# NCERT Solutions for Class 8 Maths Chapter 11 Mensuration| PDF Download

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**Chapter 11 Mensuration Class 8 Maths NCERT Solutions**through which you will take advantage over your classmates and boost your preparation. Chapter 11 Mensuration NCERT Solutions PDF Download is also available here which will help you in developing problem solving skills. These**Chapter 11 NCERT Solutions**are accurate and detailed and as per the latest syllabus released by CBSE.**NCERT Solutions for Class 8 NCERT Solutions**will improve the learning behaviour among students and increase their concentration. These NCERT Solutions will provide good experience and provide opportunities to learn new things.

**Exercise 11.1**

1. A square and a rectangular field with measurements as given in the figure have the same perimeter.

Which field has a larger area?

**Answer**

Given:

The side of a square = 60 m and the length of rectangular field = 80 m

According to question,

Perimeter of rectangular file = Perimeter of square field

⇒ 2(l+b) = 4 × Side

⇒ 2(80 + b) = 4 × 60

⇒ (80 + b) = 240/2

⇒ (80 + b) = 120

⇒ b = 120 - 80

⇒ b = 40 m

Hence, the breadth of the rectangular field is 40 m.

Now,

Area of Square field= (Side)

^{2}

= (60)

^{2}sq.m = 3600 sq.m

Area of Rectangular field = (length × breadth)

= 80 × 40 sq. m = 3200 sq. m

Hence, area of square field is larger.

**2. Mrs. Kaushik has a square plot with the measurement as shown in the figure. She wants to construct a house in the middle of the plot. A garden is developed around the house. Find the total cost of developing a garden around the house at the rate of Rs. 55 per m**

^{2}.

**Answer**

Side of a square plot = 25 m

∴ Area of square plot = (Side)

^{2}= (25)

^{2}= 625 m

^{2}

Length and Breadth of the house is 20 m and 15 m respectively

∴ Area of the house = (length x breadth )

= 20 × 15 = 300 m

^{2}

Area of garden = Area of square plot – Area of house

= (625 – 300) = 325 m

^{2}

∵ Cost of developing the garden around the house is Rs.55

∴ Total Cost of developing the garden of area 325 sq. m = Rs.(55 × 325)

= Rs.17,875

**2. The shape of a garden is rectangular in the middle and semi-circular at the ends as shown in the diagram. Find the area and the perimeter of this garden**

[Length of rectangle is 20 – (3.5 + 3.5 meters]

**Answer**

Given:

Total length of the diagram = 20 m

Diameter of semi circle on both the ends = 7 m

∴ Radius of semi circle = diameter/ 2 = 7/2 = 3.5 m

Length of rectangular field = [Total length - (radius of semicircle on both side)]

={20 – (3.5 + 3.5)}

= 20 – 7 = 13 m

Breadth of the rectangular field = 7 m

∴ Area of rectangular field = ( l x b)

= (13×7) ⇒ 91 m

^{2}

Area of two semi circles = 2 × 1/2 Ï€r

^{2}

= 2 × 1/2 × 22/7 × 3.5 × 3.5 = 38.5 m

^{2}

Total Area of garden = (91 + 38.5)⇒129.5 m

^{2}

Perimeter of two semi circles = 2 × Ï€r = 2 × 22/7 × 3.5 = 22 m

Hence, Perimeter of garden = (22 + 13 + 13)m = 48 m

**3. A flooring tile has the shape of a parallelogram whose base is 24 cm and the corresponding height is 10 cm. How many such tiles are required to cover a floor of area 1080 m**

^{2}? [If required you can split the tiles in whatever way you want to fill up the corners]

**Answer**

Base of flooring tile = 24 cm ⇒ 0.24 m

height of a flooring tile = 10 cm

⇒ 0.10 m [1cm = 1/100 m]

Now, Area of flooring tile= Base × Altitude

= 0.24 × 0.10 sq. m

= 0.024 m

^{2}

∴ Number of tiles required to cover the floor = Area of floor/area of one tile

= 1080/0.024

= 45000 tiles

Hence 45000 tiles are required to cover the floor.

4. An ant is moving around a few food pieces of different shapes scattered on the floor. For which food-piece would the ant have to take a longer round? Remember, circumference of a circle can be obtained by using the expression c = 2Ï€r, where r is the radius of the circle.

