NCERT Solutions Class 8th Maths Chapter 9 Mensuration
| Textbook | NCERT |
| Class | 8th |
| Subject | Mathematics |
| Chapter | 9th |
| Chapter Name | Mensuration |
| Category | Class 8th Maths Solutions |
| Medium | English |
| Source | Last Doubt |
NCERT Solutions Class 8th Maths Chapter 2 Mensuration
Chapter – 9
Mensuration
Exercise 9.2
| 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)
= 2×(60×40+40×50+50×60)
= 2×(2400+2000+3000)
= 14800 cm2 (b) Length of cubical box (l) = 50 cm
Breadth of 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
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(80×48+48×24+24×80)
= 2 (3840+1152+1920)
= 2×6912
= 13824
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 )
= 2×1+2(1×1.5+1.5×2)
= 2+2(1.5+3.0)
= 2+9.0
= 11
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 )
= 15×10+2(10×7+7×15)
= 150+2(70+105)
= 150+350
= 500
Now, Required number of cans = Area of hall/Area of one can
= 500/100 = 5
Therefore, 5 cans are required to paint the room.
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, 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? Solution: 
Given: Diameter of cylinder = 7 cm
.. Radius of cylinder (r) = 7/2 cm
And, Height of cylinder (h) = 7 cm
Lateral surface area of cylinder = 2лrh = 2 x 22/7 x 7/2 x7 = 154 cm2
Now, lateral surface area of cube = 4t² = 4x(7)² = 4 x 49 = 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? Solution: 
Radius of cylindrical tank, r = 7 m
Height of cylindrical tank , h = 3 m
Total surface area of cylindrical tank = 2πr(h+r)
= 2×(22/7)×7(3+7)
= 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)
= 2(128+33)
= 322
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. 
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
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
= 1,98,00,000cm2
= 1980 m2 [∵ 1 m2= 10,000 cm2]
Hence the area of the road is 1980 m2.
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
= 1,98,00,000cm2
= 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
= 2×(22/7)×7×16
= 704
Hence, the area of the label is 704 cm2. |
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