NCERT Solutions Class 8 Maths Chapter 9 Mensuration

Mensuration is the ninth chapter of Class 8 Maths. It covers various topics, including an introduction, area of polygon, solid shapes, surface area of cube, cuboid and cylinder, volume of cube, cuboid and cylinder, volume and capacity.

The chapter begins with an introduction to mensuration in section 9.1, followed by significant topics in sections 9.2, 9.3, 9.4, and 9.5.

1.0Download NCERT Class 8 Maths Chapter 9 Mensuration Solutions

ALLEN'S Experts lucidly curated the class 8 mensuration solution to improve the students' problem-solving abilities. For a more precise idea about Mensuration NCERT solutions, students can download the below NCERT Solutions class 8 Maths chapter 9 pdf solution.

2.0NCERT Solutions Class 8 Maths Chapter 9 : Important Topics

Introduction: Overview of basic mensuration concepts and their applications.

Area of Polygon: Calculating the area of various polygonal shapes.

Solid Shapes: Understanding 3D figures and their properties.

Surface Area of Cube, Cuboid, and Cylinder: Finding the surface area of these 3D shapes.

Volume of Cube, Cuboid, and Cylinder: Calculating the volume of solids.

Volume and Capacity: Differentiating between volume and capacity and solving related problems.

3.0NCERT Solutions for Class 8 Maths Chapter 9 : All Exercises

4.0NCERT Questions with Solutions for Class 8 Maths Chapter 9 - Detailed Solutions

Exercise : 9.1

• The shape of the top surface of a table is trapezium. Find its area if its parallel sides are 1 m and 1.2 m and perpendicular distance between them is 0.8 m .

• Sol. Area of top surface of a table = Area of the trapezium $=\frac{1}{2}\times ($ sum of parallel sides $)\times ($ distance between them) $=\left[\frac{1}{2}\times (1+1.2)\times 0.8\right]{\mathrm{m}}^2$ $=\left(\frac{1}{2}\times 2.2\times 0.8\right){\mathrm{m}}^2$ $=(1.1\times 0.8){\mathrm{m}}^2=0.88{\mathrm{m}}^2$
• The area of a trapezium is $34{\mathrm{cm}}^2$ and the length of one of the parallel sides is 10 cm and its height is 4 cm . Find the length of other parallel side. Sol. Let the required side be xcm . Then, area of the trapezium $=\left[\frac{1}{2}\times (10+x)\times 4\right]{\mathrm{cm}}^2=2(10+\mathrm{x}){\mathrm{cm}}^2$ But, the area of the trapezium $=34{\mathrm{cm}}^2$ (given) $\therefore 2(10+\mathrm{x})=34$ $\Rightarrow 10+\mathrm{x}=17$ $\Rightarrow \mathrm{x}=17-10=7$ Hence, the other side $=7\mathrm{cm}$
• Length of the fence of a trapezium shaped field $ABCD$ is 120 m . If $\mathrm{BC}=48\mathrm{m},\mathrm{CD}=17$ m and $\mathrm{AD}=40\mathrm{m}$, find the area of this field. Side $AB$ is perpendicular to the parallel sides AD and BC .

• Sol. Let ABCD be the given trapezium in which $\mathrm{BC}=48\mathrm{m},\mathrm{CD}=17\mathrm{m}$ and $\mathrm{AD}=40\mathrm{m}$

• Through D, draw DL $\perp$ BC. Now, BL = AD = 40 m and $\mathrm{LC}=\mathrm{BC}-\mathrm{BL}=(48-40)\mathrm{m}=8\mathrm{m}$ Applying Pythagoras theorem in right $△$ DLC, we have ${\mathrm{DL}}^2={\mathrm{DC}}^2-{\mathrm{LC}}^2=17^2-8^2=289-64=225$ $\Rightarrow \mathrm{DL}=\sqrt{225}=15\mathrm{m}$ Now, area of the trapezium ABCD $=\frac{1}{2}\times (BC+AD)\times DL$ $=\frac{1}{2}\times (48+40)\times 15{\mathrm{m}}^2$ $=(44\times 15){\mathrm{m}}^2=660{\mathrm{m}}^2$.
• The diagonal of a quadrilateral shaped field is 24 m and the perpendiculars dropped on it from the remaining opposite vertices are 8 m and 13 m . Find the area of the field.

• Sol. Let ABCD be the given quadrilateral in which $\mathrm{BE}\perp \mathrm{AC}$ and $\mathrm{DF}\perp \mathrm{AC}$. It is given that $\mathrm{AC}=24\mathrm{m},\mathrm{BE}=8\mathrm{m}$ and $\mathrm{DF}=13\mathrm{m}$.