(a)

(b)

(c)

(c)

**Answer**

(a) Radius = Diameter/2 = 2.8/2 = 1.4 cm

Circumference of semi circle = Ï€r = 22/7 × 1.4 = 4.4 cm

Total distance covered by the ant= (Circumference of semi circle + Diameter)

=( 4.4 + 2.8 )cm = 7.2 cm

(b) Diameter of semi circle = 2.8 cm

Radius = Diameter/2 = 2.8/2 = 1.4cm

Circumference of semi circle = Ï€r = 22/7 × 1.4 = 4.4 cm

Total distance covered by the ant = (1.5 + 2.8 + 1.5 + 4.4) 10.2 cm

(c) Diameter of semi circle = 2.8 cm

Radius = Diameter/2 = 2.8/2 = 1.4 cm

Circumference of semi circle = Ï€r = 22/7 × 1.4 = 4.4 cm

Total distance covered by the ant= (2 + 2 + 4.4) = 8.4 cm

Hence for figure (b) food piece, the ant would take a longer round.

Page No. 177

**Exercise 11.2**

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

**Answer**

Parallel side of the trapezium AB =1m , CD = 1.2 m and height (h) of the trapezium (AM) = 0.8 m

Area of top surface of the table = 1/2 (sum of parallel sides) Height

= 1/2 × (AB + CD) × AM

= 1/2 × ( 1 + 1.2) × 0.8

= 1/2 × 2.2 × 0.8

= 0.88 m

Thus surface area of the table is 0.88 m

= 1/2 × (AB + CD) × AM

= 1/2 × ( 1 + 1.2) × 0.8

= 1/2 × 2.2 × 0.8

= 0.88 m

^{2}Thus surface area of the table is 0.88 m

^{2}**2. The area of a trapezium is 34 cm**

^{2}and the length of one of the parallel sides is 10 cm and its height is 4 cm.Find the length of the other parallel side.

**Answer**

Let the length of the other parallel side be = b cm

Length of one parallel side = 10 am and height (h) = 4 cm

Area of trapezium = 1/2 (sum of parallel sides) Height

⇒ 34 = 1/2 ( a + b)h

⇒ 34 = 1/2 (10 + b) × 4

⇒ 34 = ( 10 + b) × 2

⇒ 34 = 20 + 2b

⇒ 34 - 20 = 2b

⇒ 14 = 2b

⇒ 7 = b

⇒ b = 7

*Hence another required parallel side is 7 cm.*

*3.*Length of the fence of a trapezium shaped field ABCD is 120 m. If BC = 48 m, CD = 17 m and AD = 40 m, find the area of this field. Side AB is perpendicular to the parallel sides AD and BC.

**Answer**

Given: BC = 48 m, CD = 17 m,

AD = 40 m and perimeter = 120 m

*∵ Perimeter of trapezium ABCD = Sum of all sides*

120 =(AB + BC + CD + DA)

120 = AB + 48 + 17 + 40

120 = AB + 105

(120 – 105) = AB

AB = 15 m

Now Area of the field = 1/2 × ( Sum of parallel sides) × Height

= 1/2 × (BC + AD) × AB

= 1/2 × (48 + 40) × 15 m

^{2}

= 1/2 ×(88) × 15 m

^{2}

= 1/2 × (1320) m

^{2}

= 660 m

^{2}

*Hence area of the field ABCD is 660*m

^{2}

**4. 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.**

**Answer**

Here h

_{1}= 13 m, h

_{2}= 8 m and AC = 24 m

*Area of quadrilateral ABCD = Area of*△

*ABC + Area of*△

*ADC*

= 1/2 b × h

_{1}+ 1/2 b × h

_{2}

= 1/2 b ( h

_{1}+ h

_{2})

= 1/2 × 24 ( 13 + 8) m

^{2}

= 1/2 × 24 (21) m

^{2}

= 12 × 21 m

^{2}

= 252 m

^{2}

*Hence required area of the field*

*is 252*m

^{2}

5. The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.

**Answer**

**Given: d**

_{1}=7.5 cm and d

_{2}= 12 cm

*Area of rhombus =12*×

*(Product of digonals)*

= 12 × (d

_{1}× d

_{2})

= 12×(7.5 × 12) cm

^{2}

= 45 cm

^{2}

*Hence area of rhombus is 45*cm

^{2}

**6. Find the area of a rhombus whose side is 6 cm and whose altitude is 4 cm. If one of the diagonals is 8 cm long, find the length of the other diagonal.**

