Our ncert solutions for Exercise 13.1 Class 9 maths 13. Surface Areas and Volumes - ncert solutions - Toppers Study is the best material for English Medium students cbse board and other state boards students.

# Exercise 13.1 Class 9 maths 13. Surface Areas and Volumes - ncert solutions - Toppers Study

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## Exercise 13.1 Class 9 maths 13. Surface Areas and Volumes - ncert solutions - Toppers Study

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### You can Find Mathematics solution Class 9 Chapter 13. Surface Areas and Volumes

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### Exercise 13.1 class 9 Mathematics Chapter 13. Surface Areas and Volumes

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- Exercise 13.1 Class 9 Maths 13. Surface Areas And Volumes - Ncert Solutions - Toppers Study
- Class 9 Ncert Solutions
- Solution Chapter 13. Surface Areas And Volumes Class 9
- Solutions Class 9
- Chapter 13. Surface Areas And Volumes Exercise 13.1 Class 9

## 13. Surface Areas and Volumes

### | Exercise 13.1 |

## Exercise 13.1 Class 9 maths 13. Surface Areas and Volumes - ncert solutions - Toppers Study

**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 Ţ 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 m^{2}

Cost of 1m^{2 }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 m ^{2}.**

**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 m^{2}

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 m ^{2} is 15000, find the height of the hall.**

**[Hint : Area of the four walls = Lateral surface area.]**

**Solution:**

Perimeter = 250m

10m^{2 }painted = Rs. 15000

Area of four walls = 15000/250 = 1500m^{2}

Area of four walls = lateral surface area

1500cm^{2 }= 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)

⇒ 937.50cm^{2}

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 = 4a^{2}

⇒ 4 × 10^{2}

⇒ 400cm^{3}

Lateral surface area of cuboid = 2(l + b) × h

⇒ 2(12.5 + 10) × 8

⇒ 2(22.5) × 8

⇒ 45 × 8

⇒ 360cm^{2}

Area of cube = 400 – 360

⇒ 40 cm^{2}

Cube has greater lateral surface area by 40cm^{2}

Total surface area of cube = 6a^{2}

⇒ 6 × 10^{2}

⇒ 600cm^{2}

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 cm^{2}

Cuboid has greater total surface area by 10m^{2}

**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)

⇒ 42250cm^{2}

(ii) tape needed for all 12 edges

4(l + b + h)

⇒ 4(30 + 25 + 25)

⇒ 4 × 80

⇒ 320cm^{2}

**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 cm ^{2}, 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)

⇒ 1450cm^{2}

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)

⇒ 630cm^{2}

Total cardboard for both boxes

⇒ 1450 + 630

⇒ 2080cm^{2}

Cardboard for 250 boxes = 250 × 2080

⇒ 520000 cm^{2}

Laps 5% = 5% of 520000

⇒ × 520000

⇒ 26000cm^{2}

Cardboard for purchasing = 520000 +26000

⇒ 546000cm^{2}

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

⇒ 47m^{2}

##### Other Pages of this Chapter: 13. Surface Areas and Volumes

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