Class 10 Maths Surface Areas and Volumes Notes

12.1 Introduction to Surface Areas and Volumes

In this chapter, we learn to calculate the surface areas and volumes of 3D objects (solids). Understanding these concepts is essential for solving problems involving real-world shapes, such as containers, pipes, or buildings. The chapter introduces the surface area and volume of various solids, both individually and when combined.

A solid is a 3D object, and it can be made up of simpler shapes like cylinders, cones, spheres, and cubes. The combination of these shapes leads to real-world objects that require us to find their total surface area or volume.

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12.2 Surface Area of a Combination of Solids

When solids are combined, we need to calculate the total surface area by considering each individual surface involved in the combination. For example, a cone on top of a cylinder or a sphere inside a hemisphere.

Key concepts:

• The surface area of solids like cylinders, cones, and spheres is calculated separately, and then we subtract or add the areas of overlapping regions (if any).
• For a combination of solids, the surface area is the total area that will be exposed to the outside. This includes lateral surface areas and the areas of the bases that are not covered.

Formulas for Surface Areas:

• Surface area of a cylinder: $$\text{Surface Area} = 2\pi r^2 + 2\pi rh$$Surface Area=2πr2+2πrh Where $r$r is the radius, and $h$h is the height.
• Surface area of a cone: $$\text{Surface Area} = \pi r(r + l)$$Surface Area=πr(r+l) Where $r$r is the radius and $l$l is the slant height.
• Surface area of a sphere: $$\text{Surface Area} = 4\pi r^2$$Surface Area=4πr2 Where $r$r is the radius.

Example:
A cone is placed on top of a cylinder. The radius of both the cone and cylinder is $7 \, \text{cm}$7cm, and the height of the cylinder is $10 \, \text{cm}$10cm, while the slant height of the cone is $12 \, \text{cm}$12cm. Find the total surface area of the combination.

Solution:
The total surface area of the combination is the surface area of the cylinder (excluding the top base) plus the surface area of the cone.

• Surface area of the cylinder (excluding the top base): $$2\pi rh = 2\pi \times 7 \times 10 = 140\pi \, \text{cm}^2$$2πrh=2π×7×10=140πcm2
• Surface area of the cone: $$\pi r (r + l) = \pi \times 7 \times (7 + 12) = 133\pi \, \text{cm}^2$$πr(r+l)=π×7×(7+12)=133πcm2

Thus, the total surface area of the combination is:$$140\pi + 133\pi = 273\pi \, \text{cm}^2 \approx 858.2 \, \text{cm}^2$$140π+133π=273πcm2≈858.2cm2

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12.3 Volume of a Combination of Solids

To calculate the volume of a combination of solids, we simply find the volume of each individual solid and add or subtract them based on the configuration of the solids.

Key concepts:

• The volume is the amount of space inside a solid.
• When solids are combined, we need to find the individual volumes and account for any space that overlaps or is not part of the overall solid.

Formulas for Volumes:

• Volume of a cylinder: $$\text{Volume} = \pi r^2 h$$Volume=πr2h Where $r$r is the radius and $h$h is the height.
• Volume of a cone: $$\text{Volume} = \frac{1}{3} \pi r^2 h$$Volume=31​πr2h Where $r$r is the radius and $h$h is the height.
• Volume of a sphere: $$\text{Volume} = \frac{4}{3} \pi r^3$$Volume=34​πr3 Where $r$r is the radius.

Example:
A cylinder and a cone are placed together. The cylinder has a radius of $5 \, \text{cm}$5cm and height of $10 \, \text{cm}$10cm, and the cone has the same radius and a height of $6 \, \text{cm}$6cm. Find the total volume of the combination.

Solution:
The total volume of the combination is the volume of the cylinder plus the volume of the cone.

• Volume of the cylinder: $$\pi r^2 h = \pi \times 5^2 \times 10 = 250\pi \, \text{cm}^3$$πr2h=π×52×10=250πcm3
• Volume of the cone: $$\frac{1}{3} \pi r^2 h = \frac{1}{3} \pi \times 5^2 \times 6 = 50\pi \, \text{cm}^3$$31​πr2h=31​π×52×6=50πcm3

Thus, the total volume of the combination is:$$250\pi + 50\pi = 300\pi \, \text{cm}^3 \approx 942.48 \, \text{cm}^3$$250π+50π=300πcm3≈942.48cm3

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12.4 Summary

In this chapter, we learned the following:

• Surface Area of a Combination of Solids: We calculate the surface area of individual solids and then combine the results. It’s important to subtract the areas of bases that are not exposed to the outside.
• Volume of a Combination of Solids: The volume is the space occupied by the solids. When solids are combined, the total volume is simply the sum of their individual volumes.

MCQs Based on the “Surface Areas and Volumes” Chapter:

1. The formula to calculate the volume of a cylinder is:

a) $\pi r^2 h$πr2h
b) $\frac{1}{3} \pi r^2 h$31​πr2h
c) $4\pi r^3$4πr3
d) $\pi r^2$πr2

Answer: a) $\pi r^2 h$πr2h

2. The total surface area of a combination of a cone and a cylinder is:

a) The surface area of the cylinder and cone added together
b) The surface area of the cone only
c) The surface area of the cylinder only
d) The surface area of the cone subtracted from the cylinder

Answer: a) The surface area of the cylinder and cone added together

3. If the radius of a sphere is 7 cm, the volume of the sphere is:

a) $\frac{4}{3} \pi (7)^3$34​π(7)3 cm³
b) $\pi (7)^3$π(7)3 cm³
c) $\frac{1}{3} \pi (7)^2$31​π(7)2 cm³
d) $4\pi (7)^2$4π(7)2 cm³

Answer: a) $\frac{4}{3} \pi (7)^3$34​π(7)3 cm³

4. If the radius of a cone is 5 cm and its height is 12 cm, the volume of the cone is:

a) $\frac{1}{3} \pi (5)^2 \times 12$31​π(5)2×12 cm³
b) $\pi (5)^2 \times 12$π(5)2×12 cm³
c) $5\pi (12)^2$5π(12)2 cm³
d) $\frac{1}{2} \pi (5)^2 \times 12$21​π(5)2×12 cm³

Answer: a) $\frac{1}{3} \pi (5)^2 \times 12$31​π(5)2×12 cm³

5. The volume of a combination of a cylinder and a cone is:

a) The sum of the individual volumes of the cylinder