**Answer**

*Rhombus is also a kind of Parallelogram.*

∴

*Area of rhombus= Base*×

*Altitude*

= (6 × 4) cm

^{2}

= 24 cm

^{2}

Also

*Area of rhombus = 12× (*d

_{1}×

*d*

_{2}

*)*

*⇒ 24 = 1/2*×( 8 × d

_{2}

*)*

⇒ 24 = 4 d

_{2}

⇒ 244 cm = d

_{2}

⇒ d

_{2}= 6 cm

*Hence, the length of the other diagonal is 6 cm.*

**7. 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 m2 is ₹4.**

**Answer**

**Here, d**

_{1}= 45 cm and d

_{2}= 30 cm

*∵ Area of one tile*

*= 12*

*×*

*(*d

_{1}

*×*d

_{1}

*)*

= 12

*×*(45

*×*30 )

= 12

*×*(350)

= 675 cm

^{2}

*So, the area of one tile is 675 cm*

^{2}

*Area of 3000 tiles = 675*×

*3000 cm*

^{2}= 2025000 cm

^{2}

= 2025000100∗100 m

^{2}

[

*1 cm = 1/100 m, Here cm*

^{2}= cm×cm = 1/100×1/100 m^{2}]= 202.50 m

^{2}

∵ Cost of polishing the floor per sq. meter = Rs. 4

∴ Cost of polishing the floor per 202.50 sq. meter = Rs. 4 × 202.50 = Rs. 810

*Hence the total cost of polishing the floor is Rs. 810*.

**8. Mohan wants to buy a trapezium shaped field. Its side along the river is parallel to and twice the sidealong the road. If the area of this field is 10500**

*m*and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river.

^{2}**Answer**

Given: Perpendicular distance (h) AM = 100 m

Area of the trapezium shaped field = 10500 m

^{2}

Let side along the road AB= x m

side along the river CD = 2x m

∴ Area of the trapezium field = 12× (AB + CD) x AM

10500 = 1/2 ( x + 2x) × 100

10500 = 3x × 50

3x = 10500/50

x = 10500/(50 × 3)

x = 70 m

*Hence the side along the river = 2m = (2*×

*70) = 140 m.*

**9. 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.**

**Answer**

*Given: Octagon having eight equal sides, each 5 m.*

*Construction: Join HC and GD It will divide the octagon into two equal trapezium.*

*And AM is perpendicular on HC and EN is perpendicular on GD*

*Area of trap. ABCD = Area of trap. GDFE ---(1)*

Area of two trapeziums = (area of trap. ABCH + area of trap. GDFE)

= (area of trap. ABCH + area of trap. ABCH) (by statement 1).

= (2×area of trap. ABCH )

= (2×1/2×(sum of parallel sides×height)

= (2×1/2×(AB + CH) x AM)

=(11 + 5)×4 m

^{2}

=(16) × 4

= 64 m

^{2}

*Area of rectangle (HCDG ) = length*×

*breadth*

= HC×HG = 11×5 = 55 m

^{2}

∴

*Total area of octagon = Area of 2 Trapezium + Area of Rectangle*

= 64 m

^{2}+ 55 m

^{2}= 119 m

^{2}

**10. There is a pentagonal shaped park as shown in the figure. For finding its are a 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?

**Answer**

First way:

*By Jyoti’s diagram*,

*Area of pentagon = Area of trapezium ABCP + Area of trapezium AEDP*

= 1/2 (AP + BC) x CP + 1/2 (ED + AP) × DP

= 1/2 (30 + 15 ) x CP + 1/2(15 + 30) × DP

= 1/2 (30 + 15) (CP + DP)

= 1/2 × 45 × CD

= 1/2 × 45 × 15

= 337.5 m

^{2}

Second way:

*By Kavita’s diagram*

Here, a perpendicular AM drawn to BE. AM = 30 – 15 = 15 m

*Area of pentagon = Area of*△

*ABE + Area of square BCDE*

={12×15×15}+(15 × 15) m

^{2}

= (112.5 + 225.0) m

^{2}

= 337.5 m

^{2}

*Hence total area of pentagon shaped park = 337.5 m*

^{2}11. Diagram of the adjacent picture frame has outer dimensions = 24 cm × 28 cm and inner dimensions 16 cm × 20 cm. Find the area of each section of theframe, if the width of each section is same.

**Answer**

*Here two of given figures (I) and (II) are similar in dimensions. And also figures (III) and (IV) are similar in dimensions.*

∴

*Area of figure (I) = Area of trapezium*

= 1/2 (a+b) × h = 1/2 (28 + 20) × 4

= 1/2 × 48 × 4 = 96 cm

^{2}

*Also, Area of figure (II) = 96 cm*

^{2}*Now Area of figure (III)*

*Area of trapezium = 1/2 (a + b)*× h

= 1/2 (24 + 16) × 4

= 1/2 × 40 × 4

= 80 cm

^{2}

*Also Area of figure (IV) = 80 cm*

^{2}
Page No. 186

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

(a) Length of cuboidal box (l)= 60 cm

Breadth of cuboidal box (b) = 40 cm

Height of cuboidal box (h) = 50 cm

∴ Total surface area of cuboidal box = 2(lb+bh+hl)

= 2 (60 × 40 + 40 × 50 + 50 × 60) cm

= 2 (2400 + 2000 + 3000) cm2

= 2 × 7400 cm

= 14800 cm

(b) Length of the cube is 50 cm

∴ Total surface area of cuboidal box =6(side)

= 6 (50)

= 6 (2500) cm

= 15000 cm

Thus, the cuboidal box (a) requires the lesser amount of material.

