GSEB Class 9 Maths Solutions Chapter 13 Surface Areas and Volumes Exercise 13.1

Question 1. A plastic box 1.5 m long, 1.25 m and 65 cm deep is to be made. It is opened at the top. Ignoring the thickness of the plastic sheet, determine
1. The area of the sheet required for making the box.
2. The cost of sheet for it, if a sheet measuring 1 m² costs Rs. 20
Answer:
1. For the plastic box:
Length \( l = 1.5 \) m
Breadth \( b = 1.25 \) m
Height \( h = 65 \) cm \( = 0.65 \) m
The area of the sheet needed for making the box (open at the top) is calculated as:
Area \( = lb + 2(bh + hl) \)
\( = (1.5)(1.25) + 2\{(1.25)(0.65) + (0.65)(1.5)\} \)
\( = 1.875 + 2\{0.8125 + 0.975\} \)
\( = 1.875 + 2(1.7875) \)
\( = 1.875 + 3.575 \)
\( = 5.45 \) m²
So, the area of the sheet required for making this box is \( 5.45 \) m².
2. The cost of the sheet, if 1 m² costs Rs. 20:
Cost \( = 5.45 \times 20 \)
\( = Rs. 109 \)
Therefore, the total cost for the sheet will be Rs. 109.
In simple words: First, figure out the length, width, and height of the box. Since the box is open at the top, calculate the area of the bottom and the four sides. Then, multiply this total area by the given cost per square meter to find the overall cost.

Exam Tip: Remember to convert all dimensions to the same unit (meters or centimeters) before starting calculations to avoid errors. Pay attention to whether the box is open or closed when calculating the surface area.

Question 2. The length, breadth, and height of a room are 5 m, 4 m, and 3m respectively. Find the cost of whitewashing the walls of the room and the ceiling at the rate of Rs. 7.50 perm?
Answer:
Given dimensions for the room:
Length \( l = 5 \) m
Breadth \( b = 4 \) m
Height \( h = 3 \) m
The area of the walls of the room is calculated as:
Area of walls \( = 2(l + b)h \)
\( = 2(5 + 4)3 \)
\( = 2(9)3 \)
\( = 54 \) m²
The area of the ceiling is calculated as:
Area of ceiling \( = lb \)
\( = (5)(4) \)
\( = 20 \) m²
The total area to be whitewashed (walls and ceiling) is:
Total area \( = 54 \) m² \( + 20 \) m² \( = 74 \) m²
The cost of whitewashing the walls and the ceiling at a rate of Rs. 7.50 per m² is:
Cost \( = 74 \times 7.50 \)
\( = Rs. 555 \)
Thus, the total cost to whitewash the room's walls and ceiling is Rs. 555.
In simple words: First, find the area of the four walls and the area of the ceiling. Add these two areas together to get the total area. Then, multiply this total area by the cost per square meter to find the total whitewashing cost.

Exam Tip: Always remember to include the ceiling area when "whitewashing the room and the ceiling" is specified. Double-check that all measurements are in the same units before calculating.

Question 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:
Let the length, breadth, and height of the rectangular hall be \( l \) m, \( b \) m, and \( h \) m respectively.
The perimeter of the floor is given as \( 250 \) m.
Perimeter \( = 2(l + b) = 250 \)

\( \implies l + b = 125 \)
The total cost of painting the four walls is Rs. 15000, and the rate of painting is Rs. 10 per m².
The area of the four walls (lateral surface area) can be found using the total cost and rate:
Area of the four walls \( = \frac {Total \ cost \ of \ painting}{Rate \ of \ painting/m^2} \)
\( = \frac {Rs. 15000}{Rs. 10/m^2} \)
\( = 1500 \) m²
We know the formula for the area of four walls:
Area of four walls \( = 2(l + b)h \)
So, \( 2(l + b)h = 1500 \)
Substitute the value of \( (l + b) \) from the perimeter:
\( 2(125)h = 1500 \)
\( 250h = 1500 \)
\( h = \frac {1500}{250} \)
\( h = 6 \) m
Hence, the height of the hall is \( 6 \) m.
In simple words: First, use the perimeter to find the sum of the room's length and breadth. Next, calculate the area of the four walls by dividing the total painting cost by the cost per square meter. Finally, use the formula for the area of four walls and the sum of length and breadth to figure out the hall's height.

