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NCERT Solutions class 9 Mathematics 13. Surface Areas and Volumes Exercise 13.1

Detailed NCERT Solutions for 9 Mathematics 13. Surface Areas and Volumes to simplify learning. Understand chapters clearly and practice with free solutions for better results.

NCERT Solutions class 9 Mathematics 13. Surface Areas and Volumes Exercise 13.1

NCERT Solutions class 9 Mathematics 13. Surface Areas and Volumes Exercise 13.1

Detailed NCERT Solutions for 9 Mathematics 13. Surface Areas and Volumes to simplify learning. Understand chapters clearly and practice with free solutions for better results.

9 Mathematics Chapter 13. Surface Areas and Volumes - Exercise 13.1

Preparing for exams becomes easier with Exercise 13.1. Whether you are studying for board exams or mid-term exams, 9 Mathematics Chapter 13. Surface Areas and Volumes solutions provide quick revising points, well-structured answers, and additional practice material to help you score better.

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13. Surface Areas and Volumes

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  Þ 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

 

Other Pages of this Chapter:

📘 Why Exercise 13.1 are Important?

Exercise 13.1 are created by experts to give step-by-step explanations. Around 60–70% of exam questions are based on NCERT concepts. Our 9 Mathematics Chapter 13. Surface Areas and Volumes solutions help you understand the core concepts and practice effectively.

✍️ Quick Revising Points as Notes in Page-1

Revision is the key to exam success. Our notes for 9 Mathematics highlight important formulas, key definitions, and exam-ready points from Chapter 13. Surface Areas and Volumes. These quick revision notes make last-minute preparation easy.

📚 NCERT Exercise Solutions

Every NCERT chapter ends with exercises, and solving them is crucial. Our Exercise 13.1 include complete solutions for 9 Mathematics Chapter 13. Surface Areas and Volumes exercises. With step-by-step answers, you gain clarity and confidence to attempt similar exam questions.

📝 Additional Important Questions & Answers

To boost your preparation, we also provide additional important questions with answers. These are prepared from previous year board papers, sample papers, and important concepts of Chapter 13. Surface Areas and Volumes. Practicing these ensures you are well-prepared for both board and mid-term exams.

🎯 Useful for Board & Mid-Term Exams

Our Exercise 13.1 are useful for both board exams and mid-term exams. For 9 Mathematics, we provide notes, exercises, and important Q&A so that you can revise smartly and write perfect answers in exams.

🌟 Final Words

In short, Exercise 13.1 for 9 Mathematics Chapter 13. Surface Areas and Volumes are a complete study package. With quick revising points, NCERT exercises, and additional important questions, you can prepare effectively for exams. Make these solutions your study companion and excel in your academic journey.

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