The questions provided in NCERT books are prepared in accordance with CBSE, thus holding higher chances of appearing in CBSE final examination. CBSE Class 8 NCERT solutions clear concepts ensuring to stay with students in exams. Get detailed answers for NCERT Solutions of Class 8 Maths Chapter 11, exercise 11.3 prepared by BYJU’S subject experts based on the latest CBSE guidelines. Download free NCERT Solutions for Maths Chapter 11- Mensuration and improve fundamentals.
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 adjoining figure. Which box requires the 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)
= 14800 cm2
(b) Length of cubical box (l) = 50 cm
Breadth of cubicalbox (b) = 50 cm
Height of cubicalbox (h) = 50 cm
Total surface area of cubical box = 6(side)2
Surface area of the cubical box is 15000 cm2
From the result of (a) and (b), cuboidal box requires the lesser amount of material to make.
2. A suitcase with 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 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 (3840+1152+1920)
Total surface area of suitcase box is 13824 cm2
Area of Tarpaulin cloth = Surface area of suitcase
l×b = 13824
l ×96 = 13824
l = 144
Required tarpaulin for 100 suitcases = 144×100 = 14400 cm = 144 m
Hence tarpaulin cloth required to cover 100 suitcases is 144 m.
3. Find the side of a cube whose surface area is 600cm^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 side cannot be negative, the measure of each side of a 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?
Solution:Length of cabinet, l = 2 m, Breadth of cabinet, b = 1 m and Height of cabinet, h = 1.5 m
Surface area of cabinet = lb+2(bh+hl )
Required surface area of cabinet is 11m2.
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^2of area is painted. How many cans of paint will she 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 )
Now, Required number of cans = Area of hall/Area of one can
= 500/100 = 5
Therefore, 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 areas?
Diameter of cylinder = 7 cm (Given)
Radius of cylinder, r = 7/2 cm
Height of cylinder, h = 7 cm
Lateral surface area of cylinder = 2πrh
= 2×(22/7)×(7/2)×7 = 154
So, Lateral surface area of cylinder is154 cm2
Now, lateral surface area of cube = 4 (side)2=4×72 = 4×49 = 196
Lateral surface area of cube is 196 cm2
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. How much sheet of metal is required?
Radius of cylindrical tank, r = 7 m
Height of cylindrical tank , h = 3 m
Total surface area of cylindrical tank = 2πr(h+r)
= 44×10 = 440
Therefore, 440 m2 metal sheet is required.
8. The lateral surface area of a hollow cylinder is 4224cm2. It is cut along its height and formed a rectangular sheet of width 33 cm. Find the perimeter of rectangular sheet?
Solution: Lateral surface area of hollow cylinder = 4224 cm2
Height of hollow cylinder, h = 33 cm and say r be the radius of the hollow cylinder
Curved surface area of hollow cylinder = 2πrh
4224 = 2×π×r×33
r = (4224)/(2π×33)
r = 64/π
Now, Length of rectangular sheet, l = 2πr
l = 2 π×(64/π) = 128 (using value of r)
So the length of the rectangular sheet is 128 cm.
Also, Perimeter of rectangular sheet = 2(l+b)
The 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, d = 84 cm
Radius of road roller, r = d/2 = 84/2 = 42 cm
Length of road roller, h = 1 m = 100 cm
Formula for Curved surface area of road roller = 2πrh
= 2×(22/7)×42×100 = 26400
Curved surface area of road roller is 26400 cm2
Again, Area covered by road roller in 750 revolutions = 26400×750cm2
= 1980 m2 [∵ 1 m2= 10,000 cm2]
Hence the area of the road is 1980 m2.
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?
Solution:Diameter of cylindrical container , d = 14 cm
Radius of cylindrical container, r = d/2 = 14/2 = 7 cm
Height of cylindrical container = 20 cm
Height of the label, say h = 20–2–2 (from the figure)
= 16 cm
Curved surface area of label = 2πrh
Hence, the area of the label is 704 cm2.
Chapter 11 Mensuration, exercise 11.3 is about identifying solid shapes, the surface area of 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 practice NCERT Class 8 Maths Solutions and improve your skills. Students will find it extremely easy to understand the questions and how to go about solving them.