2. A suitcase with measures 80 cm × 48 cm × 24 cm is to be covered with a tarpaulin cloth. How many meters of tarpaulin of width 96 cm is required to cover 100 such suitcases?

Given:

Length of suitcase box (l) = 80 cm, Breadth of suitcase box (b) = 48 cm

And Height of cuboidal box (h) = 24 cm

∴ Total surface area of suitcase box= 2(lb+bh+hl)

= 2 (80 × 48 + 48 × 24 + 24 × 80) cm

= 2 (3840 + 1152 + 1920)

= 2 × 6912 = 13824 cm

Area of Tarpaulin cloth = Surface area of suitcase

⇒ l × b = 13824

⇒ l × 96 = 13824

⇒ l = 13824/96

= 144 cm

Required tarpaulin for 100 suitcases = (144 × 100) cm

= 14400 cm

= 144 m [ 1cm = 100 m]

Thus, 144 m tarpaulin cloth required to cover 100 suitcases.

3.Find the side of a cube whose surface area id 600 cm

Here, Surface area of cube = 600 cm

⇒ 6l

⇒ l

⇒ l = √100 cm

⇒ l = 10 cm

Hence the side of cube is 10 cm

Breadth of cabinet (b) = 1 m

Height of cabinet (h) = 1.5 m

∴ Surface area of cabinet = {Area of Base of cabinet (Cuboid) + Area of four walls}

= lb+2(l+b)h

={2×1 + 2(1+2) 1.5} m

= 2 + 2 (3) 1.5 m

= 2+6 (1.5) m

= (2 + 9.0) m

= 11 m

Hence required surface area of cabinet is 11 m

5. Daniel is paining 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 m

Length of wall (l) = 15 m

Breadth of wall (b) = 10 m

Height of wall (h) = 7 m

∴ Total Surface area of classroom=(Area of Base of ceiling (Cuboid) + Area of four walls)

= lb+2(l+b)h

= {15 × 10 + 2 (10 +15) (7)} m

= {150 + 2 (25) (7)} m

= (150 + 350) m

= 500 m

Area of one can is 100 m

Now Required number of cans = Area of hall/ Area of one can = 500/100 = 5 cans

Hence 5 cans are required to paint the room.

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

Diameter of cylinder = 7 cm

∴ Radius of cylinder (r) = 7/2 cm

Height of cylinder (h) = 7 cm

Lateral surface area of cylinder = 2Ï€rh

= 2 × 22/7 × 7/2 × 7

= 154 cm

Now lateral surface area of cube = 4(Side)

= (4×49) cm

= 196 cm

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. Howmuch sheet of metal is required?

Height of cylindrical tank (h) = 3 m

= 2 × 22/7 × 7(3 + 7) m

= 44 × 10 m

= 440 m

Hence 440 m

8. The lateral surface area of a hollow cylinder is 4224 cm

Lateral surface area of hollow cylinder = 4224 cm

Height of hollow cylinder = 33 cm

Curved surface area of hollow cylinder = 2Ï€rh

⇒ 4224 = 2 × 22/7 × r × 33

⇒ r = (4224 × 7)/(2 × 22 × 33)

= (64 × 7)/22 cm

Now Length of rectangular sheet = 2Ï€r

l = 2 × 22/7 × (64 × 7)/22

= 128 cm

= 2 (128 + 33)

= 2 × 161

= 322 cm

Hence perimeter of rectangular 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 1 m.

Diameter of road roller = 84 cm

∴ Radius of road roller (r) = d/2 = 84/2

= 42 cm

Length of road roller (h) = 1 m = 100 cm

Curved surface area of road roller = 2Ï€rh

= 2 × 22/7 × 42 × 100

= 26400 cm

∴ Area covered by road roller in 750 revolutions = 26400 × 750 cm

= 1,98,00,000 cm

= 1980 m2 [∵ 1 m

Thus, the area of the road is 1980 m

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 figure). If the label is placed 2 cm from top and bottom, what is the area of the label?

