Given: Diameter of cylindrical container = 14 cm ∴ Radius of cylindrical container (r) = d/2=14/2=7cm Height of cylindrical container = 20 cm Height of the label (h) = 20 – 2 – 2 = 16 cm Curved surface area of label = 2πrh= (2)22/7x7x16 = 704 cm² Hence, the area of the label of 704 cm² Class 8 MathsRead more
Given: Diameter of cylindrical container = 14 cm
∴ Radius of cylindrical container (r) = d/2=14/2=7cm
Height of cylindrical container = 20 cm
Height of the label (h) = 20 – 2 – 2 = 16 cm
Curved surface area of label = 2πrh= (2)22/7x7x16 = 704 cm²
Hence, the area of the label of 704 cm²
Class 8 Maths Chapter 11 Exercise 11.3 Solution in Video
Given: Diameter of road roller = 84 cm ∴ Radius of road roller (r) = d/2=84/2=42cm Length of road roller (h) = 1 m = 100 cm Curved surface area of road roller = 2πrh=(2)22/7x42x100 = 26400 cm² ∴ Area covered by road roller in 750 revolutions = 26400 x 750 = 1,98,00,000 cm² = 1980 m² [∵ 1m² = 10,000Read more
Given: Diameter of road roller = 84 cm
∴ Radius of road roller (r) = d/2=84/2=42cm
Length of road roller (h) = 1 m = 100 cm
Curved surface area of road roller = 2πrh=(2)22/7x42x100 = 26400 cm²
∴ Area covered by road roller in 750 revolutions = 26400 x 750
= 1,98,00,000 cm²
= 1980 m²
[∵ 1m² = 10,000 cm²]
Hence, the area of the road is 1980 m².
Class 8 Maths Chapter 11 Exercise 11.3 Solution in Video
Given: Lateral surface area of hollow cylinder = 4224 cm² And Height of hollow cylinder = 33 cm Curved surface area of hollow cylinder = 2𝜋rh ⇒ 4224=2x22/7xrx33 ⇒ r= 4224x7/2x22x33=64x7/22 cm Now Length of rectangular sheet = 2𝜋r ⇒ l=2x22/7x64x7/22=128cm Perimeter of rectangular sheet = 2(l+b) = 2(1Read more
Given: Lateral surface area of hollow cylinder = 4224 cm²
And Height of hollow cylinder = 33 cm
Curved surface area of hollow cylinder = 2𝜋rh
⇒ 4224=2×22/7xrx33
⇒ r= 4224×7/2x22x33=64×7/22 cm
Now Length of rectangular sheet = 2𝜋r
⇒ l=2×22/7x64x7/22=128cm
Perimeter of rectangular sheet = 2(l+b)
= 2(128+33)=2×161 = 322cm
Hence, the perimeter of rectangular sheet is 322 cm.
Class 8 Maths Chapter 11 Exercise 11.3 Solution in Video
Given: Radius of cylindrical tank (r) = 7 m Height of cylindrical tank (h) = 3 m Total surface area of cylindrical tank = 2𝜋r(h+r) = 2x22/7x7(3+7) = 44x10 = 440cm² Hence, 440 m² metal sheet is required. Class 8 Maths Chapter 11 Exercise 11.3 Solution in Video for more answers vist to: https://www.tiRead more
Given: 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 = 440cm²
Hence, 440 m² metal sheet is required.
Class 8 Maths Chapter 11 Exercise 11.3 Solution in Video
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 = 2rh𝜋 = 2x22/7x7/2x7=154cm² Now lateral surface area of cube = 4l²=4x(7)²=4x49=196cm² Hence, the cube has larger lateral surface area. Class 8 Maths Chapter 11 ExeRead more
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 = 2rh𝜋 = 2×22/7×7/2×7=154cm²
Now lateral surface area of cube = 4l²=4x(7)²=4×49=196cm²
Hence, the cube has larger lateral surface area.
Class 8 Maths Chapter 11 Exercise 11.3 Solution in Video
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 x 10 + 2 (10 x 7 + 7 x 15) = 150 + 2 (70 + 105) = 150 + 350 = 500 m² Now Required number of cans = Area of hall/Area of one can =500/100=5cms Hence, 5 cansRead more
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 x 10 + 2 (10 x 7 + 7 x 15)
= 150 + 2 (70 + 105) = 150 + 350 = 500 m²
Now Required number of cans = Area of hall/Area of one can =500/100=5cms
Hence, 5 cans are required to paint the room.
