CBSE Class 9 Maths Chapter 13: Surface Areas and Volumes NCERT Solutions

Given the dimensions of the plastic box:

Length (l) = 1.5 m

Width (b) = 1.25 m

Depth (height, h) = 65 cm = 0.65 m

Since the box is open at the top, we need to calculate the area of the base and the four sides.

1. Area of the sheet required: The surface area of an open-top box is given by the formula: Area = Area of base + Area of four walls. Area = $lb\; +\; 2(bh\; +\; hl)$ Substituting the given values: Area = $(1.5\; \backslash times\; 1.25)\; +\; 2((1.25\; \backslash times\; 0.65)\; +\; (0.65\; \backslash times\; 1.5))$ Area = $1.875\; +\; 2(0.8125\; +\; 0.975)$ Area = $1.875\; +\; 2(1.7875)$ Area = $1.875\; +\; 3.575$ Area = $5.45\; \backslash text\{\; m\}^2$ Therefore, 5.45 m² of plastic sheet is required.
2. Cost of the sheet: Cost of 1 m² of the sheet = Rs 20. Cost of 5.45 m² of the sheet = $5.45\; \backslash times\; 20$ Cost = Rs 109. Hence, the cost of the sheet required is Rs 109.

Given the dimensions of the room:

Length (l) = 5 m

Width (b) = 4 m

Height (h) = 3 m

We need to find the cost of white washing the four walls and the ceiling. The area to be white washed is the sum of the area of the four walls and the area of the ceiling.

Area of the four walls = 2h(l + b)

Area of the ceiling = lb

Total area to be white washed = 2h(l + b) + lb

Substituting the given values:

Total Area = [2 \times 3 (5 + 4) + (5 \times 4)] \text{ m}^2

Total Area = [6 \times 9 + 20] \text{ m}^2

Total Area = [54 + 20] \text{ m}^2

Total Area = 74 \text{ m}^2

The rate of white washing is Rs 7.50 per m².

Total cost of white washing = Total Area × Rate per m²

Total Cost = 74 \times 7.50

Total Cost = Rs 555.

Hence, the cost of white washing the walls and the ceiling of the room is Rs 555.

Let the length, breadth, and height of the rectangular hall be l, b, and h respectively.

Given the perimeter of the floor = 250 m.

Perimeter of a rectangle is given by 2(l + b).

So, 2(l + b) = 250 \text{ m}.

The cost of painting the four walls is Rs 15000 at a rate of Rs 10 per m².

Area of the four walls = \frac{\text{Total Cost}}{\text{Rate per m}^2}

Area of the four walls = \frac{15000}{10} = 1500 \text{ m}^2.

The formula for the area of the four walls of a rectangular hall is 2h(l + b).

So, we have the equation:

2h(l + b) = 1500

We know that 2(l + b) = 250. Substituting this value into the equation:

h \times 250 = 1500

Now, we can solve for h:

h = \frac{1500}{250}

h = 6 \text{ m}

Hence, the height of the hall is 6 m.

Given the dimensions of one brick:

Length (l) = 22.5 cm

Width (b) = 10 cm

Height (h) = 7.5 cm

First, calculate the total surface area of one brick. The formula for the total surface area of a cuboid is 2(lb + bh + hl).

Total surface area of 1 brick = 2((22.5 \times 10) + (10 \times 7.5) + (7.5 \times 22.5)) \text{ cm}^2

Total surface area = 2(225 + 75 + 168.75) \text{ cm}^2

Total surface area = 2(468.75) \text{ cm}^2

Total surface area = 937.5 \text{ cm}^2

Now, convert this area to square meters, as the paint coverage is given in m².

Since 1 m = 100 cm, then 1 m² = (100 cm)² = 10000 cm².

Area in m² = \frac{937.5}{10000} \text{ m}^2 = 0.09375 \text{ m}^2.

The total area that can be painted is 9.375 m².

Number of bricks that can be painted = \frac{\text{Total area to be painted}}{\text{Surface area of one brick}}

Number of bricks = \frac{9.375}{0.09375}

Number of bricks = 100.

Therefore, 100 bricks can be painted out of the container.

Given:

Edge of the cubical box (a) = 10 cm

Dimensions of the cuboidal box:

Length (l) = 12.5 cm

Width (b) = 10 cm

Height (h) = 8 cm

(i) Comparison of Lateral Surface Areas:

Lateral surface area of the cubical box = 4a^2

Lateral surface area (cube) = 4 \times (10)^2 = 4 \times 100 = 400 \text{ cm}^2.

Lateral surface area of the cuboidal box = 2h(l + b)

Lateral surface area (cuboid) = 2 \times 8 (12.5 + 10) = 16 \times 22.5 = 360 \text{ cm}^2.

Comparing the two, the cubical box has a greater lateral surface area.

Difference = Lateral surface area (cube) - Lateral surface area (cuboid)

Difference = 400 - 360 = 40 \text{ cm}^2.

So, the cubical box has a greater lateral surface area by 40 cm².

(ii) Comparison of Total Surface Areas:

Total surface area of the cubical box = 6a^2

Total surface area (cube) = 6 \times (10)^2 = 6 \times 100 = 600 \text{ cm}^2.

Total surface area of the cuboidal box = 2(lb + bh + hl)

Total surface area (cuboid) = 2((12.5 \times 10) + (10 \times 8) + (8 \times 12.5)) \text{ cm}^2

Total surface area (cuboid) = 2(125 + 80 + 100) \text{ cm}^2

Total surface area (cuboid) = 2(305) = 610 \text{ cm}^2.

Comparing the two, the cubical box has a smaller total surface area.

Difference = Total surface area (cuboid) - Total surface area (cube)

Difference = 610 - 600 = 10 \text{ cm}^2.

So, the cubical box has a smaller total surface area by 10 cm².

Given the dimensions of the greenhouse (a cuboid):

Length (l) = 30 cm

Width (b) = 25 cm

Height (h) = 25 cm

(i) Area of the glass:

The greenhouse is made entirely of glass panes, including the base. This means we need to calculate the total surface area of the cuboid.

Total surface area = 2(lb + bh + hl)

Total surface area = 2((30 \times 25) + (25 \times 25) + (25 \times 30)) \text{ cm}^2

Total surface area = 2(750 + 625 + 750) \text{ cm}^2

Total surface area = 2(2125) \text{ cm}^2

Total surface area = 4250 \text{ cm}^2.

Therefore, the area of the glass used is 4250 cm².

(ii) Length of tape needed for all the 12 edges:

A cuboid has 12 edges. These consist of 4 edges of length l, 4 edges of length b, and 4 edges of length h.

Total length of tape needed = Sum of the lengths of all 12 edges

Total length = 4l + 4b + 4h = 4(l + b + h)

Total length = 4(30 + 25 + 25) \text{ cm}

Total length = 4(80) \text{ cm}

Total length = 320 cm.

Hence, 320 cm of tape is needed for all the 12 edges.