6 and take the square root.

$\Rightarrow \dfrac{{600}}{6} = {a^2} $

$\Rightarrow 100 = {a^2} $

$\Rightarrow \sqrt {100}  = \sqrt {{a^2}}  $

$\Rightarrow  \pm 10 = a $ 

Since the length cannot be negative, then, only the positive value will be considered. Thus, $a = 10$.

The side of the square is 10 cm.

4. Rukhshar painted the outside of the cabinet of measure $1\,{\text{m}} \times 2\,{\text{m}} \times 1.5\,{\text{m}}$. How much surface area did she cover if she painted all except the bottom of the cabinet?

Ans: The length, $l$ of the cabinet is 2 m. The breadth,$b$ of the cabinet is 1 m. The height, $h$ of the cabinet is $1.5$ m.

The area of the cabinet that is to be painted will be the lateral surface area added to the area of the top.

Thus, $A = 2 \times h \times \left( {l + b} \right) + l \times b$.

Substitute $l$ as 2, $b$ as 2 m, $h$ as $1.5$ and simplify.

$\Rightarrow A = 2 \times 1.5 \times \left( {2 + 1} \right) + 2 \times 1 $

$\Rightarrow A = 2 \times 1.5 \times \left( 3 \right) + 2 $

$\Rightarrow A = 9 + 2 $

$\Rightarrow A = 11\,{{\text{m}}^2} $ 

Therefore, the total area to be painted is 11 square meter.

5. Daniel is painting the walls and ceiling of a cuboidal hall with length, breadth, and height of 15 m, 10 m, and 7 m respectively. From each can of paint 100 ${{\text{m}}^{\text{2}}}$ of area is painted. How many cans of paint will she need to paint the room?

Ans: The length, $l$ of the wall is 15 m. The breadth,$b$ of the cabinet is 10 m. The height, $h$ of the cabinet is 7 m.

The area of the hall will be the sum of the area of the wall and area of the ceiling.

Thus, $A = 2 \times h \times \left( {l + b} \right) + l \times b$.

$\Rightarrow A = 2 \times 7 \times \left( {15 + 10} \right) + 15 \times 10 $

$\Rightarrow A = 14 \times \left( {25} \right) + 150 $

$\Rightarrow A = 350 + 150 $

$\Rightarrow A = 500\,{{\text{m}}^2} $ 

The surface area that is needed to be painted from each can is 100 ${{\text{m}}^{\text{2}}}$.

Let the number required to paint $500$${{\text{m}}^{\text{2}}}$ be $n$.

$\Rightarrow n = \dfrac{{{\text{total area to be painted}}}}{{{\text{area needs to be painted of each can}}}} $

$ \Rightarrow n = \dfrac{{{\text{500}}}}{{100}} $

$ \Rightarrow n = 5 $ 

Hence, to paint the walls and to paint the ceiling of the cuboidal hall, 5 cans are required.

6. Describe how the two figures below are alike and how they are different. Which box has larger lateral surface area?

Ans: The given figures have the same height.

The two figures have a difference which is that one figure is a cylinder and the other figure is a cube.

The formula of lateral surface area of the cube is $A = 4{a^2}$. For the given cube, $a$ is 7 cm. 

$\Rightarrow {A_{{\text{cube}}}} = 4 \times {\left( 7 \right)^2} $

$\Rightarrow {A_{{\text{cube}}}} = 4 \times 49 $

$ \Rightarrow {A_{{\text{cube}}}} = 196\,{\text{c}}{{\text{m}}^2} $ 

The formula of lateral surface area of the cylinder is $A = 2\pi rh$. Substitute $h$ as 7, $r$ as 7 in the formula.

$ \Rightarrow {A_{{\text{cylinder}}}} = 2 \times \dfrac{{22}}{7} \times \dfrac{7}{2} \times 7 $

$\Rightarrow {A_{{\text{cylinder}}}} = 2 \times 11 \times 7 $

 $\Rightarrow {A_{{\text{cylinder}}}} = 154\,{\text{c}}{{\text{m}}^2} $ 

Hence, ${A_{{\text{cylinder}}}} < {A_{{\text{cube}}}}$. 

Therefore, the cube has a 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?

Ans: The formula of total surface area of the cylinder is $T = 2\pi r\left( {r + h} \right)$. Substitute $r$ as 7, and $h$ as 3 in the formula . Take $\pi  = \dfrac{{22}}{7}$.

$\Rightarrow T = 2 \times \dfrac{{22}}{7} \times 7\left( {7 + 3} \right) $

$\Rightarrow T = 2 \times \dfrac{{22}}{7} \times 7 \times 10 $

$\Rightarrow T = 2 \times 22 \times 10 $

$\Rightarrow T = 440\,{{\text{m}}^2} $ 

Hence, $440\,{{\text{m}}^2}$ of sheet of metal is required to make the cylindrical tank.

8. The lateral surface area of the hollow cylinder is 4224 square centimeter. It is cut along its height and formed a rectangular sheet of width 33 cm. Find the perimeter of the rectangular sheet.

Ans: A hollow cylinder is cut along its height to form a rectangular sheet whose width is 33 cm.The area of the cylinder will be equal to the area of the rectangular sheet. Assume the length of the sheet is $l$.

$\Rightarrow 4224 = 33 \times l$

Divide both sides by 33.

