NCERT Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.1

NCERT Solutions Class 9 Maths
Chapter – 13 (Surface Areas and Volumes)

The NCERT Solutions in English Language for Class 9 Mathematics Chapter – 13 Surface Areas and Volumes Exercise 13.1 has been provided here to help the students in solving the questions from this exercise.

Chapter 13: Surface Areas and Volumes

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 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 the sheet for it, if a sheet measuring 1m² costs Rs. 20.

Answer – Given :
Length, l = 1.5 m
Breadth, b = 1.25 m
Height, h = 65 cm = 65/100 m = 0.65 m
NCERT Class 9 Solutions Maths
(i) Box is to be open at the top.
Area of sheet required = 2lh + 2bh + lb
= lb + 2(l + b)h
= (1.5 m × 1.25m) + 2 × (1.25 m + 1.5 m) × 0.65 m
= 1.875 m² + 2 × 2.75 m × 0.65 m
= 1.875 m² + 3.575 m²
= 5.45 m²

(ii) Cost of sheet per m² area = Rs.20.
Therefore, the cost of the sheet = Rate of the sheet × Area of the sheet
= ₹ 20/ m² × 5.45 m²
= ₹109

2. The length, breadth and height of a room are 5 m, 4 m and 3 m, respectively. Find the cost of whitewashing the walls of the room and ceiling at the rate of Rs 7.50 per m².
Answer – Given :
Length (l) of room = 5 m
Breadth (b) of room = 4 m
Height (h) of room = 3 m
Surface area of 5 faces = Area of the 4 walls and ceiling = lb + 2(l + b)h
lb + 2(l + b)h = (5 m × 4 m) + 2 × (5 m + 4 m) × 3 m
= 20 m² + 2 × 9 m × 3 m
= 20 m² + 54 m²= 74 m²
The cost of whitewashing the walls of the room and ceiling = Rate × Area
= ₹ 7.50 / m² × 74 m²= ₹ 555
Thus, the cost of whitewashing the walls of the room and the ceiling is ₹555.

3. 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 m² is Rs.15000, find the height of the hall.
[Hint: Area of the four walls = Lateral surface area]
Answer – Given :
The rate of painting is ₹10 / m²
Perimeter of the floor = 250 m
The cost of painting the four walls is ₹15000.
Perimeter of the floor of hall = 2(l + b) = 250 m ——————- (i)
Area of four walls = 2lh + 2bh = 2(l + b)h
Now, Area of four walls = 15000/10 m² = 1500 m²2(l + b)h = 1500 m² [From equation (i)]
250 m × h = 1500 m²h = 1500 m²/250 m = 6 m
Thus, the height of the hall is 6 m.

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?
Answer – Given : Dimensions of the brick 22.5cm × 10cm × 7.5cm
The area which can be painted with the paint available in the container = 9.375m².
The area of each brick to be painted = 2(lb + bh + hl)
= 2 × (22.5 cm × 10 cm + 10 cm × 7.5 cm + 7.5 cm × 22.5 cm)
= 2 × (225 cm² + 75 cm² + 168.75 cm²)
= 2 × 468.75 cm²= 937.5 cm²Number of bricks that can be painted = The area which can be painted with the paint available in the container / The area of each brick
= 9.375 m²/ 937.5 cm²= (9.375 × 10000 cm²) / 937.5 cm² [since 1m² = 10000cm²]
= 100
Thus, the number of bricks that can be painted out of the container is 100.

5. A cubical box has each edge 10 cm, and another cuboidal box is 12.5cm 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?

Answer –
Edge length of the cube, a = 10 cm
Length of the cuboid, l = 12.5 cm
Breadth of the cuboid, b = 10 cm
Height of the cuboid, h = 8 cm

(i) Lateral surface area of the cube = 4a²= 4 × (10 cm)²= 4 × 100 cm²= 400 cm²Lateral surface area of the cuboid = (l + b)h
= 2 × (12.5 cm + 10 cm) × 8 cm
= 2 × 22.5 cm × 8 cm
= 360 cm²We see that, the cubical box has a greater lateral surface area by (400 cm² – 360 cm²) = 40 cm²

(ii) Total surface area of the cube = 6a²= 6 × (10 cm)²= 6 × 100 cm²= 600 cm²Total surface area of the cuboid = 2(lb + bh + hl)
= 2 × (12.5 cm × 10 cm + 10 cm × 8 cm + 8 cm × 12.5 cm)
= 2 × (125 cm² + 80 cm² + 100 cm²)
= 2 × 305 cm²= 610 cm²Cubical box has a smaller total surface area by (610 cm² – 600 cm²) = 10 cm²

Thus, the cubical box has a greater lateral surface area by 40 cm² and the cubical box has a smaller total surface area by 10 cm².