Class 8 Maths Chapter 11 Exercise 11.3 Solution in Video
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 x 48 + 48 x 24 + 24 x 80) = 2 (3840 + 1152 + 1920) = 2 x 6912 = 13824 cm² Area of Tarpaulin cloth = Surface area of suitcasRead more
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 x 48 + 48 x 24 + 24 x 80)
= 2 (3840 + 1152 + 1920)
= 2 x 6912 = 13824 cm²
Area of Tarpaulin cloth = Surface area of suitcase
⇒ lxb = 13824
⇒ lx96 = 13824
⇒ l= 13824×96=144cm
Required tarpaulin for 100 suitcases = 144 x 100 = 14400 cm = 144 m
Hence, the tarpaulin cloth required to cover 100 suitcases is 144 m.
Class 8 Maths Chapter 11 Exercise 11.3 Solution in Video
Class 8 Maths Chapter 11 Exercise 11.4 Solution in Video
(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 x 40 + 40 x 50 + 50 x 60) = 2 (2400 + 2000 + 3000) = 2 x 7400 = 14800 cm² (b) Given: Length of cuboidal box (l)= 50 cm BreadRead more
(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 x 40 + 40 x 50 + 50 x 60)
= 2 (2400 + 2000 + 3000)
= 2 x 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 x 50 + 50 x 50 + 50 x 50)
= 2 (2500 + 2500 + 2500)
= 2 x 7500 = 15000 cm²
Hence, the cuboidal box (a) requires the lesser amount of material to make, since
surface area of box (a) is less than that of box (b).
Class 8 Maths Chapter 11 Exercise 11.3 Solution in Video
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?
Given: Diameter of cylindrical container = 14 cm ∴ Radius of cylindrical container (r) = d/2=14/2=7cm Height of cylindrical container = 20 cm Height of the label (h) = 20 – 2 – 2 = 16 cm Curved surface area of label = 2πrh= (2)22/7x7x16 = 704 cm² Hence, the area of the label of 704 cm² Class 8 MathsRead more
Given: Diameter of cylindrical container = 14 cm
∴ Radius of cylindrical container (r) = d/2=14/2=7cm
Height of cylindrical container = 20 cm
Height of the label (h) = 20 – 2 – 2 = 16 cm
Curved surface area of label = 2πrh= (2)22/7x7x16 = 704 cm²
Hence, the area of the label of 704 cm²
Class 8 Maths Chapter 11 Exercise 11.3 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-11/
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.
Given: Diameter of road roller = 84 cm ∴ Radius of road roller (r) = d/2=84/2=42cm Length of road roller (h) = 1 m = 100 cm Curved surface area of road roller = 2πrh=(2)22/7x42x100 = 26400 cm² ∴ Area covered by road roller in 750 revolutions = 26400 x 750 = 1,98,00,000 cm² = 1980 m² [∵ 1m² = 10,000Read more
Given: Diameter of road roller = 84 cm
∴ Radius of road roller (r) = d/2=84/2=42cm
Length of road roller (h) = 1 m = 100 cm
Curved surface area of road roller = 2πrh=(2)22/7x42x100 = 26400 cm²
∴ Area covered by road roller in 750 revolutions = 26400 x 750
= 1,98,00,000 cm²
= 1980 m²
[∵ 1m² = 10,000 cm²]
Hence, the area of the road is 1980 m².
Class 8 Maths Chapter 11 Exercise 11.3 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-11/
The lateral surface area of a hollow cylinder is 4224 cm². It is cut along its height and formed a rectangular sheet of width 33 cm. Find the perimeter of rectangular sheet?
