Given: Base area of cuboid = 180 cm² and Volume of cuboid = 900 cm³ We know that, Volume of cuboid = lxbxh ⇒ 900 = 180xh [∵Base area = lxb=180(given)] ⇒ h=900/180=5m Hence, the height of cuboid is 5 m. Class 8 Maths Chapter 11 Exercise 11.4 Solution in Video for more answers vist to: https://www.tiwRead more
Given: Base area of cuboid = 180 cm² and Volume of cuboid = 900 cm³
We know that, Volume of cuboid = lxbxh
⇒ 900 = 180xh [∵Base area = lxb=180(given)]
⇒ h=900/180=5m
Hence, the height of cuboid is 5 m.
Class 8 Maths Chapter 11 Exercise 11.4 Solution in Video
Yes, we can say that volume of cylinder B is greater, since radius of cylinder B is greater than that of cylinder A (and square of radius gives more value than previous). Diameter of cylinder A = 7 cm ⇒ Radius of cylinder A = 7/2cm And Height of cylinder A = 14 cm ∴ Volume of cylinder A = πr²h = 22/Read more
Yes, we can say that volume of cylinder B is greater, since radius of cylinder B is
greater than that of cylinder A (and square of radius gives more value than
previous).
Diameter of cylinder A = 7 cm
⇒ Radius of cylinder A = 7/2cm
And Height of cylinder A = 14 cm
∴ Volume of cylinder A = πr²h = 22/7×7/2×7/2×14
= 539 cm³
Now Diameter of cylinder B = 14 cm
⇒ Radius of cylinder B = 14/2 = 7cm
And Height of cylinder B = 7 cm
∴ Volume of cylinder A = πr²h = 22/7x7x7x7
= 1078 cm³
Total surface area of cylinder A = πr(2h+r)
[∵ It is open from top]
22/7×7/2(2×14+7/2)=11x(28+7/2)
= 11×63/2=346.5 cm²
Total surface area of cylinder B = 𝜋r(2h+r) [∵It is open from top]
= 22/7×7(2×7+7)
22x (14+7) = 22×21 = 462cm²
Yes, cylinder with greater volume also has greater surface area.
Class 8 Maths Chapter 11 Exercise 11.4 Solution in Video
Here, 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 x 1 + 2 (1 x 1.5 + 1.5 x 2) = 2 + 2 (1.5 + 3.0) = 2 + 9.0 = 11 m2 Hence, the required surface area of cabinet is 11 m². Class 8 Maths Chapter 11 Exercise 11.3Read more
Here,
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 x 1 + 2 (1 x 1.5 + 1.5 x 2)
= 2 + 2 (1.5 + 3.0) = 2 + 9.0 = 11 m2
Hence, the required surface area of cabinet is 11 m².
Class 8 Maths Chapter 11 Exercise 11.3 Solution in Video
Here Surface area of cube = 600 cm² ⇒ 6l² = 600 ⇒ l²=100 ⇒ l = 10 cm Hence the side of cube is 10 cm Class 8 Maths Chapter 11 Exercise 11.3 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-11/
Here Surface area of cube = 600 cm²
⇒ 6l² = 600 ⇒ l²=100 ⇒ l = 10 cm
Hence the side of cube is 10 cm
Class 8 Maths Chapter 11 Exercise 11.3 Solution in Video
We find area when a region covered by a boundary, such as outer and inner surface area of a cylinder, a cone, a sphere and surface of wall or floor. When the amount of space occupied by an object such as water, milk, coffee, tea, etc. Then we have to find out volume of the object. (a) Volume (b) SurRead more
We find area when a region covered by a boundary, such as outer and inner surface
area of a cylinder, a cone, a sphere and surface of wall or floor.
When the amount of space occupied by an object such as water, milk, coffee, tea, etc. Then we have to find out volume of the object.
(a) Volume (b) Surface are (c) Volume
Class 8 Maths Chapter 11 Exercise 11.4 Solution in Video
Find the height of a cuboid whose base area is 180 cm² and volume is 900 cm³?
