13. Surface Areas and Volumes Mathematics Exercise - 13.1 class 9 Maths in English - CBSE Study
NCERT Solutions for Class 9 Mathematics are carefully prepared according to the latest CBSE syllabus and NCERT textbooks to help students understand every concept clearly. These solutions cover all important 13. Surface Areas and Volumes with detailed explanations and step-by-step answers for better exam preparation. Each Exercise 13.1 is explained in simple language so that students can easily grasp the fundamentals and improve their academic performance. The study material is designed to support daily homework, revision practice, and final exam preparation for Class 9 students. With accurate answers, concept clarity, and structured content, these NCERT solutions help learners build confidence and score higher marks in their examinations. Whether you are revising a specific topic or preparing an entire chapter, this resource provides reliable and syllabus-based guidance for complete success in Mathematics.
Class 9 English Medium Mathematics All Chapters:
13. Surface Areas and Volumes
1. Exercise 13.1
EXERCISE 13.1
1. A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is opened 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:
L = 1.5m, B = 1.25m, H = 65cm ? 0.65m
Surface area of cuboid = 2(l + b) × h + lb
⇒ 2(1.5 × 1.25) × 0.65 + 1.5 × 1.25
⇒ 2 × 2.75 × 0.65 × 1.875
⇒ 3.575 × 1.875
⇒ 5.45 m2
Cost of 1m2 sheet = 5.45 × 20
⇒ 109.00
⇒ Rs.109
2. 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 7.50 per m2.
Solution:
L = 5m, B = 4m, h = 3m
Surface area of cuboid = 2(l + b) × h + lb
⇒ 2(5 + 4) × 3 + 5 ×4
⇒ 2 × 9 × 3 + 20
⇒ 54 + 20
⇒ 74 m2
Cost of painting = 74 × 7.50
⇒ Rs. 555.70
3. The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of 10 per m2 is 15000, find the height of the hall.
[Hint : Area of the four walls = Lateral surface area.]
Solution:
Perimeter = 250m
10m2 painted = Rs. 15000
Area of four walls = 15000/250 = 1500m2
Area of four walls = lateral surface area
1500cm2 = 2(l + b) × h
1500 = 250m × h
h = 1500/250
h = 6m
4. 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:
L = 22.5, B = 10cm, H = 7.5cm
Surface area of cuboid = 2(lb + bh + hl)
⇒ 2(22.5 × 10 + 10 × 7.5 + 7.5 × 22.5)
⇒ 2(225 + 75 + 168.75)
⇒ 2(468.75)
⇒ 937.50cm2
No. of bricks =
⇒ 100 bricks
5. A cubical box has each edge 10 cm and another cuboidal box is 12.5 cm long, 10 cm wide and 8 cm high.
(i) Which box has the greater lateral surface area and by how much?
(ii) Which box has the smaller total surface area and by how much?
Solution:
Cube = a = 10cm, cuboid = l = 12.5cm, b = 10cm, h = 8cm
Lateral surface area of cube = 4a2
⇒ 4 × 102
⇒ 400cm3
Lateral surface area of cuboid = 2(l + b) × h
⇒ 2(12.5 + 10) × 8
⇒ 2(22.5) × 8
⇒ 45 × 8
⇒ 360cm2
Area of cube = 400 – 360
⇒ 40 cm2
Cube has greater lateral surface area by 40cm2
Total surface area of cube = 6a2
⇒ 6 × 102
⇒ 600cm2
Total surface area of cuboid = 2(lb + bh + hl)
⇒ 2(12.5 × 10 +10 × 8 + 8 × 12.5)
⇒ 2(125 + 80 + 100)
⇒ 2(305)
⇒ 610
Area of cube = area of cuboid
600 = 610
⇒ 610 – 600
⇒ 10 cm2
Cuboid has greater total surface area by 10m2
6. A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30 cm long, 25 cm wide and 25 cm high.
(i) What is the area of the glass?
(ii) How much of tape is needed for all the 12 edges?
Solution:
(i) area of glass = total surface area of cuboid
2(lb + bh + hl)
⇒ 2(30 × 25 + 25 × 25 + 25 × 30)
⇒ 2(750 + 625 + 750)
⇒ 2(2125)
⇒ 42250cm2
(ii) tape needed for all 12 edges
4(l + b + h)
⇒ 4(30 + 25 + 25)
⇒ 4 × 80
⇒ 320cm2
7. Shanti Sweets Stall was placing an order for making cardboard boxes for packing their sweets. Two sizes of boxes were required. The bigger of dimensions 25 cm × 20 cm × 5 cm and the smaller of dimensions 15 cm × 12 cm × 5 cm. For all the overlaps, 5% of the total surface area is required extra. If the cost of the cardboard is 4 for 1000 cm2, find the cost of cardboard required for supplying 250 boxes of each kind.
Solution:
L = 25cm, B = 20cm, H = 5cm
Total surface area of cuboid = 2(lb + bh + hl)
⇒ 2(25 × 20 + 20 × 5 + 5 × 25)
⇒ 2(500 + 100 + 125)
⇒ 2(725)
⇒ 1450cm2
L = 15cm, B = 12cm, H = 5cm
Total surface area of cuboid = 2(lb + bh + hl)
⇒ 2(15 × 12 + 12 × 5 + 5 ×15)
⇒ 2(180 + 60 + 75)
⇒ 2(315)
⇒ 630cm2
Total cardboard for both boxes
⇒ 1450 + 630
⇒ 2080cm2
Cardboard for 250 boxes = 250 × 2080
⇒ 520000 cm2
Laps 5% = 5% of 520000
⇒ × 520000
⇒ 26000cm2
Cardboard for purchasing = 520000 +26000
⇒ 546000cm2
Cost of cardboard = 546000 ×
⇒ 546 × 4
⇒ Rs. 2184
8. Parveen wanted to make a temporary shelter for her car, by making a box-like structure with tarpaulin that covers all the four sides and the top of the car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small, and therefore negligible, how much tarpaulin would be required to make the shelter of height 2.5 m, with base dimensions 4 m × 3 m?
Solution:
L = 4m, B = 3m, H = 2.5m
Tarpaulin required for shelter = lb + 2bh + 2hl
⇒ 4 × 3 + 2 × 3 × + 2 × × 4
⇒ 12 + 15 + 20
⇒ 47m2
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