NCERT Grade 8-Mensuration-Answers

NCERT Solutions for Class 8 Maths

Find 100% accurate solutions for NCERT Class VIII Science. All questions have been explained and solved step-by-step, to help you understand thoroughly. Free Download option available!

Sign Up to Download as PDF

1.    Given: The side of a square = 60 m

And the length of rectangular field = 80 m According to question,

Perimeter of rectangular field = Perimeter of square field

⇒ 2 (l + b) = 4 × side

⇒ 2 (80 + b) = 4 × 60                    ⇒                    (160 + 21) = 240

⇒ 2h = 240 – 160                            ⇒                     2h =80

⇒ b = 40 m

Now          Area of Square field = (Side)2 = (60)2 = 3600 m2

And         Area of Rectangular field = length × breadth = 80 × 40 = 3200 m2

Hence, area of square field is larger.

2.    Side of a square plot = 25 m

∴ Area of square plot = (Side)2 = (25)2 = 625 m2

Length of the house = 20 m and

Breadth of the house = 15 m

∴ Area of the house = length × breadth = 20 × 15 = 300 m2

Area of garden = Area of square plot – Area of house = 625 – 300 = 325 m2

∵ Cost of developing the garden per sq. m = 55

∴ Cost of developing the garden 325 sq. m = 55 × 325 = 17,875

Hence total cost of developing a garden around is ₹ 17,875.

3.    Given: Total length = 20 m

Diameter of semi circle = 7 m

∴ Radius of semi circle  = 3.5 m

Length of rectangular field = 20 – (3.5 + 3.5) = 20 – 7 = 13 m

Breadth of the rectangular field = 7 m

∴ Area of rectangular field = lxb = 13 x 7 = 91 m2

Area of two semi circles 

Area of garden = 91 + 38.5 = 129.5 m2

Now          Perimeter of two semi circles 

And     Perimeter of garden = 22 + 13 + 13 = 48 m

4.   Given:    Base of flooring tile = 24 cm = 0.24 m

Corresponding height of a flooring tile = 10 cm = 0.10 m

Now       Area of flooring tile = Base × Altitude = 0.24 × 0.10 = 0.024 m2

∴ Number of tiles required to cover the floor 

Hence 45000 tiles are required to cover the floor.

5.    (a) Radius 

Circumference of semi circle 

Total distance covered by the ant = Circumference of semi circle + Diameter = 4.4 + 2.8 = 7.2 cm

(b) Diameter of semi circle = 2.8 cm

∴ Radius 

Circumference of semi circle 

(c) Diameter of semi circle = 2.8 cm

∴ Radius 

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.

6.   Here one parallel side of the trapezium (a) = 1 m

And second side (b) = 1.2 m and height (h) = 0.8 m

Area of top surface of the table 

Hence surface area of the table is 0.88 m2

7.    Let the length of the other parallel side be

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

Area of trapezium 

                        ⇒                        34 = (10 + b) × 2

⇒ 34 = 20+ 2b                            ⇒                        34 – 20 = 2b

⇒ 14 = 2b                                     ⇒                         7 = b

⇒ b = 7

Hence another required parallel side is 7 cm.

8.   Given: BC = 48 m, CD = 17 m, AD = 40 m and perimeter = 120 m

∵ Perimeter of trapezium ABCD = AB + BC + CD + DA

⇒ 120 = AB + 48 + 17 + 40

⇒ 120 = AB = 105

⇒ AB = 120 – 105 = 15 m

Now Area of the field 

Hence area of the field ABCD is 660 m2

9.   Here h1 = 13 m, h2 = 8 m and AC = 24 m

Area of quadrilateral ABCD = Area of Δ ABC + Area of Δ ADC

Hence required area of the field is 252 m2

10.  Given: d, =7.5 cm and d2 = 12 cm

We know that,

Area of rhombus 

Hence area of rhombus is 45 cm2.