Given: Base area of cuboid = 180 cm² and Volume of cuboid = 900 cm³ We know that, Volume of cuboid = lxbxh ⇒ 900 = 180xh [∵Base area = lxb=180(given)] ⇒ h=900/180=5m Hence, the height of cuboid is 5 m. Class 8 Maths Chapter 11 Exercise 11.4 Solution in Video for more answers vist to: https://www.tiwRead more
Given: Base area of cuboid = 180 cm² and Volume of cuboid = 900 cm³
We know that, Volume of cuboid = lxbxh
⇒ 900 = 180xh [∵Base area = lxb=180(given)]
⇒ h=900/180=5m
Hence, the height of cuboid is 5 m.
Class 8 Maths Chapter 11 Exercise 11.4 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-11/
Diameter of cylinder A is 7 cm and the height is 14 cm. Diameter of cylinder B is 14 cm and height is 7 cm. Without doing any calculations can you suggest whose volume is greater? Verify it by finding the volume of both the cylinders. Check whether the cylinder with greater volume also has greater surface area.
Yes, we can say that volume of cylinder B is greater, since radius of cylinder B is greater than that of cylinder A (and square of radius gives more value than previous). Diameter of cylinder A = 7 cm ⇒ Radius of cylinder A = 7/2cm And Height of cylinder A = 14 cm ∴ Volume of cylinder A = πr²h = 22/Read more
Yes, we can say that volume of cylinder B is greater, since radius of cylinder B is
greater than that of cylinder A (and square of radius gives more value than
previous).
Diameter of cylinder A = 7 cm
⇒ Radius of cylinder A = 7/2cm
And Height of cylinder A = 14 cm
∴ Volume of cylinder A = πr²h = 22/7×7/2×7/2×14
= 539 cm³
Now Diameter of cylinder B = 14 cm
⇒ Radius of cylinder B = 14/2 = 7cm
And Height of cylinder B = 7 cm
∴ Volume of cylinder A = πr²h = 22/7x7x7x7
= 1078 cm³
Total surface area of cylinder A = πr(2h+r)
[∵ It is open from top]
22/7×7/2(2×14+7/2)=11x(28+7/2)
= 11×63/2=346.5 cm²
Total surface area of cylinder B = 𝜋r(2h+r) [∵It is open from top]
= 22/7×7(2×7+7)
22x (14+7) = 22×21 = 462cm²
Yes, cylinder with greater volume also has greater surface area.
Class 8 Maths Chapter 11 Exercise 11.4 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-11/
Rukshar painted the outside of the cabinet of measure 1 m x 2 m x 1.5 m. How much surface area did she cover if she painted all except the bottom of the cabinet?
Here, 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 x 1 + 2 (1 x 1.5 + 1.5 x 2) = 2 + 2 (1.5 + 3.0) = 2 + 9.0 = 11 m2 Hence, the required surface area of cabinet is 11 m². Class 8 Maths Chapter 11 Exercise 11.3Read more
Here,
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 x 1 + 2 (1 x 1.5 + 1.5 x 2)
= 2 + 2 (1.5 + 3.0) = 2 + 9.0 = 11 m2
Hence, the required surface area of cabinet is 11 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/
Find the side of a cube whose surface area id 600 cm².
Here Surface area of cube = 600 cm² ⇒ 6l² = 600 ⇒ l²=100 ⇒ l = 10 cm Hence the side of cube is 10 cm Class 8 Maths Chapter 11 Exercise 11.3 Solution in Video for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-11/
Here Surface area of cube = 600 cm²
⇒ 6l² = 600 ⇒ l²=100 ⇒ l = 10 cm
Hence the side of cube is 10 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/
Given a cylindrical tank, in which situation will you find surface are and in which situation volume. (a) To find how much it can hold. (b) Number of cement bags required to plaster it. (c) To find the number of smaller tanks that can be filled with water from it.
We find area when a region covered by a boundary, such as outer and inner surface area of a cylinder, a cone, a sphere and surface of wall or floor. When the amount of space occupied by an object such as water, milk, coffee, tea, etc. Then we have to find out volume of the object. (a) Volume (b) SurRead more
We find area when a region covered by a boundary, such as outer and inner surface
area of a cylinder, a cone, a sphere and surface of wall or floor.
When the amount of space occupied by an object such as water, milk, coffee, tea, etc. Then we have to find out volume of the object.
(a) Volume (b) Surface are (c) Volume
Class 8 Maths Chapter 11 Exercise 11.4 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-11/