Diameter of cylindrical container = 14 cm

∴ Radius of cylindrical container (r) = d/2 = 14/2 = 7 cm

Height of cylindrical container = 20 cm

Height of the label (h) = (20 – 2 – 2) = 16 cm

Curved surface area of label = 2Ï€rh

= 2×22/7×7×16

= 704 cm

Hence the area of the label of 704 cm

Page No. 191

1. 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.

(a) Volume (

(b) Surface area

(c) Volume

Diameter of cylinder A = 7 cm

⇒ Radius(r) of cylinder A = 7/2 cm and Height(h) of cylinder A = 14 cm

∴ Volume of cylinder A = Ï€r

= 22/7 × 7/2 ×7/2 × 14

= 539 cm

Now Diameter of cylinder B = 14 cm

⇒ Radius of cylinder B = 14/2 = 7 cm and Height of cylinder B = 7 cm

∴ Volume of cylinder A = Ï€r

= 22/7 × 7 ×7 ×7 cm

= 1078 cm

Since the cylinder A and cylinder B is open from upper end then it will exclude from the Total surface area

Total surface area of cylinder A = ( Area of lower end circle + curved surface area of cyliner)

= Ï€r

= Ï€r(r+2h)

= (22/7 × 7/2) (7/2 + 2×14)

= 11(7/2+28)

= 11(31.5) cm

Total surface area of cylinder B = Ï€r(2h + r)

= 22/7 × 7(2×7 + 7)

= 22 × (14 + 7)

= 22 × 21 = 462 cm

Yes, cylinder with greater volume also has greater surface area.

Let the Length, breadth and height of the cuboid be l, b, h.

Base area of cuboid = 180 cm

L x B = 180 cm

Volume of cuboid = l ×b × h

Volume of cuboid = 900 cm

(lb)h = 900 (From eq. 1)

(180) h = 900

h = 900/180

= 5 m

Hence the height of cuboid is 5 m.

Given: Length of cuboid (l) = 60 cm,

Breadth of cuboid (b) = 54 cm and

Height of cuboid (h) = 30 cm

We know that, Volume of cuboid = l × b ×h

= (60 × 54 × 30) cm

And Volume of cube = (Side)

= 6 × 6 × 6 cm

∴ Number of small cubes = (Volume of cuboid)/( Volume of cube)

= (60×54×30)/(6×6× 6)

= 450

Hence required number of small cubes are 450.

Given: Volume of cylinder = 1.54 m

∴ Radius (r) = d/2 = 140/2 = 70 cm

= 70/100 m = 0.7m [1cm=1/100m]

Volume of cylinder = Ï€r

⇒ 1.54 = 22/7 × 0.7 × 0.7 × h

⇒ h = (1.54 × 7)/(22 × 0.7 × 0.7)

⇒ h = ( 154 × 7 × 10 × 10)/( 22 × 7 × 7 × 100)

= 1 m

Hence height of the cylinder is 1 m.

Given: Radius of cylindrical tank (r) = 1.5 m

Height of cylindrical tank (h) = 7 m

Volume of cylindrical tank = Ï€r

= 22/7 × 1.5 × 1.5 × 7

= 49.5 m

= 49.5 × 1000 liters [∵1m

= 49500 liters

Hence required quantity of milk is 49500 liters that can be stored in the tank.

(i) how many times will its surface area increase?

(ii) how many times will its volume increase?

Let, l units be the edge of the cube.

Surface area = 6l

and Volume of the cube = l

When its edge is doubled = 2l

(i) 6(2l)

= 4 (Surface area) {∵6l

The surface area of the new cube will be 4 times that of the original cube.

(ii) Volume of cube (V) = l

When edge of cube is doubled = 2l, then

Volume of new cube = (2l)

Volume of new cube = 8× Volume of cube

Hence, volume will increase 8 times.

8. Water is pouring into a cuboidal reservoir at the rate of 60 liters per minute. If the volume of reservoir is 108m

= 108×1000 litres [1m

=108000 litres

Since water is pouring into reservoir @ 60 litres per minute and in

Time taken to fill the reservoir = 108000/(60×60) hours

= 30 hours

Hence, 30 hours it will take to fill the reservoir.

**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?

**Answer**(a) Length of cuboidal box (l)= 60 cm

Breadth of cuboidal box (b) = 40 cm

Height of cuboidal box (h) = 50 cm

∴ Total surface area of cuboidal box = 2(lb+bh+hl)

= 2 (60 × 40 + 40 × 50 + 50 × 60) cm

^{2}= 2 (2400 + 2000 + 3000) cm2

= 2 × 7400 cm

^{2}= 14800 cm

^{2}(b) Length of the cube is 50 cm

∴ Total surface area of cuboidal box =6(side)

^{2}= 6 (50)

^{2}cm^{2}= 6 (2500) cm

^{2}= 15000 cm

^{2}Thus, the cuboidal box (a) requires the lesser amount of material.

