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 walls.

1. Area of the sheet required (Surface area of the open box): The surface area of an open box is the sum of the area of the base and the area of the four vertical walls. Area = Area of base + Area of four walls Area = $lb\; +\; 2(bh\; +\; hl)$ Substitute the given values: Area = $(1.5\; \backslash times\; 1.25)\; +\; 2((1.25\; \backslash times\; 0.65)\; +\; (0.65\; \backslash times\; 1.5))$ m² Area = $1.875\; +\; 2(0.8125\; +\; 0.975)$ m² Area = $1.875\; +\; 2(1.7875)$ m² Area = $1.875\; +\; 3.575$ m² Area = $5.45$ m² Therefore, 5.45 m² of plastic sheet is required.
2. Cost of the sheet: The cost of 1 m² of the sheet is Rs 20. The total area of the sheet required is 5.45 m². Total cost = Area × Cost per m² Total cost = $5.45\; \backslash times\; 20$ Rs Total cost = Rs 109 Hence, the cost of the sheet is Rs 109.

Given the dimensions of the room:

Length (l) = 5 m

Breadth (b) = 4 m

Height (h) = 3 m

We need to find the cost of white washing the 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 to be white washed = Area of four walls + Area of ceiling

Area = 2h(l + b) + lb

Substitute the given values:

Area = [2 \times 3 (5 + 4) + (5 \times 4)] m²

Area = [6 \times 9 + 20] m²

Area = [54 + 20] m²

Area = 74 m²

The rate of white washing is Rs 7.50 per m².

Total cost = Area × Rate

Total cost = 74 \times 7.50 Rs

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.

The formula for the perimeter of a rectangle is 2(l + b).

So, 2(l + b) = 250 m.

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

The area of the four walls = Total Cost / Rate per m²

Area of four walls = \frac{15000}{10} m²

Area of four walls = 1500 m².

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

So, we have 2h(l + b) = 1500 m².

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

h \times (2(l + b)) = 1500

h \times 250 = 1500

Now, solve for h:

h = \frac{1500}{250}

h = 6 m

Hence, the height of the hall is 6 m.

Given the dimensions of a brick:

Length (l) = 22.5 cm

Breadth (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) cm²

Total surface area of 1 brick = 2(225 + 75 + 168.75) cm²

Total surface area of 1 brick = 2(468.75) cm²

Total surface area of 1 brick = 937.5 cm².

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

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

Total surface area of 1 brick in m² = \frac{937.5}{10000} m² = 0.09375 m².

The total area that can be painted with the available paint is 9.375 m².

Number of bricks that can be painted = Total paintable area / Surface area of one brick

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

Number of bricks = 100

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

Given dimensions: For the cubical box: edge (a) = 10 cm For the cuboidal box: Length (l) = 12.5 cm, Width (b) = 10 cm, Height (h) = 8 cm

1. Comparison of lateral surface areas: Lateral surface area of the cubical box = $4a^2$ Lateral surface area = $4\; \backslash times\; (10)^2$ cm² = $4\; \backslash times\; 100$ cm² = $400$ cm². Lateral surface area of the cuboidal box = $2h(l\; +\; b)$ Lateral surface area = $2\; \backslash times\; 8\; (12.5\; +\; 10)$ cm² Lateral surface area = $16\; \backslash times\; 22.5$ cm² = $360$ cm². Comparing the two, the cubical box has a greater lateral surface area. Difference = Lateral surface area of cube - Lateral surface area of cuboid Difference = $400\; -\; 360$ cm² = $40$ cm². So, the cubical box has a greater lateral surface area by 40 cm².
2. Comparison of total surface areas: Total surface area of the cubical box = $6a^2$ Total surface area = $6\; \backslash times\; (10)^2$ cm² = $6\; \backslash times\; 100$ cm² = $600$ cm². Total surface area of the cuboidal box = $2(lb\; +\; bh\; +\; hl)$ Total surface area = $2(12.5\; \backslash times\; 10\; +\; 10\; \backslash times\; 8\; +\; 8\; \backslash times\; 12.5)$ cm² Total surface area = $2(125\; +\; 80\; +\; 100)$ cm² Total surface area = $2(305)$ cm² = $610$ cm². Comparing the two, the cubical box has a smaller total surface area. Difference = Total surface area of cuboid - Total surface area of cube Difference = $610\; -\; 600$ cm² = $10$ cm². So, the cubical box has a smaller total surface area by 10 cm².

Given the dimensions of the greenhouse (which is a cuboid): Length (l) = 30 cm Width (b) = 25 cm Height (h) = 25 cm

1. Area of the glass: Since the greenhouse is made entirely of glass panes, including the base, we need to calculate the total surface area of the cuboid. Total surface area = $2(lb\; +\; bh\; +\; hl)$ Total surface area = $2(30\; \backslash times\; 25\; +\; 25\; \backslash times\; 25\; +\; 25\; \backslash times\; 30)$ cm² Total surface area = $2(750\; +\; 625\; +\; 750)$ cm² Total surface area = $2(2125)$ cm² Total surface area = $4250$ cm². Therefore, the area of the glass required is 4250 cm².
2. Length of tape needed for all the 12 edges: A cuboid has 12 edges: 4 edges of length (l), 4 edges of breadth (b), and 4 edges of height (h). Total length of tape needed = Sum of the lengths of all 12 edges Total length = $4l\; +\; 4b\; +\; 4h$ Total length = $4(l\; +\; b\; +\; h)$ Substitute the given values: Total length = $4(30\; +\; 25\; +\; 25)$ cm Total length = $4(80)$ cm Total length = $320$ cm. Hence, 320 cm of tape is needed for all the 12 edges.