RD Sharma Solutions Class 9 Chapter 18 consists of various topics which are important from exam perspective. Below are some topics you will be seeing here:
Surface area: The surface area of a solid object is a measure of the total area that the
RD Sharma Solutions Class 9 Chapter 18 consists of various topics which are important from exam perspective. Below are some topics you will be seeing here:
Surface area: The surface area of a solid object is a measure of the total area that the surface of the object occupies.
Volume: Volume is the quantity of three-dimensional space enclosed by closed surface i.e, the space that a substance or shape occupies or contains. The SI unit of volume is ‘cubic meter’.
The volume of a cuboid: A cuboid is a three-dimensional shape, so to find the volume of cuboid we need to know 3 measurements. i.e. Length, Width and Height.
The volume is found using the formula : Volume = Length * Height * Width
Surface Area of a Cuboid: The total surface area of a cuboid is the sum of the areas of its six faces.
Formula : 2 (length.width + width.length + height.length)
Surface Area of a Cube: The surface area of a cube is the area of the six squares that cover it. Formula: 6(Side)2
The formula for Volume of a cube: (Side)3
Our RD Sharma Solutions for Class 9 are designed according to the syllabus prescribed by the Central Board of Secondary Education and in accordance with the guidelines given by CCE. Here students will find three exercises of Class 9 Maths Chapter 18 – Surface Area and Volume of Cuboid and Cube. Click on the below link to access all chapter 18 exercises.
Download PDF of RD Sharma Solutions for Class 9 Maths Chapter 18 Surface Area and Volume of Cuboid and Cube
Access Answers to Maths RD Sharma Chapter 18 Surface Area and Volume of Cuboid and Cube
Exercise 18.1 Page No: 18.14
Question 1: Find the lateral surface area and total surface area of a cuboid of length 80 cm, breadth 40 cm and height 20 cm.
Given, Dimensions of a cuboid:
Length (l) = 80 cm
Breadth (b) = 40 cm
Height (h) = 20 cm
We know that, Total Surface Area = 2[lb + bh + hl]
By substituting the values, we get
Therefore, Total Surface Area = 11200 cm2
Lateral Surface Area = 2[l + b]h
Thus, Lateral Surface Area is 4800 cm2.
Question 2: Find the lateral surface area and total surface area of a cube of edge 10 cm.
Side of a Cube = 10 cm (Given)
Formula for Cube Lateral Surface Area = 4 side2
Cube Lateral Surface Area = 4(10 x 10)
= 400 cm2
Total Surface Area = 6 Side2
= 600 cm2
Question 3: Find the ratio of the total surface area and lateral surface area of a cube.
Total Surface Area of the Cube (TSA) = 6 Side2
Lateral surface area of the Cube (LSA) = 4 Side2
Ratio of TSA and LSA = (6 Side2)/ (4 Side2) = 3/2 or 3:2.
Question 4: Mary wants to decorate her Christmas tree. She wants to place the tree on a wooden block covered with colored paper with a picture of Santa Claus on it. She must know the exact quantity of paper to buy for this purpose. If the box has length, breadth, and height as 80 cm, 40 cm and 20 cm respectively. How many square sheets of paper of side 40 cm would she require?
The dimensions of the wooden block are:
Length (l) = 80cm
Breadth (b) = 40cm
Height (h) = 20cm
Surface Area of the wooden box = 2[lb + bh + hl]
Surface Area of the wooden box is 11200 cm2
The Area of each sheet of the paper = 40×40 cm2 = 1600 cm2
The total number of sheets required = (Surface area of the box )/(Area of one sheet of paper)
Therefore, Marry would require 7 sheets. Answer!!
Question 5: The length, breadth, and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of Rs 7.50 m2.
Formula: Total Area to be washed = lb + 2(l + b)h ….(1)
Where, l = length, b = breadth and h = height.
Length = l = 5 m
Breadth = b = 4 m
Height = h = 3 m
Total area to be white washed = (5×4) + 2(5 + 4)3
Total area to be white washed is 74 m2
Now, cost of white washing 1 m2 is Rs. 7.50 (Given)
Therefore, the cost of white washing 74 m2 = (74 x 7.50)
= Rs. 555
Question 6: Three equal cubes are placed adjacently in a row. Find the ratio of a total surface area of the new cuboid to that of the sum of the surface areas of the three cubes.
Let breadth of the cuboid = a
Then, length of the new cuboid = 3a and
Height of the new cuboid = a
Total surface area of the new cuboid (TSA) = 2(lb+bh+hl)
= 2(3a x a + a x a + a x 3a)
Total Surface area of three cubes = 3 x (6 side2 )
= 3 x 6a2
Therefore, ratio of a total surface area of the new cuboid to that of the sum of the surface areas of the three cubes = 14a2/18a2 = 7/9 or 7:9
Therefore, required ratio is 7:9. Answer.
Question 7: A 4 cm cube is cut into 1 cm cubes. Calculate the total surface area of all the small cubes.
Edge of the cube = 4 cm (Given)
Volume of the cube = Side3 = 43 = 64
Volume of the cube is 64cm3
Edge of the cube = 1 cm3
So, Total number of small cubes = 64cm3/1cm3 = 64
And, total surface area of all the c