2. A suitcase with measures 80 cm × 48 cm × 24 cm is to be covered with a tarpaulin cloth. How many meters of tarpaulin of width 96 cm is required to cover 100 such suitcases?

**Answer**Given:

Length of suitcase box (l) = 80 cm, Breadth of suitcase box (b) = 48 cm

And Height of cuboidal box (h) = 24 cm

∴ Total surface area of suitcase box= 2(lb+bh+hl)

= 2 (80 × 48 + 48 × 24 + 24 × 80) cm

^{2}= 2 (3840 + 1152 + 1920)

= 2 × 6912 = 13824 cm

^{2}Area of Tarpaulin cloth = Surface area of suitcase

⇒ l × b = 13824

⇒ l × 96 = 13824

⇒ l = 13824/96

= 144 cm

Required tarpaulin for 100 suitcases = (144 × 100) cm

= 14400 cm

= 144 m [ 1cm = 100 m]

Thus, 144 m tarpaulin cloth required to cover 100 suitcases.

3.Find the side of a cube whose surface area id 600 cm

^{2}

**Answer**Here, Surface area of cube = 600 cm

^{2}⇒ 6l

^{2}= 600 cm^{2}⇒ l

^{2}= 100 cm^{2}⇒ l = √100 cm

⇒ l = 10 cm

Hence the side of cube is 10 cm

**4. Rukshar painted the outside of the cabinet of measure 1 m × 2 m × 1.5 m. How much surface area did she cover if she painted all except the bottom of the cabinet?**

**Answer****Length of cabinet (l) = 2 m**

Breadth of cabinet (b) = 1 m

Height of cabinet (h) = 1.5 m

∴ Surface area of cabinet = {Area of Base of cabinet (Cuboid) + Area of four walls}

= lb+2(l+b)h

={2×1 + 2(1+2) 1.5} m

^{2}= 2 + 2 (3) 1.5 m

^{2}= 2+6 (1.5) m

^{2}= (2 + 9.0) m

^{2}= 11 m

^{2}Hence required surface area of cabinet is 11 m

^{2}5. Daniel is paining 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 m

^{2}of area is painted. How many cans of paint will she need to paint the room?**Answer**Length of wall (l) = 15 m

Breadth of wall (b) = 10 m

Height of wall (h) = 7 m

∴ Total Surface area of classroom=(Area of Base of ceiling (Cuboid) + Area of four walls)

= lb+2(l+b)h

= {15 × 10 + 2 (10 +15) (7)} m

^{2}= {150 + 2 (25) (7)} m

^{2}= (150 + 350) m

^{2}= 500 m

^{2}Area of one can is 100 m

^{2}Now Required number of cans = Area of hall/ Area of one can = 500/100 = 5 cans

Hence 5 cans are required to paint the room.

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

**Answer**Diameter of cylinder = 7 cm

∴ Radius of cylinder (r) = 7/2 cm

Height of cylinder (h) = 7 cm

Lateral surface area of cylinder = 2Ï€rh

= 2 × 22/7 × 7/2 × 7

= 154 cm

^{2}Now lateral surface area of cube = 4(Side)

^{2 }= 4(7)^{2}cm^{2}= (4×49) cm

^{2}= 196 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. Howmuch sheet of metal is required?

**Answer****Radius of cylindrical tank (r) = 7 m**

Height of cylindrical tank (h) = 3 m

*Total surface area of cylindrical tank = (Curved surface area + Area of upper end (circle)+ Area of Lower (circle) end)**= (2Ï€rh+Ï€r*^{2}+Ï€r^{2})*= (2Ï€rh+2Ï€r*^{2})*=2Ï€r(h+r)*= 2 × 22/7 × 7(3 + 7) m

^{2}= 44 × 10 m

^{2}= 440 m

^{2}Hence 440 m

^{2}metal sheet is required.8. The lateral surface area of a hollow cylinder is 4224 cm

^{2}. It is cut along its height and formed a rectangular sheet of width 33 cm. Find the perimeter of rectangular sheet?**Answer**Lateral surface area of hollow cylinder = 4224 cm

^{2}Height of hollow cylinder = 33 cm

Curved surface area of hollow cylinder = 2Ï€rh

⇒ 4224 = 2 × 22/7 × r × 33

⇒ r = (4224 × 7)/(2 × 22 × 33)

= (64 × 7)/22 cm

Now Length of rectangular sheet = 2Ï€r

l = 2 × 22/7 × (64 × 7)/22

= 128 cm

*Perimeter of rectangular sheet = 2(l + b)*= 2 (128 + 33)

= 2 × 161

= 322 cm

Hence perimeter of rectangular 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 1 m.