• Now, area of quad. ABCD $=$ area of $△ABC+$ area of $△ACD$ $=\frac{1}{2}\times \mathrm{AC}\times \mathrm{BE}+\frac{1}{2}\times \mathrm{AC}\times \mathrm{DF}$ $=\left(\frac{1}{2}\times 24\times 8+\frac{1}{2}\times 24\times 13\right){\mathrm{m}}^2$ $=(12\times 8+12\times 13){\mathrm{m}}^2$ $=(96+156){\mathrm{m}}^2=252{\mathrm{m}}^2$
• The diagonals of a rhombus are 7.5 cm and 12 cm . Find its area. Sol. Area of a rhombus $=\frac{1}{2}\times$ (product of diagonals) $=\left(\frac{1}{2}\times 7.5\times 12\right){\mathrm{cm}}^2=45{\mathrm{cm}}^2$
• Find the area of a rhombus whose side is 5 cm and whose altitude is 4.8 cm . If one of its diagonals is 8 cm long, find the length of the other diagonal. Sol. Let ABCD be a rhombus of side 5 cm and whose altitude $\mathrm{DE}=4.8\mathrm{cm}$. Also, one of its diagonals, $\mathrm{BD}=8\mathrm{cm}$. Area of the rhombus ABCD $=2\times$ Area $(△ABD)=2\times \frac{1}{2}\times AB\times DE$ $=(5\times 4.8){\mathrm{cm}}^2=24{\mathrm{cm}}^2$ Also, area of the rhombus ABCD $=\frac{1}{2}\times \mathrm{AC}\times \mathrm{BD}$ $\Rightarrow \frac{1}{2}\times \mathrm{AC}\times 8=24$ $\Rightarrow \mathrm{AC}=6\mathrm{cm}$

• Hence, the other diagonal is 6 cm .
• The floor of a building consists of 3000 tiles which are rhombus shaped and each of its diagonals are 45 cm and 30 cm in length. Find the total cost of polishing the floor, if the cost per ${\mathrm{m}}^2$ is ₹ 4 . Sol. Area of the floor $=3000\times$ Area of one tile $=3000\times \frac{1}{2}\times 45\times 30{\mathrm{cm}}^2$ $=1500\times 45\times 30{\mathrm{cm}}^2$ $=\frac{1500\times 45\times 30}{100\times 100}{\mathrm{m}}^2=202.5{\mathrm{m}}^2$ Cost of polishing the floor ₹ 4 per m ${}^2$ $=₹(4\times 202.5)=₹810$
• Mohan wants to buy a trapezium shaped field. It's side along the river is parallel to and twice the side along the road. If the area of this field is $10500{\mathrm{m}}^2$ and the perpendicular distance between the two parallel sides is 100 m , find the length of the side along the river.

• Sol. Let the parallel sides of the trapezium shaped field be x m and 2 x m . Then, its area. $=\frac{1}{2}(\mathrm{x}+2\mathrm{x})\times 100{\mathrm{m}}^2$ $=50\times 3\mathrm{x}{\mathrm{m}}^2=150\mathrm{x}{\mathrm{m}}^2$ But, it is given that the area of the field is $10500{\mathrm{m}}^2$. $\therefore 150\mathrm{x}=10500$ $\Rightarrow \mathrm{x}=\frac{10500}{150}=70$ $\therefore$ The length of the side along the river is $2\times 70$, i.e., 140 metres.
• Top surface of a raised platform is in the shape of a regular octagon as shown in the figure. Find the area of the octagonal surface.

• Sol. Area of the octagonal surface ABCDEFGH = Area (trap. ABCH) + Area (rect. HCDG) + Area (trap. GDEF)

• $=\left[\frac{1}{2}(5+11)\times 4+11\times 5+\frac{1}{2}(11+5)\times 4\right]{\mathrm{m}}^2$ $=(32+55+32){\mathrm{m}}^2=119{\mathrm{m}}^2$
• There is a pentagonal shaped park as shown in the figure. For finding its area Jyoti and Kavita divided it in two different ways. Find the area of this park using both ways. Can you suggest some other way of finding its area? Sol. Taking Jyoti's diagram : Area of the pentagonal shaped park

• $=2\times$ Area of trapezium ABEF $=2\times \frac{1}{2}\times (15+30)\times \frac{15}{2}{\mathrm{m}}^2$ $=\left(45\times \frac{15}{2}\right){\mathrm{m}}^2=337.5{\mathrm{m}}^2$ Taking Kavita's diagram : Area of the pentagonal shaped park $=$ Area (square ACDF) + Area ( $△\mathrm{DEF}$ ) $=15\times 15+\frac{1}{2}\times 15\times 15$ $=225+\frac{225}{2}=225+112.5=337.5{\mathrm{m}}^2$

• Diagram of the adjacent picture frame has outer dimensions $24\mathrm{cm}\times 28\mathrm{cm}$ and inner dimensions $16\mathrm{cm}\times 20\mathrm{cm}$. Find the area of each section of the frame, if the width of each section is same.