11.  Since rhombus is also a kind of parallelogram.

Area of rhombus = Base × Altitude = 6 × 4 = 24 cm2

Also       Area of rhombus 

Hence the length of the other diagonal is 60 cm.

12.  Here, d1 = 45 cm and d2 = 30 cm

∵ Area of one tile 

∴ Area of 3000 tiles    = 675 × 3000 = 2025000 cm2

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

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

Hence the total cost of polishing the floor is ₹ 810.

13.  Given: Perpendicular distance (h) = 100 m

Area of the trapezium shaped field = 10500 m2

Let side along the road be x m and side along the river = 2 × m

Area of the trapezium field 

Hence the side along the river = 2x = 2 × 70 = 140 m.

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

Construction: Divided the octagon in 3 figures, two trapeziums whose parallel and perpendicular sides are 11 m and 4 m respectively and third figure is rectangle having length and breadth 11 m and 5 m respectively.

Now   Area of two trapeziums 

And    Area of rectangle = length × breadth = 11 × 5 = 55 m2

∴ Total area of octagon = 64 + 55 = 119 m2

15.  First way:  By Jyoti’s diagram,

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

Second way:  By Kavita’s diagram

Here, a perpendicular AM drawn to BE.

AM = 30 – 15 = 15 m

Area of pentagon = Area of Δ ABE + Area of square BCDE

Hence total area of pentagon shaped park = 337.5 m2.

16.  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 

Also    Area of figure (II) = 96 cm2

Now    Area of figure (III) = Area of trapezium 

Also    Area of figure (IV) = 80 cm2

17.  (a) Given: 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)

= 2 (2400 + 2000 + 3000)

= 2 × 7400 = 14800 cm2.

(b) Given: Length of cuboidal box (l) = 50 cm

Breadth of cuboidal box (b) = 50 cm Height of cuboidal box (h) = 50 cm

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

= 2(50 × 50+ 50 × 50 + 50 × 50)

= 2 (2500 + 2500 + 2500)

= 2 × 7500 = 15000 cm2

Hence cuboidal box (a) requires the lesser amount of materal to make, since surface area of box (a) is less than that of box (b).

18.  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)

= 2 (3840 + 1152 + 1920)

= 2 × 6912 = 13824 cm2

Area of Tarpaulin cloth = Surface area of suitcase

Required tarpaulin for 100 suitcases = 144 × 100 = 14400 cm = 144 m

Hence tarpaulin cloth required to cover 100 suitcases is 144 m.

19.  Here Surface area of cube = 600 cm2

⇒ 6l2 = 600                    ⇒             l2 = 100          ⇒            l = 10 cm

Hence the side of cube is 10 cm

20.  Here, Length of cabinet (1) = 2 m, Breadth of cabinet (b) = 1 m

And Height of cabinet (h) = 1.5 m

∴ Surface area of cabinet = lb + 2 (bh + hl)

= 2 × 1 + 2 (1 × 1.5 + 1.5 × 2)

= 2 + 2 (1.5 + 3.0)

= 2 + 9.0

= 11 m2

Hence required surface area of cabinet is 11 m2.

21.  Here, Length of wall (l) = 15 m, Breadth of wall (b) = 10 m And Height of wall (h) = 7 m

Total Surface area of classroom = lb + 2 (bh + hl)

= 15 × 10 + 2 (10 × 7 + 7 × 15)

= 150 + 2 (70 + 105)

= 150 + 350

= 500 m2

Now  Required number of cans 

Hence 5 cans are required to paint the room.

22.  Given: Diameter of cylinder = 7 cm

Radius of cylinder (r) = 2 – 7 cm

And Height of cylinder (h) = 7 cm

Lateral surface area of cylinder 

Now    lateral surface area of cube = 4l2 = 4 × (7)2 = 4 × 49 = 196 cm2

Hence the cube has larger lateral surface area.

23.  Given: Radius of cylindrical tank (r) = 7 m

Height of cylindrical tank (h) = 3 m

Total surface area of cylindrical tank 

Hence 440 m2 metal sheet is required.