**Answer**Diameter of road roller = 84 cm

∴ Radius of road roller (r) = d/2 = 84/2

= 42 cm

Length of road roller (h) = 1 m = 100 cm

Curved surface area of road roller = 2Ï€rh

= 2 × 22/7 × 42 × 100

= 26400 cm

^{2}∴ Area covered by road roller in 750 revolutions = 26400 × 750 cm

^{2}= 1,98,00,000 cm

^{2}= 1980 m2 [∵ 1 m

^{2}= 10,000 cm^{2}]Thus, the area of the road is 1980 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 figure). If the label is placed 2 cm from top and bottom, what is the area of the label?

**Answer**Diameter of cylindrical container = 14 cm

∴ Radius of cylindrical container (r) = d/2 = 14/2 = 7 cm

Height of cylindrical container = 20 cm

Height of the label (h) = (20 – 2 – 2) = 16 cm

Curved surface area of label = 2Ï€rh

= 2×22/7×7×16

= 704 cm

^{2}Hence the area of the label of 704 cm

^{2}.Page No. 191

**Exercise 11.4**1. 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.

**Answer**(a) Volume (

*it is measure of the amount of space inside of a solild figures)*(b) Surface area

*(the outside part or uppermost layer of the soild figures)*(c) Volume

**2. 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 it by finding the volume of both the cylinders. Check whether the cylinder with greater volume also has greater surface area.**

**Answer****Yes, we can say that volume of cylinder B is greater, Because radius of cylinder B is greater than that of cylinder A.**

Diameter of cylinder A = 7 cm

⇒ Radius(r) of cylinder A = 7/2 cm and Height(h) of cylinder A = 14 cm

∴ Volume of cylinder A = Ï€r

^{2}h= 22/7 × 7/2 ×7/2 × 14

= 539 cm

^{3}Now Diameter of cylinder B = 14 cm

⇒ Radius of cylinder B = 14/2 = 7 cm and Height of cylinder B = 7 cm

∴ Volume of cylinder A = Ï€r

^{2}h= 22/7 × 7 ×7 ×7 cm

^{3}= 1078 cm

^{3}Since the cylinder A and cylinder B is open from upper end then it will exclude from the Total surface area

Total surface area of cylinder A = ( Area of lower end circle + curved surface area of cyliner)

= Ï€r

^{2}+ 2Ï€rh)= Ï€r(r+2h)

= (22/7 × 7/2) (7/2 + 2×14)

= 11(7/2+28)

= 11(31.5) cm

^{2}= 346.5 cm^{2}Total surface area of cylinder B = Ï€r(2h + r)

= 22/7 × 7(2×7 + 7)

= 22 × (14 + 7)

= 22 × 21 = 462 cm

^{2}Yes, cylinder with greater volume also has greater surface area.

**3. Find the height of a cuboid whose base area is 180 cm**

^{2}and volume is 900 cm^{3}?**Answer**Let the Length, breadth and height of the cuboid be l, b, h.

*Base of the cuboid is form a Recatangle*so,that the Base(Reactangle) Area is (Length x Breadth)Base area of cuboid = 180 cm

^{2}L x B = 180 cm

^{2}--- (1)Volume of cuboid = l ×b × h

Volume of cuboid = 900 cm

^{2}(lb)h = 900 (From eq. 1)

(180) h = 900

h = 900/180

= 5 m

Hence the height of cuboid is 5 m.

**4. A cuboid is of dimensions 60 cm × 54 cm × 30 cm. How many small cubes with side 6 cm can be placed in the given cuboid?**

**Answer**Given: Length of cuboid (l) = 60 cm,

Breadth of cuboid (b) = 54 cm and

Height of cuboid (h) = 30 cm

We know that, Volume of cuboid = l × b ×h

= (60 × 54 × 30) cm

^{3}And Volume of cube = (Side)

^{3}= 6 × 6 × 6 cm

^{3}∴ Number of small cubes = (Volume of cuboid)/( Volume of cube)

= (60×54×30)/(6×6× 6)

= 450

Hence required number of small cubes are 450.

