AP Board Class 9 Maths Chapter 10 Surface Areas is all about surface areas and volumes. In this chapter, students will mainly learn what is surface area and volume and how to find the surface areas and volumes of 3-dimensional objects like;

- Cones
- Cylinders
- Spheres

## What Is Surface Area?

Surface area is defined as the sum of the total area or region covered by a geometric solid. In simple terms, surface area is the amount of space covered by the outside of a two or three-dimensional shape. It is measured in square meters.

Whereas, volume is the sum of how much space a figure takes. Volume is calculated to find the capacity of a figure. It is measured in cubic units.

### Cylinder

Any solid which has two circular ends along with a curved surface area is said to be a cylinder.

### Cone

Cone is a geometrical figure having a circle as the base and a vertex at the top.

### Sphere

In general terms, a figure that is three-dimensional and circular in shape is called as a sphere. For example a ball. When we consider the geometry aspect, a sphere is defined as the set of points that are equidistant from a fixed point in space.

Students can check out some of the chapter questions along with their solutions below;

**Question 1:** Deepak wanted to make a temporary cover for his bike. He used some tarpaulin to form a box-like structure and cover all the four sides of his bike. However, let us assume that the stitching to be done is very small. So how many tarpaulins would Deepak require to make the cover of height 2 in having 3m × 5m as base dimensions?

**solutions:** Here, height (h) = 2 m

Base dimension = 3 m × 5 m

=Length (1) = 3 m and Breadth (b) = 5 m

âˆµThe structure is like a cuboid.

∴ The surface area of the cuboid, excluding the base.

= [Lateral surface area] + [Area of ceiling]

= [2 (l + b)h] + [lb]

= [2 (3 + 5) × 2] + [3 × 5] m^{2}

= [36] + [15] m^{2}

= 51 m^{2}

Thus, 51 m^{2} tarpaulin would be required.

**Question 2: ** Find the surface area of a sphere with diameter 20 cm.

**solutions:**

Here, Diameter = 20 cm

\(Radius=\frac{20}{2} = 10 cm^{2}\) \(Surface Area = 4\pi r^{2} = 4\times\frac{22}{10}\times \left (10 \right )^{2} cm^{2}\) \(= 4\times\frac{22}{10}\times 10 \times 10 cm^{2}\)88 X 10 cm^{2}

= 880 cm^{2}

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