The topics discussed in this chapter are as follows:
- Surface Area and Volume of Cuboid
- Surface Area and Volume of Cube
- Surface Area and Volume of Cylinder
- Surface Area and Volume of Right Circular Cone
- Surface Area and Volume of Sphere
- Surface Area and Volume of Hemisphere
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NCERT Solutions for Chapter 13: Surface Areas and Volume
Also Check
Chapter 1: Real Numbers
Chapter 2: Polynomials
Chapter 3: Pair of Linear Equations in Two Variables
Chapter 4: Quadratic Equations
Chapter 5: Arithmetic Progression
Chapter 6: Triangles
Chapter 7: Coordinate Geometry
Chapter 8: Introduction to Trigonometry
Chapter 9: Some Applications of Trigonometry
Chapter 10: Circle
Chapter 11: Constructions
Chapter 12: Areas Related to Circles
Chapter 14: Statistics
Chapter 15: Probability
Check out Frequently Asked Questions (FAQs) for Chapter 13: Surface Areas and Volume
If two cubes of edge 5 cm each are joined end to end, find the surface area of the resulting cuboid.
Total length (l) = 5 + 5 = 10 cm
Breadth (b) = 5 cm, Height (h) = 5 cm
Surface Area = 2 (lb + bh + lh)
= 2(10 × 5 + 5 × 5 + 5 × 10) = 2 × 125 = 250 cm2
Breadth (b) = 5 cm, Height (h) = 5 cm
Surface Area = 2 (lb + bh + lh)
= 2(10 × 5 + 5 × 5 + 5 × 10) = 2 × 125 = 250 cm2
A solid piece of iron in the form of a cuboid of dimension 49 cm × 33 cm × 24 cm is melted to form a solid sphere. Find the radius of sphere.
Volume of iron piece = Volume of the sphere formed
= 49 × 33 × 24 = 4/3 πr2
r = 21 cm
= 49 × 33 × 24 = 4/3 πr2
r = 21 cm
A mason constructs a wall of dimensions 270 cm × 300 cm × 350 cm with the bricks each of size 22.5 cm × 11.25 cm × 8.75 cm and it is assumed that space is covered by the mortar. Find the number of bricks used to construct the wall.
Space occupied with bricks = 7/8 × volume of the wall
= 7/8 × 270 × 300 × 350
= 7/8 × 270 × 300 × 350
Volume and surface area of a solid hemisphere are numerically equal. What is the diameter of hemisphere?
Volume of hemisphere = Surface area of hemisphere
= 2/3πr2 = 3πr2 = , units r = 9/2 units
= 2/3πr2 = 3πr2 = , units r = 9/2 units
The slant height of the frustum of a cone is 5 cm. If the difference between the radii of its two circular ends is 4 cm, find the height of the frustum.
Let r and R be radii of the circular ends of the frustum of the cone.
Then, R – r = 4, l = 5
We know, l2 = (R – r)2 + h2
⇒ 52 = 42 + h2 or h2 = 25 – 16 = 9
⇒ h = 3 cm
Then, R – r = 4, l = 5
We know, l2 = (R – r)2 + h2
⇒ 52 = 42 + h2 or h2 = 25 – 16 = 9
⇒ h = 3 cm
