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
(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/
There 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/