# RD Sharma Chapter 18 Class 9 Maths Exercise 18.2 Solutions

RD Sharma Chapter 18 Class 9 Maths Exercise 18.2 Solutions can be comprehensive for a greater exercise on surface area and exercises related to Cuboid volume. This exercise is considered the most effective method. This is because of the strategic and stepwise approach to solving the questions.

It can also help the students obtain a broader understanding of different topics around the surface area and problems related to the Cuboid and answer questions on Cubes. To be specific about this exercise, it is more focused on the cuboid volume calculations and exercises around cubes.

Learn about Class 9 Chapter 18 (Surface Area And Volume Of A Cuboid And Cube)

## Download RD Sharma Chapter 18 Class 9 Maths Exercise 18.2 Solutions PDF

Solutions for Class 9 Maths Chapter 18 Surface Area and Volume of Cuboid and Cube Exercise 18.2

## What are the Topics Covered Under RD Sharma Chapter 18 Class 9 Maths Exercise 18.2 Solutions?

Students are primarily taught about the following topics as part of Class 9 Maths Chapter 18 Cube Exercise 18.2.

• It provides exercises around the surface area of a Cuboid
• It covers the topic around the calculation of the surface area of a cube
• It covers the topics around calculations of lateral surface Area of a Cuboid
• It covers the topics around the Lateral Surface Area of a Cub

Surface Area of a Cuboid

Cuboid comes with six rectangular faces, the surface area of which is calculated upon adding the areas of all six faces. It can be calculated using the formula SA= 2lw+2lh+2hw, where l means length, w means width, and h means the prism’s height.

Surface Area of a Cube

The surface area is calculated through formula A=6a2, where A is the area, and ‘a’ is the length of the edge.

Lateral Surface Area of a Cuboid

The lateral surface area of Cuboid has calculated through the formula Lateral Surface area = 2 height (length + breadth)

Lateral Surface Area of a Cube

The lateral surface area is calculated through formula 4a2, where ‘a’ is the length of each side.

## Important Definitions RD Sharma Chapter 18 Class 9 Maths Exercise 18.2 Solutions

### Surface Area of a Cuboid

Cuboid comes with six rectangular faces, the surface area of which is calculated upon adding the areas of all six faces. It can be calculated using the formula SA= 2lw+2lh+2hw, where l means length, w means width, and h means the prism’s height.

### Surface Area of a Cube

The surface area is calculated through formula A = 6a2, where A is the area, and ‘a’ is the length of the edge.

### Lateral Surface Area of a Cuboid

The lateral surface area of Cuboid has calculated through the formula Lateral Surface area = 2 height (length + breadth)

### Lateral Surface Area of a Cube

The lateral surface area is calculated through formula 4a2, where ‘a’ is the length of each side.

### Examples of RD Sharma Chapter 18 Class 9 Maths Exercise 18.2 Solutions

Ques: A cuboidal water tank measures 5 meters long, 6 meters wide, and 4.5 meters deep. How many liters (l) of water can it hold?

Solution:

Dimensions of the cuboidal tank-

= Length = l = 5m

= Breadth = b = 6m

= Height = h = 4.5m

= Volume of the cuboidal water tank = l x b x h

By equating the values, we get-

= Volume, V = 5× 6× 4.5

= 135

Therefore, Volume of the cuboidal water tank is 135 m3

Converting into liters:

= As 1 m3 = 1000 liters

= So, 135m3 = (135 × 1000) liters

= 135000 liters

Hence, the tank can accommodate 1,35,000 liters of water.

Ques: A cuboidal vessel is 10 meters long and 8 meters wide. How high must it be created to hold 380 cubic meters of a liquid?

Solution:

Dimensions of the cuboidal vessel:

Length = l = 10 m

Breadth = b = 8 m

vessel volume = 380 m3

Let ‘h’ be the height of cuboidal vessel.

Volume of cuboidal vessel = lbh

lbh = 380 m3

or 10× 8× h = 380

or h = (380)/ (10×8)

or h = 4.75

Hence, height of the vessel should be 4.75 m.

Ques: Find the cost of digging a cuboidal hole 8 meters long, 6 meters broad, and 3 meters deep at the rate of Rs 30 per m3.

Solution:

Dimensions of the cuboidal hole-

= Length = l = 8 m

= Breadth = b = 6 m

= Height or depth = h = 3 m

= Volume of the Cuboidal pit = l x b x h

= 8× 6× 3

= 144m3

Volume of the Cuboidal hole is 144 m3

Now, finding the cost-

= Cost of digging 1 m3 hole = Rs. 30 (from question)

= Cost of digging 144 m3 holeit = 144 x 30 = Rs. 4320