$\Rightarrow \dfrac{{4224}}{{33}} = l $

$\Rightarrow 128\,{\text{cm}} = l $ 

The rectangular sheet has a length of 128 cm.

The perimeter of the rectangular sheet will be $2\left( {l + b} \right)$.

Substitute $l$ as 128, and $b$ as 33.

$\Rightarrow {\text{perimeter}} = 2\left( {128 + 33} \right) $

$\Rightarrow {\text{perimeter}} = 2\left( {161} \right) $

$ \Rightarrow {\text{perimeter}} = 322{\text{cm}} $ 

Hence, the perimeter of the sheet is 322 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.

Ans: The distance covered by the roller in 1 revolution will be equal to its lateral surface area. The formula of lateral surface area of the cylinder is $A = 2\pi rh$.

The diameter is 84 cm. The radius will be $\dfrac{{84}}{2} = 42\,{\text{cm}}$.

The SI units of the road roller are different. Convert 42 cm into meters.

$ \Rightarrow 42\,{\text{cm = }}\dfrac{{42}}{{100}}{\text{m}}$

Let the area covered by the road roller in 1 revolution be $L$.

$ \Rightarrow L = 2 \times \dfrac{{22}}{7} \times \dfrac{{42}}{{100}} \times 1 $

 $  \Rightarrow L = 2 \times 22 \times \dfrac{6}{{100}} \times 1 $

 $  \Rightarrow L = \dfrac{{264}}{{100}}{{\text{m}}^2} $ 

Now, find the distance covered in 750 revolutions by multiplying the lateral surface area by 750.

$\Rightarrow L' = 750 \times \dfrac{{264}}{{100}}{{\text{m}}^2} $

$\Rightarrow L' = 1980{{\text{m}}^2} $ 

Hence, the road has an area of 1980 square centimeter.

10. A company packages its milk powder in a cylindrical container whose base has a diameter of 14cm and height 20cm. 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 area of the label.

Ans: The height of the label will be the difference of 20 cm and 2 cm.

$\Rightarrow 20 - 2 - 2 = 20 - 4 $

$= 16\,\,{\text{cm}} $ 

The diameter of the label is 14 cm.

The radius is half of the diameter. So,

$\Rightarrow r = \dfrac{{14}}{2} $

 $\Rightarrow r = 7{\text{cm}} $ 

The area of the label is of the form of a cylinder. The radius is 7 cm and height 16 cm.

$A = 2 \times \pi  \times 7 \times 16 $

$  A = 2 \times \dfrac{{22}}{7} \times 7 \times 16 $

$  A = 2 \times 22 \times 16 $

$  A = 704\,{\text{c}}{{\text{m}}^2} $ 

The area of the label is 704 square cm.

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration (Ex 11.3) Exercise 11.3

Important Topics Covered in Exercise 11.3 of Class 8 Maths NCERT Solutions 

NCERT Solutions Class 8 Maths Exercise 11.3 are mainly based on the concepts of surface areas of solids or 3D shapes like cubes, cuboids, and cylinders. The total surface area is defined as the sum of areas of all the faces of a solid. Surface area formulas are derived from the plane figures. So, students should have an in-depth knowledge of the shapes studied in the earlier exercises to solve these problems. 

The key skill that students need to have to solve the sums given in this exercise is that of visualising shapes. For example, a sum might ask a student to cover only the four faces of a cube-shaped room without including the roof and floor, so students have to modify the formula to get the desired result. This exercise consists of questions on how to solve such problems. 

Below are some important formulas discussed in this exercise.

1. Total surface area of a cuboid = 2(lb + bh + hl). A cuboid is made up of six faces that are rectangular in shape. As the area of a rectangle is length * breadth, this concept is used to derive the surface area formula. So, first, we need to find out the area of each rectangular plate and then sum it up to get the desired result.

2. Total surface area of a cube = 6a2. Here, ‘a’ is the length of the side. A cube consists of six squares. So, first, we need to find out the area of each square and then sum it up six times to get the desired result.

3. Total surface area of a cylinder = 2πr (r + h). The formula for the curved surface area of a cylinder is 2πrh (multiplying height by the perimeter of a circle). It is also enclosed by 2 circles on the top and bottom given by πr2. We need to sum this up to get the desired formula.

Opting for the NCERT solutions for Ex 11.3 Class 8 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 11.3 Class 8 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Chapter wise NCERT Solutions for Class 8 Maths

• Chapter 1 - Rational Numbers

• Chapter 2 - Linear Equations in One Variable

• Chapter 3 - Understanding Quadrilaterals

• Chapter 4 - Practical Geometry

• Chapter 5 - Data Handling

• Chapter 6 - Squares and Square Roots

• Chapter 7 - Cubes and Cube Roots

• Chapter 8 - Comparing Quantities

• Chapter 9 - Algebraic Expressions and Identities

• Chapter 10 - Visualising Solid Shapes

• Chapter 11 - Mensuration

• Chapter 12 - Exponents and Powers

• Chapter 13 - Direct and Inverse Proportions

• Chapter 14 - Factorisation

• Chapter 15 - Introduction to Graphs

• Chapter 16 - Playing with Numbers

Class 8 Maths Chapter 11 Includes:

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 8 students who are thorough with all the concepts from the Subject Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 8 Maths Chapter 11 Exercise 11.3 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 8 Maths Chapter 11 Exercise 11.3, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it. 

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