**5. Find the height of the cylinder whose volume if 1.54 m**

^{3}and diameter of the base is 140 cm.**Answer**Given: Volume of cylinder = 1.54 m

^{3}and Diameter of cylinder = 140 cm∴ Radius (r) = d/2 = 140/2 = 70 cm

= 70/100 m = 0.7m [1cm=1/100m]

Volume of cylinder = Ï€r

^{2}h⇒ 1.54 = 22/7 × 0.7 × 0.7 × h

⇒ h = (1.54 × 7)/(22 × 0.7 × 0.7)

⇒ h = ( 154 × 7 × 10 × 10)/( 22 × 7 × 7 × 100)

= 1 m

Hence height of the cylinder is 1 m.

**6. 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 liters that can be stored in the tank.**

**Answer**Given: Radius of cylindrical tank (r) = 1.5 m

Height of cylindrical tank (h) = 7 m

Volume of cylindrical tank = Ï€r

^{2}h= 22/7 × 1.5 × 1.5 × 7

= 49.5 m

^{3}= 49.5 × 1000 liters [∵1m

^{3}= 1000 liters]= 49500 liters

Hence required quantity of milk is 49500 liters that can be stored in the tank.

**7. 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?

**Answer**Let, l units be the edge of the cube.

Surface area = 6l

^{2}and Volume of the cube = l

^{3}When its edge is doubled = 2l

(i) 6(2l)

^{2}= 6(4l^{2}) = 4(6l^{2})= 4 (Surface area) {∵6l

^{2 }= Surface area of the cube}The surface area of the new cube will be 4 times that of the original cube.

(ii) Volume of cube (V) = l

^{3}When edge of cube is doubled = 2l, then

Volume of new cube = (2l)

^{3}= 8l^{3}Volume of new cube = 8× Volume of cube

Hence, volume will increase 8 times.

8. Water is pouring into a cuboidal reservoir at the rate of 60 liters per minute. If the volume of reservoir is 108m

^{3}, find the number of hours it will take to fill the reservoir.

**Answer****Volume of reservoir = 108m**

^{3}= 108×1000 litres [1m

^{3}=1000 l]=108000 litres

Since water is pouring into reservoir @ 60 litres per minute and in

Time taken to fill the reservoir = 108000/(60×60) hours

= 30 hours

Hence, 30 hours it will take to fill the reservoir.

**Go Back To NCERT Solutions for Class 8 Maths**

## NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

In Chapter 11 Mensuration Maths NCERT Solutions we are dealing with problems related to perimeter and area of other plane closed figures like quadrilaterals. We will also learn about surface area and volume of solids such as cube, cuboid and cylinder.

• A square, a rectangle, a trapezium, a rhombus, a parallelogram, a triangle, a circle, etc., are plane figures and the surfaces enclosed by their boundaries are called areas.

• The area of a square = Side × Side

• The area of a rectangle = Length × Breadth

• The area of a circle = Ï€r

^{2}[where r is the radius]• The area of triangle = 1/2 × Base × Altitude

• The area of a parallelogram = Base × Height

Below you will find exercisewise

**NCERT Solutions of Chapter 11 Class 8 Maths**that is very useful in building basic knowledge and concepts given in the chapter. You will get to learn about a lot of topics which you'll find interesting.Here you will get accurate and detailed NCERT Solutions of Class 8 Maths prepared by Studyrankers experts that will be useful in scoring great marks in the exams and completing your homework on time.

### NCERT Solutions for Class 8 Maths Chapters:

**FAQ on Chapter**

**11 Mensuration**

#### How many exercises in Chapter 11 Mensuration Class 8 Maths?

There are four exercises in Chapter 11 Mensuration NCERT Solutions which will be helpful in understanding the basic concepts embedded in the questions. These solutions can help you in solving supplementary Maths Books.

#### If the lateral surface of the cylinder is 500 cm² and its height is 10 cm, then find the radius of its base.

The lateral surface area is A =2Ï€rh. The curved surface area is A = 500 cm

^{2}and its height is 10 cm, henceA =2Ï€rh

500 = 2 × 3.14 × r × 10

500 = 62.8r

r = 500/62.8

= 7.96

Therefore the radius of the cylinder is 7.96 cm.

#### A horse is tethered by a rope 10 m long at a point. Find the area of the region where it can graze (Ï€ = 3.14)

The area of the region the horse can graze is circular with a radius equal to the length of the rope.

Area of the circle is Ï€r

^{2}= 3.14 × 10

^{2}= 3.14 × 100

=314

Hence the area of the region the horse can graze is 314cm

^{2}.#### What are right circular cylinders?

The cylinder which has congruent circular faces that are parallel to each other. The line segment joining the center of circular faces is perpendicular to the base. Such cylinders are known as right circular cylinders.