• Sol. Width of each section $=4\mathrm{cm}$. Area of trapezium ABQP = Area of trapezium RCDS $=\frac{1}{2}\times 4\times (16+24)\mathrm{cm}=80{\mathrm{cm}}^2$ Area of trapezium $BQRC=$ Area of trapezium APSD $=\frac{1}{2}\times 4\times (28+20){\mathrm{cm}}^2$ $=96{\mathrm{cm}}^2$

Exercise: 9.2

• 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.
• 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.
• 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 $=10\mathrm{cm}$.
• 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 $=2\mathrm{bh}+2\ell \mathrm{h}+\ell \mathrm{b}$ $=(2\times 1\times 1.5+2\times 2\times 1.5+2\times 1){\mathrm{m}}^2$ $=(3+6+2){\mathrm{m}}^2=11{\mathrm{m}}^2$
• 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 $=\frac{500}{100}=5$.
• 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.
• 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 $$\begin{gathered} =\text{ Total surface area of the cylinder } \\ =\left(2\pi r\mathrm{rh}+2\pi {\mathrm{r}}^2\right)\text{ sq. units. } \\ =\left(2\times \frac{22}{7}\times 7\times 3+2\times \frac{22}{7}\times 7\times 7\right){\mathrm{m}}^2 \\ =(132+308){\mathrm{m}}^2=440{\mathrm{m}}^2 \end{gathered}$$
• 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) $=[2(128+33)]\mathrm{cm}$ $=(2\times 161)\mathrm{cm}$ $=322\mathrm{cm}$
• 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 $=\left(750\times \frac{264}{100}\right){\mathrm{m}}^2=1980{\mathrm{m}}^2$
• 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 $=\left(2\times \frac{22}{7}\times 7\times 16\right){\mathrm{cm}}^2$ $=704{\mathrm{cm}}^2$

Exercise : 9.3

• Given a cylindrical tank, in which situation will you find surface area and in which situation volume. (a) To find how much it can hold. (b) Number of cement bags required to plaster it. (c) To find the number of smaller tanks that can be filled with water from it. Sol. (a) Volume (b) Surface area (c) Volume
• Diameter of cylinder $A$ is 7 cm , and the height is 14 cm . Diameter of cylinder $B$ is 14 cm and height is 7 cm . Without doing any calculations can you suggest whose volume is greater? Verify by finding the volume of both the cylinders. Check whether the cylinder with greater volume also has greater surface area?

