Surface areas and Volumes Class 10 ||Maths|| Chapter 12 NCERT Notes

Surface Areas and Volumes Class 10 ||Maths|| Chapter 12 NCERT Notes

Surface Areas and Volumes of Different Solids

1. Cuboid

A cuboid is a three-dimensional figure with six rectangular faces. It has three dimensions: length (l), breadth (b), and height (h).

• Total Surface Area (TSA):
  $\text{TSA}=2(lb+bh+hl)$

• Lateral Surface Area (LSA) (excluding top and bottom surfaces):
  $\text{LSA}=2h(l+b)$

• Volume:
  $V=l\times b\times h$

2. Cube

A cube is a special case of a cuboid where all sides are equal, i.e., length = breadth = height = a.

• Total Surface Area (TSA):
  $\text{TSA}=6a^2$

• Lateral Surface Area (LSA):
  $\text{LSA}=4a^2$

• Volume:
  $V=a^3$

3. Cylinder

A cylinder has two parallel circular bases and a curved surface. The radius of the base is r and the height is h.

• Curved Surface Area (CSA):
  $\text{CSA}=2\pi rh$

• Total Surface Area (TSA) (including top and bottom):
  $\text{TSA}=2\pi r(r+h)$

• Volume:
  $V=\pi r^{2}h$

4. Cone

A cone has a circular base and a single vertex. The radius of the base is r, and the slant height is l (with height h).

• Curved Surface Area (CSA):
  $\text{CSA}=\pi rl$

• Total Surface Area (TSA) (including base):
  $\text{TSA}=\pi r(l+r)$

• Volume:
  $V=\frac{1}{3}\pi r^{2}h$

• Relation between slant height, height, and radius (Pythagoras theorem):
  $l^2=h^2+r^2$

5. Sphere

A sphere is a perfectly round three-dimensional object where every point on the surface is equidistant from the center. The radius is r.

• Surface Area:
  $\text{SA}=4\pi r^2$

• Volume:
  $V=\frac{4}{3}\pi r^3$

6. Hemisphere

A hemisphere is half of a sphere, having both a curved surface and a circular base.

• Curved Surface Area (CSA):
  $\text{CSA}=2\pi r^2$

• Total Surface Area (TSA) (including the base):
  $\text{TSA}=3\pi r^2$

• Volume:
  $V=\frac{2}{3}\pi r^3$

7. Frustum of a Cone

A frustum is formed when a cone is cut parallel to its base and the top portion is removed. The base radii are r₁ and r₂, and the slant height is l.

• Curved Surface Area (CSA):
  $\text{CSA}=\pi l(r_1+r_2)$

• Total Surface Area (TSA):
  $\text{TSA}=\pi [r_1^2+r_2^2+l(r_1+r_2)]$

• Volume:
  $V=\frac{1}{3}\pi h(r_1^2+r_2^2+r_1{r}_2)$

Detailed Formula Breakdown and Key Concepts

Curved Surface Area (CSA) vs. Total Surface Area (TSA):

• CSA is the area of only the curved part of the solid, excluding the bases.
• TSA includes both the curved area and the area of the bases (for shapes like cylinders, cones, and hemispheres).

Volume:

• Volume measures the capacity of a solid or how much space it occupies.

• The formulas for volume vary depending on the shape, and each solid's formula involves different geometric dimensions (radius, height, slant height, etc.).

Frustum:

The frustum of a cone is a truncated cone, and its volume and surface area are derived from both the base and top radii. This concept is particularly useful in real-life applications like calculating the volume of objects like buckets, lampshades, etc.

Applications and Problem Solving

The chapter often involves word problems that apply these formulas in real-life contexts. Here’s a step-by-step approach to solving these problems:

1. Understand the Shape:
   Identify the shape involved (cuboid, cylinder, sphere, etc.) and its dimensions from the problem statement.

2. Select the Appropriate Formula:
   Based on whether you're asked for surface area or volume, use the correct formula from the formulas provided above.

3. Substitute the Given Values:
   Input the given dimensions (e.g., radius, height, length) into the formula.

4. Solve and Simplify:
   Carry out the necessary calculations, paying attention to units (e.g., square units for area, cubic units for volume).

Key Examples from NCERT

Example 1: Surface Area of a Cylinder

Given a cylinder with radius r = 7 cm and height h = 10 cm, find the total surface area.

• TSA of Cylinder:
  $\text{TSA}=2\pi r(r+h)$
  Substituting the values,
  $\text{TSA}=2\times 3.14\times 7(7+10)$
  $\text{TSA}=439.6{\text{cm}}^2$

Example 2: Volume of a Cone

Find the volume of a cone with a radius of r = 5 cm and height h = 12 cm.

• Volume of Cone:
  $V=\frac{1}{3}\pi r^{2}h$
  $V=\frac{1}{3}\times 3.14\times 5^2\times 12$
  $V=314{\text{cm}}^3$

Example 3: Frustum of a Cone

A frustum of a cone has base radii of r₁ = 6 cm and r₂ = 4 cm, and a height h = 8 cm. Find its volume.

• Volume of Frustum:
  $V=\frac{1}{3}\pi h(r_1^2+r_2^2+r_1{r}_2)$
  Substituting the values,
  $V=\frac{1}{3}\times 3.14\times 8(6^2+4^2+6\times 4)$
  $V=603.2{\text{cm}}^3$

Practical Applications of Surface Areas and Volumes:

1. Architecture and Construction:
   Surface area and volume calculations help in determining the amount of material required for construction (e.g., paint for walls, concrete for pillars).

2. Packaging Industry:
   Volumes are used to design containers, bottles, and boxes to ensure optimal space usage.

3. Real-life Objects:
   Many household items, from water tanks to vases, rely on the principles of these formulas for their design.

That’s a detailed overview of Surface Areas and Volumes for Class 10! If you need more practice questions, a deeper dive into specific problems, or further clarification on any topic, feel free to ask! ✒️📐📊