6. A small indoor greenhouse (herbarium) is made entirely of glass panes (including the base) held together with tape. It is 30cm long, 25 cm wide, and 25 cm high.
(i) What is the area of the glass?
(ii) How much tape is needed for all 12 edges?

Answer – Given:
Length, l = 30 cm
Breadth, b = 25 cm
Height, h = 25 cm

(i) The area of the glass = 2(lb + bh + hl)
= 2 × (30 cm × 25 cm + 25 cm × 25 cm + 25 cm × 30 cm)
= 2 × (750 cm² + 625 cm² + 750 cm²)
= 2 × 2125 cm²= 4250 cm²The total surface area of the glass is 4250 cm²

(ii) Length of the tape needed for all the 12 edges = 4(l + b + h)
= 4 × (30 cm + 25 cm + 25 cm)
= 4 × 80 cm
= 320 cm
Therefore, 320 cm tape is required for all 12 edges.

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×20cm×5cm, and the smaller of dimensions 15cm×12cm×5cm. For all the overlaps, 5% of the total surface area is required extra. If the cost of the cardboard is Rs. 4 for 1000 cm², find the cost of cardboard required for supplying 250 boxes of each kind.

Answer –
For Bigger Box:
Let the length, breadth and height of the bigger box be L, B and H respectively.
Length, L = 25 cm
Breadth, B = 20 cm
Height, H = 5 cm
The area of the card board = 2(LB + BH + HL)
= 2 × (25cm × 20cm + 20cm × 5cm + 5cm × 25cm)
= 2 × (500cm² + 100cm² + 125cm²)
= 2 × 725 cm²
= 1450 cm²
For all the overlaps, 5% of the total surface area is required extra. Therefore,
Overlap area = 5% of 1450 cm²
= $\frac{5}{100}$ × 1450cm²
= 72.5 cm²Net area of the card board required for each bigger box = 1450 cm² + 72.5 cm² = 1522.5 cm²
We can now find the area of 250 such boxes and the total cost of the cardboard at ₹4 per 1000 cm^².
Area of card board required for 250 such boxes = 250 × 1522.5 cm² = 380625 cm²
The total cost of the cardboard at ₹4 per 1000 cm² = (4/1000) × 380625 = ₹1522.50

For Smaller Box:
Let the length, breadth and height of the smaller box be l, b and h respectively.
Length, l = 15 cm
Breadth, b = 12 cm
Height, h = 5 cm
The area of the cardboard = 2(lb + bh + hl)
= 2 × (15 cm × 12 cm + 12 cm × 5 cm + 5 cm × 15 cm)
= 2 × (180 cm² + 60 cm² + 75 cm²)
= 2 × 315 cm²
= 630 cm²
For all the overlaps, 5% of the total surface area is required extra. Therefore,
Overlap area = 5% of 630cm²
= $\frac{5}{100}$ × 630 cm²
= 31.5 cm²
Net area of the cardboard required for smaller box = 630 cm² + 31.5 cm² = 661.5 cm²
Area of cardboard required for 250 such boxes = 250 × 661.5 cm² =165375 cm²
The total cost of the cardboard at ₹4 per 1000 cm² = ₹ (4/1000) × 165375 = ₹661.50
Cost of cardboard required for supplying 250 boxes of each kind = ₹1522.50 + ₹661.50 = ₹2184

8. Praveen wanted to make a temporary shelter for her car by making a box-like structure with a tarpaulin that covers all 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 2.5m, with base dimensions 4m×3m?

Answer – Let the length, breadth and height of the shelter be l, b and h respectively.
Length, l = 4 m
Breadth, b = 3 m
Height, h = 2.5 m
The area of the tarpaulin required to make the shelter = lb + 2(l + b)h
= (4 m × 3 m) + 2 × (4 m + 3 m) × 2.5 m
= 12 m² + 2 × 7 m × 2.5 m
= 12 m² + 35 m²
= 47 m²
Hence, 47 m² of tarpaulin will be required.

NCERT Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.1