Class 9 Maths Chapter 13 Exercise 13.1 - Surface Areas and Volumes NCERT Solutions
Get a solid understanding of Class 9 Maths Chapter 13 with our easy-to-follow NCERT Solutions for Exercise 13.1. This exercise focuses on Surface Areas and Volumes, which is important for doing well in your exams. Our Class 9 Chapter 13 Maths Exercise 13.1 Solutions explain each problem in simple, step-by-step language, making it easy for you to understand.
These solutions are created by a CBSE Maths expert, so you can trust that you’re getting clear and accurate help with Surface Areas and Volumes problems.
Q1. A plastic box 1.5 m long, 1.25 m wide and 65 cm deep, is to be made. It is to be open at the top. Ignoring the thickness of the plastic sheet, determine:
(i) The area of the sheet required for making the box.
(ii) The cost of sheet for it, if a sheet measuring 1m2 costs Rs. 20.
Solution:
(i) The area of the sheet required for making the box = The surface area of the box which is open at the top
The length of the box = 1.5 m, breadth = 1.25 m and height = 65 cm = 0.65 m
The surface area of the box which is open at the top, S =2(lb + bh + hl) – lb =lb + 2h(l+b)
S = lb + 2h(l+b)
S =1.5×1.25 + 2×0.65(1.5+ 1.25)
S = 1.875 + 1.30(2.75)
S = 1.875 + 3.575
S =5.45
The area of the sheet required for making the box is 5.45 cm²
The cost of sheet measuring 1m2 costs Rs. 20
Therefore the cost of sheet is
20 × 5.45 = Rs 109
Q2. 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 ceiling at the rate of Rs 7.50 per m².
Solution:
Surface area four walls and ceiling = 2h(l+b) + lb
l= 5 m, b = 4 m,h = 3m
Surface area of four walls = 2×3(5 +4) +5×4= 6× 9+20 = 54+20= 74m²
The rate of white washing the walls of the room is Rs 7.50 per m²
The cost of white washing the walls of the room is = 74 × 7.50 = Rs 555
Q3.The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of Rs.10 per m2is Rs.15000, find the height of the hall.
Solution:
Let the height of the hall is h
The perimeter of rectangular hall is = 2 (Length + Breadh) =250 m
2(l+b) = 250
Total cost of painting the four walls = Rs 15000
The rate of painting the walls = Rs 10 /m²
Surface area of four walls = 15000/10 = 1500 m²
2h(l+b) = 1500
h 2(l+b) =1500
h× 250 =1500
h = 1500/250 =6
Therefore the height of the hall is 6 m.
Q4.The paint in a certain container is sufficient to paint an area equal to 9.375 m². How many bricks of dimensions 22.5 cm×10 cm×7.5 cm can be painted out of this container?
Solution:
Let the number of bricks to be painted are x
It is given that the paint in a certain container is sufficient to paint an area equal to 9.375 m²
The length ,breadth and height of of a brick is 22.5 cm×10 cm×7.5 cm
Total surface area of a brick is =2(lb +bh +hl)
TSA of a brick = 2(22.5 ×10 +10 ×7.5 + 7.5×22.5)
= 2(225 +75 + 168.75)
=2×468.75 =937.5 cm² = 937.5/10000 =0.09375 m²
The number of bricks to be painted = 9.375 /0.09375 = 100
Understanding Class 9 Maths Chapter 13 Exercise 13.1 is important for getting the hang of Surface Areas and Volumes. Our simple, step-by-step solutions for Class 9 Maths Chapter 13 Exercise 13.1 will help you work through each problem easily. With these solutions, Surface Areas and Volumes will feel much easier to learn. Be sure to download the PDF for Class 9 Chapter 13 Maths Exercise 13.1, so you can practice anytime and keep these key concepts handy. Keep practicing, and you’ll do great in your exams!
NCERT Solutions of class 9 maths
| Chapter 1- Number System | Chapter 9-Areas of parallelogram and triangles |
| Chapter 2-Polynomial | Chapter 10-Circles |
| Chapter 3- Coordinate Geometry | Chapter 11-Construction |
| Chapter 4- Linear equations in two variables | Chapter 12-Heron’s Formula |
| Chapter 5- Introduction to Euclid’s Geometry | Chapter 13-Surface Areas and Volumes |
| Chapter 6-Lines and Angles | Chapter 14-Statistics |
| Chapter 7-Triangles | Chapter 15-Probability |
| Chapter 8- Quadrilateral |