Given: Lateral surface area of hollow cylinder = 4224 cm² And Height of hollow cylinder = 33 cm Curved surface area of hollow cylinder = 2𝜋rh ⇒ 4224=2x22/7xrx33 ⇒ r= 4224x7/2x22x33=64x7/22 cm Now Length of rectangular sheet = 2𝜋r ⇒ l=2x22/7x64x7/22=128cm Perimeter of rectangular sheet = 2(l+b) = 2(1Read more
Given: Lateral surface area of hollow cylinder = 4224 cm²
And Height of hollow cylinder = 33 cm
Curved surface area of hollow cylinder = 2𝜋rh
⇒ 4224=2×22/7xrx33
⇒ r= 4224×7/2x22x33=64×7/22 cm
Now Length of rectangular sheet = 2𝜋r
⇒ l=2×22/7x64x7/22=128cm
Perimeter of rectangular sheet = 2(l+b)
= 2(128+33)=2×161 = 322cm
Hence, the perimeter of rectangular sheet is 322 cm.
Class 8 Maths Chapter 11 Exercise 11.3 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-11/
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?
Given: Radius of cylindrical tank (r) = 7 m Height of cylindrical tank (h) = 3 m Total surface area of cylindrical tank = 2𝜋r(h+r) = 2x22/7x7(3+7) = 44x10 = 440cm² Hence, 440 m² metal sheet is required. Class 8 Maths Chapter 11 Exercise 11.3 Solution in Video for more answers vist to: https://www.tiRead more
Given: 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 = 440cm²
Hence, 440 m² metal sheet is required.
Class 8 Maths Chapter 11 Exercise 11.3 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-11/
Describe how the two figures below are alike and how they are different. Which box has larger lateral surface area?
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 = 2rh𝜋 = 2x22/7x7/2x7=154cm² Now lateral surface area of cube = 4l²=4x(7)²=4x49=196cm² Hence, the cube has larger lateral surface area. Class 8 Maths Chapter 11 ExeRead more
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 = 2rh𝜋 = 2×22/7×7/2×7=154cm²
Now lateral surface area of cube = 4l²=4x(7)²=4×49=196cm²
Hence, the cube has larger lateral surface area.
Class 8 Maths Chapter 11 Exercise 11.3 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-11/
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 m² of area is painted. How many cans of paint will she need to paint the room?
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 x 10 + 2 (10 x 7 + 7 x 15) = 150 + 2 (70 + 105) = 150 + 350 = 500 m² Now Required number of cans = Area of hall/Area of one can =500/100=5cms Hence, 5 cansRead more
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 x 10 + 2 (10 x 7 + 7 x 15)
= 150 + 2 (70 + 105) = 150 + 350 = 500 m²
Now Required number of cans = Area of hall/Area of one can =500/100=5cms
Hence, 5 cans are required to paint the room.
Class 8 Maths Chapter 11 Exercise 11.3 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-11/
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?
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 x 48 + 48 x 24 + 24 x 80) = 2 (3840 + 1152 + 1920) = 2 x 6912 = 13824 cm² Area of Tarpaulin cloth = Surface area of suitcasRead more
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 x 48 + 48 x 24 + 24 x 80)
= 2 (3840 + 1152 + 1920)
= 2 x 6912 = 13824 cm²
Area of Tarpaulin cloth = Surface area of suitcase
⇒ lxb = 13824
⇒ lx96 = 13824
⇒ l= 13824×96=144cm
Required tarpaulin for 100 suitcases = 144 x 100 = 14400 cm = 144 m
Hence, the tarpaulin cloth required to cover 100 suitcases is 144 m.
Class 8 Maths Chapter 11 Exercise 11.3 Solution in Video
Class 8 Maths Chapter 11 Exercise 11.4 Solution in Video
See lessThere are two cuboidal boxes as shown in the adjoining figure. Which box requires the lesser amount of material to make?
(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 x 40 + 40 x 50 + 50 x 60) = 2 (2400 + 2000 + 3000) = 2 x 7400 = 14800 cm² (b) Given: Length of cuboidal box (l)= 50 cm BreadRead more
(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 x 40 + 40 x 50 + 50 x 60)
= 2 (2400 + 2000 + 3000)
= 2 x 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 x 50 + 50 x 50 + 50 x 50)
= 2 (2500 + 2500 + 2500)
= 2 x 7500 = 15000 cm²
Hence, the cuboidal box (a) requires the lesser amount of material to make, since
surface area of box (a) is less than that of box (b).
Class 8 Maths Chapter 11 Exercise 11.3 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-11/