24.  Given: Lateral surface area of hollow cylinder = 4224 cm2

And Height of hollow cylinder = 33 cm

Curved surface area of hollow cylinder = 2πrh

Now    Length of rectangular sheet = 2πr

Perimeter of rectangular sheet = 2 (l + b)

= 2 (128 + 33) = 2 × 161 = 322 cm

Hence perimeter of rectangular sheet is 322 cm.

25.  Given: Diameter of road roller = 84 cm

∴ Radius of road roller 

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

Curved surface area of road roller 

∵ Area covered by road roller in 750 revolutions = 26400 × 750 = 1,98,00,000 cm2

= 1980 m2            [∵ 1 m2 = 10,000 cm2]

Hence the area of the road is 1980 m2.

26.  Given: Diameter of cylindrical container = 14 cm

∴ Radius of cylindrical container 

Height of cylindrical container = 20 cm

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

Curved surface area of label 

Hence the area of the label of 704 cm2.

27.  We find area when a region covered by a boundary, such as outer and inner surface area of a cylinder, a cone, a sphere and surface of wall or floor.

When the amount of space occupied by an object such as water, milk, coffee, tea, etc., then we have to find out volume of the object.

(a) Volume          (b) Surface are            (c) Volume

28.  Yes, we can say that volume of cylinder B is greater, since radius of cylinder B is greater than that of cylinder A (and square of radius gives more value than previous).

Diameter of cylinder A = 7 cm         ⇒          Radius of cylinder A =  cm

And Height of cylinder A = 14 cm

Volume of cylinder A 

= 539 cm3

Now        Diameter of cylinder B = 14 cm

⇒ Radius of cylinder B 

And Height of cylinder B = 7 cm

∴ Volume of cylinder A =  = 1078 cm3

Total surface area of cylinder A = πr (2h + r)           [∵ It is open from top]

Total surface area of cylinder B = πr (2h + r)            [∵ It is open from top]

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

29.  Given: Base area of cuboid = 180 cm2 and Volume of cuboid = 900 cm3

We know that,

Volume of cuboid = l × b × h                                 [•.• Base area = lxb= 180 (given)]

⇒ 900 = 180 × h

Hence the height of cuboid is 5 m.

30.  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 cm3

And   Volume of cube = (Side)3 = 6 ×× 6 cm3

∴ Number of small cubes 

Hence required cubes are 450.

31.  Given: Volume of cylinder = 1.54 m3 and Diameter of cylinder = 140 cm

∴ Radius 

Volume of cylinder = πr2h

Hence height of the cylinder is 1 m.

32.  Given: Radius of cylindrical tank (r) = 5 m and Height of cylindrical tank (h) = 7 m Volume of cylindrical tank = πr2h

=  ×1.5 × 1.5 × 7 = 49.5 cm3

= 49.5 × 1000 liters                                     [∵ m3 = 1000 liters]

= 49500 liters

Hence required quantity of milk is 49500 liters.

33.  (i) Let the edge of cube be

Since, Surface area of the cube (A) = 6l2 When edge of cube is doubled, then

Surface area of the cube (A’)= 6 (2l)2 = 6 × 412 = 4 × 612

A’ = 4 × A

Hence surface area will increase four times.

(ii) Volume of cube (V) = l3

When edge of cube is doubled, then Volume of cube (V’)= (2l)3 = 8l3

V’ = 8 × V

Hence volume will increase 8 times.

34.  Given: volume of reservoir = 108 m3

Rate of pouring water into cuboidal reservoir = 60 liters/minute

∵  mwater filled in reservoir will take = 1 hour

∴ 1 m3 water filled in reservoir will take 

∴ 108 mwater filled in reservoir will take  hours = 30 hours

It will take 30 hours to fill the reservoir.

MySchoolPage connects you with exceptional, certified math tutors who help you stay focused, understand concepts better and score well in exams!

Have a Question?




Mathematics - Videos


Physics - Videos


Biology - Videos


Chemistry - Videos

Added to Cart