• Sol. The heights and diameters of these cylinders A and B are interchanged. We know that, Volume of cylinder $=\pi r^{2}h$ Radius of cylinder $\mathrm{A}=\frac{7}{2}\mathrm{cm}$ Radius of cylinder $\mathrm{B}\left(\frac{14}{2}\right)\mathrm{cm}=7\mathrm{cm}$ As the radius of cylinder B is greater, therefore, the volume of cylinder $B$ will be greater. Let us verify it by calculating the volume of both the cylinders. Volume of cylinder $\mathrm{A}=\pi {\mathrm{r}}^2\mathrm{h}$ $=\left(\frac{22}{7}\times \frac{7}{2}\times \frac{7}{2}\times 14\right){\mathrm{cm}}^3=539{\mathrm{cm}}^3$ Volume of cylinder $B=\pi r^{2}h$ $=\left(\frac{22}{7}\times 7\times 7\times 7\right){\mathrm{cm}}^3=1078{\mathrm{cm}}^3$ Volume of cylinder B is greater. $=$ Surface area of cylinder $\mathrm{A}=2\pi \mathrm{r}(\mathrm{r}+\mathrm{h})$ $=\left[2\times \frac{22}{7}\times \frac{7}{2}\left(\frac{7}{2}+14\right)\right]{\mathrm{cm}}^2$ $=\left[22\times \left(\frac{7+28}{2}\right)\right]{\mathrm{cm}}^2=\left(22\times \frac{35}{2}\right){\mathrm{cm}}^2$ $=385{\mathrm{cm}}^2$ Surface area of cylinder B $=2\pi r(r+h)$ $=\left[2\times \frac{22}{7}\times 7\times (7+7)\right]{\mathrm{cm}}^2$ $=(44\times 14){\mathrm{cm}}^2$ $=616{\mathrm{cm}}^2$ Thus, the surface area of cylinder B is also greater than the surface area of cylinder A.
• Find the height of a cuboid whose base area is $180{\mathrm{cm}}^2$ and volume is $900{\mathrm{cm}}^3$ ? Sol. Volume of the cuboid $=900{\mathrm{cm}}^3$ $\Rightarrow$ (Area of the base) $\times$ Height $=900{\mathrm{cm}}^3$ $\Rightarrow 180\times$ Height $=900$ $\Rightarrow$ Height $=\frac{900}{180}=5$ Hence, the height of the cuboid is 5 cm .
• A cuboid is of dimensions $60\mathrm{cm}\times 54\mathrm{cm}\times$ 30 cm . How many small cubes with side 6 cm can be placed in the given cuboid? Sol. Volume of cuboid $=(60\times 54\times 30){\mathrm{cm}}^3$ $=97200{\mathrm{cm}}^3$ Volume of cube $=(6\times 6\times 6){\mathrm{cm}}^3$ $=216{\mathrm{cm}}^3$ No. of cubes $=\frac{\text{ Volume of cuboid }}{\text{ Volume of cube }}$ $=\frac{97200}{216}=450$
• Find the height of the cylinder whose volume is $1.54{\mathrm{m}}^3$ and diameter of the base is 140 cm ? Sol. Let $h$ be the height of cylinder whose radius, $\mathrm{r}=\frac{140}{2}\mathrm{cm}=70\mathrm{cm}=\frac{70}{100}\mathrm{m}=0.7\mathrm{m}$ and volume $=1.54{\mathrm{m}}^3$. $\because$ volume $=\pi r^{2}h$ $\Rightarrow \pi r^2\mathrm{h}=1.54$ $\Rightarrow \frac{22}{7}\times 0.7\times 0.7\times \mathrm{h}=1.54$ $\Rightarrow \mathrm{h}=\frac{1.54\times 7}{22\times 0.7\times 0.7}=1\mathrm{m}$ Hence, the height of cylinder is 1 metre.
• A milk tank is in the form of cylinder whose radius is 1.5 m and length is 7 m . Find the quantity of milk in litres that can be stored in the tank.

• Sol. Quantity of milk that can be stored in the tank = Volume of the tank $=\pi r^{2}h$, where $r=1.5$ and $h=7\mathrm{m}$ $=\left(\frac{22}{7}\times 1.5\times 1.5\times 7\right){\mathrm{m}}^3=49.5{\mathrm{m}}^3$ $=(49.5\times 1000)$ litres $\left[\because 1{\mathrm{m}}^3=1000\mathrm{ltr}.\right]$ $=49500$ litres.
• If each edge of a cube is doubled, (i) How many times will its surface area increase? (ii) How many times will its volume increase? Sol. Let x units be the edge of the cube. Then, its surface area $=6x^2$ and its volume $=x^3$. When its edge is doubled, (i) Its surface area $=6(2\mathrm{x}{)}^2$ $=6\times 4x^2=24x^2$ $\Rightarrow$ The surface area of the new cube will be 4 times that of the original cube. (ii) Its volume $=(2\mathrm{x}{)}^3=8{\mathrm{x}}^3$ $\Rightarrow$ The volume of the new cube will be 8 times that of the original cube.
• Water is pouring into a cuboidal reservoir at the rate of 60 litres per minute. If the volume of reservoir is $108{\mathrm{m}}^3$, find the number of hours it will take to fill the reservoir.

• Sol. Volume of the reservoir $=108{\mathrm{m}}^3$ $=108\times 1000$ litres $=108000$ litres Since water is pouring into reservoir at the rate of 60 litres per minute. $\therefore$ Time taken to fill the reservoir $=\frac{108000}{60\times 60}$ hours $=30$ hours

5.0Key Features of Class 8 Maths Chapter 9 NCERT Solutions – Mensuration

• Practical Application: Helps students understand real-life measurements such as area, volume, and surface area.
• Foundation for Advanced Topics: Builds a strong base for geometry and advanced math topics in higher classes.
• Improves Problem-Solving Skills: Enhances logical thinking and analytical abilities through diverse problem-solving exercises.
• Boosts Confidence: Familiarizes students with various shapes and figures, making complex concepts easier to grasp.