The questions provided in NCERT books are prepared in accordance with the CBSE curriculum, thus holding higher chances of appearing in the annual examination. The clarity of concepts and topics covered in CBSE Class 8 NCERT solutions ensures students stay focussed on their studies. Get detailed answers for NCERT Solutions of Class 8 Maths Chapter 11, Exercise 11.3, prepared by the subject experts at BYJU’S, based on the latest CBSE guidelines. Download the free NCERT Solutions for Maths Chapter 11 – Mensuration and build strong fundamentals on the concepts covered.
NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Exercise 11.3
Access Other Exercise Solutions of Class 8 Maths Chapter 11 Mensuration
Exercise 11.1 Solutions: 5 Questions (Long answers)
Exercise 11.2 Solutions: 11 Questions (Long answers)
Exercise 11.4 Solutions: 8 Questions (2 Short answers, 6 Long answers)
Access Answers to NCERT Class 8 Maths Chapter 11 Mensuration Exercise 11.3 Page Number 186
1. There are two cuboidal boxes, as shown in the below figure. Which box requires a lesser amount of material to make?
Solution:
(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)
= 14800Â cm2
(b)Â Length of the cubical box (l) = 50 cm
The breadth of the cubical box (b) = 50 cm
Height of cubical box (h) = 50 cm
Total surface area of cubical box = 6(side)2
= 6(50×50)
= 6×2500
= 15000
Therefore, the surface area of the cubical box is 15000 cm2.
From the result of (a) and (b), the cuboidal box requires a lesser amount of material to make.
2. A suitcase that measures 80 cm x 48 cm x 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?
Solution: Length of suitcase, l = 80 cm,
The breadth of the suitcase, b = 48 cm
And the Height of the cuboidal, h = 24 cm
Total surface area of suitcase = 2(lb+bh+hl)
= 2(80×48+48×24+24×80)
= 2 (3840+1152+1920)
= 2×6912
= 13824
Therefore, the total surface area of the suitcase is 13824 cm2.
Area of Tarpaulin cloth = Surface area of the suitcase
l×b = 13824
l ×96 = 13824
l = 144
Required tarpaulin for 100 suitcases = 144×100 = 14400 cm = 144 m
Hence, the tarpaulin cloth required to cover 100 suitcases is 144 m.
3. Find the side of a cube whose surface area is 600 cm^2.
Solution: Surface area of cube = 600 cm2 (Given)
Formula for surface area of a cube = 6(side)2
Substituting the values, we get
6(side)2 = 600
(side)2 = 100
Or side = ±10
Since the side cannot be negative, the measure of each side of a cube is 10 cm.
4. Rukshar painted the outside of a cabinet with a measure of 1 m×2 m×1.5 m. How much surface area did she cover if she painted all except the bottom of the cabinet?
Solution: Length of cabinet, l = 2 m, breadth of cabinet, b = 1 m and height of cabinet, h = 1.5 m
Area painted = Total surface area of the cabinet – Area of bottom
Total surface area of the cabinet = 2(lb+bh+hl )
= 2(2×1+1×1.5+1.5×2)
= 2(2+1.5+3.0)
= 13 m2
Area of bottom = Length × Breadth
= 2 × 1
= 2 m2
Area painted = 13 – 2 = 11 m2
Therefore, the area of the cabinet painted by Rukshar is 11 m2.
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 m^2 of the area is painted. How many cans of paint will he need to paint the room?
Solution: 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
Now, the required number of cans =Â Area of hall/Area of one can
= 500/100 = 5
Therefore, 5 cans of paint 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 areas?
Solution:
Similarity
Both figures have the same length and the same height.
Difference
The first figure has circular bottom and top.
The second figure has square bottom and top.
The first figure is a cylinder and the second figure is a cube.
The diameter of the cylinder = 7 cm (Given)
The radius of the cylinder, r  = 7/2  cm
The height of the cylinder, h = 7 cm
The lateral surface area of the cylinder = 2Ï€rh
= 2×(22/7)×(7/2)×7 = 154
So, the lateral surface area of the cylinder is 154 cm2
Now, lateral surface area of the cube = 4 (side)2 = 4×72 = 4×49 = 196
The lateral surface area of the cube is 196 cm2
Hence, 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 used to build it?
Solution:
The radius of the cylindrical tank, r  = 7 m
The height of the cylindrical tank, h = 3 m
The total surface area of the cylindrical tank = 2Ï€r(h+r)
= 2×(22/7)×7(3+7)
= 44×10 = 440
Therefore, a 440 m2Â metal sheet is required.
8. The lateral surface area of a hollow cylinder is 4224 cm2. It is cut along its height and forms a rectangular sheet of width 33 cm. Find the perimeter of the rectangular sheet.
Solution: Lateral surface area of hollow cylinder = 4224Â cm2
Width of rectangular sheet = 33 cm, and say l be the length of the rectangular sheet
The lateral surface area of the cylinder = Area of the rectangular sheet
4224 = b × l
4224 = 33 × l
l = 4224/33 = 128 cm
So, the length of the rectangular sheet is 128 cm.
Also, the perimeter of the rectangular sheet =Â 2(l+b)
= 2(128+33)
= 322 cm
Thus, the perimeter of the 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 the length is 1 m.
Solution:
Diameter of road roller, d = 84 cm
Radius of road roller, r = d/2 = 84/2 = 42 cm
Length of road roller, h = 1 m = 100 cm
The formula for the curved surface area of road roller = 2Ï€rh
= 2×(22/7)×42×100 = 26400
The curved surface area of the road roller is 26400 cm2
Again, the area covered by the road roller in 750 revolutions = 26400×750cm2
= 1,98,00,000 cm2
= 1980 m2  [∵ 1 m2= 10,000 cm2]
Hence, the area of the road is 1980 m2.
10. A company packages its milk powder in a cylindrical container whose base has a diameter of 14 cm and a height of 20 cm. The company places a label around the surface of the container (as shown in the figure). If the label is placed 2 cm from the top and bottom, what is the area of the label?
Solution:
Diameter of the cylindrical container, d = 14 cm
Radius of cylindrical container, r = d/2 = 14/2 Â = 7 cm
Height of cylindrical container = 20 cm
Height of the label, say h = 20–2–2 (from the figure)
= 16 cm
The curved surface area of the label = 2Ï€rh
= 2×(22/7)×7×16
= 704
Hence, the area of the label is 704 cm2.
Chapter 11 Mensuration, Exercise 11.3 is about identifying solid shapes and the surface area of a cube, cuboid and cylinder. After practising these problems, students will be able to identify three-dimensional geometrical figures as well as calculate the area covered by them. Download and practise NCERT Class 8 Maths Solutions and improve your skills. You will find it extremely easy to understand the questions and how to go about solving them.
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