Exam Tip: Remember that the area of the four walls is equivalent to the lateral surface area of the cuboid. Carefully use the given perimeter to find the sum of length and breadth first, which simplifies finding the height.

Question 4. A certain container is sufficient to paint on an area equal to 9.375 m². How many bricks of dimensions 22.5 cm x 10 cm x 7.5 cm can be painted out of this container?
Answer:
The total area that can be painted by the container is \( 9.375 \) m².
For one brick:
Length \( l = 22.5 \) cm
Breadth \( b = 10 \) cm
Height \( h = 7.5 \) cm
First, calculate the total surface area of one brick using the formula \( 2(lb + bh + hl) \):
Total surface area of a brick \( = 2(22.5 \times 10 + 10 \times 7.5 + 7.5 \times 22.5) \)
\( = 2(225 + 75 + 168.75) \)
\( = 2(468.75) \)
\( = 937.5 \) cm²
Now, convert the brick's surface area from cm² to m² for consistency with the paint container's area:
\( 937.5 \) cm² \( = \frac {937.5}{10000} \) m² \( = 0.09375 \) m²
The number of bricks that can be painted from the container is:
Number of bricks \( = \frac {Area \ painting \ by \ 1 \ container \ of \ paint}{Surface \ area \ of \ 1 \ brick} \)
\( = \frac {9.375 \ m^2}{0.09375 \ m^2} \)
\( = 100 \)
Therefore, 100 bricks can be painted from the given container.
In simple words: First, find the total area of one brick, making sure to use the same units (like square meters) as the paint quantity. Then, divide the total area the paint can cover by the area of one brick to find out how many bricks can be painted.

Exam Tip: Always pay close attention to units! Converting all measurements to a consistent unit (like meters) at the beginning can prevent calculation errors. Remember to calculate the total surface area of a brick, not just the lateral surface area.

Question 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.
1. Which box has the greater lateral surface area and by how much?
2. Which box has the smaller total surface area and by how much?
Answer:
Let's calculate the surface areas for both boxes.

For the cubical box:
Edge length \( a = 10 \) cm
1. Lateral surface area (LSA) of the cubical box \( = 4a^2 = 4(10)^2 = 4 \times 100 = 400 \) cm²
2. Total surface area (TSA) of the cubical box \( = 6a^2 = 6(10)^2 = 6 \times 100 = 600 \) cm²

For the cuboidal box:
Length \( l = 12.5 \) cm
Breadth \( b = 10 \) cm
Height \( h = 8 \) cm
1. Lateral surface area (LSA) of the cuboidal box \( = 2(l + b)h \)
\( = 2(12.5 + 10)(8) \)
\( = 2(22.5)(8) \)
\( = 45 \times 8 \)
\( = 360 \) cm²
Comparing the lateral surface areas:
Cubical box LSA \( = 400 \) cm²
Cuboidal box LSA \( = 360 \) cm²
The cubical box has a greater lateral surface area. The difference is \( 400 - 360 = 40 \) cm².

2. Total surface area (TSA) of the cuboidal box \( = 2(lb + bh + hl) \)
\( = 2[(12.5)(10) + (10)(8) + (8)(12.5)] \)
\( = 2[125 + 80 + 100] \)
\( = 2[305] \)
\( = 610 \) cm²
Comparing the total surface areas:
Cubical box TSA \( = 600 \) cm²
Cuboidal box TSA \( = 610 \) cm²
The cubical box has a smaller total surface area. The difference is \( 610 - 600 = 10 \) cm².
In simple words: First, calculate the side area (lateral surface area) for both boxes and see which one is bigger. Then, calculate the full outside area (total surface area) for both boxes and see which one is smaller. For each comparison, find the exact difference between the two boxes.

Exam Tip: Clearly differentiate between lateral surface area (area of walls) and total surface area (area of all faces). Remember the formulas for cubes and cuboids, and organize your calculations neatly for each part of the question.

Question 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.
1. What is the area of the glass?
2. How much of tape is needed for all the 12 edges?
Answer:
Given dimensions of the herbarium:
Length \( l = 30 \) cm
Breadth \( b = 25 \) cm
Height \( h = 25 \) cm

1. The area of the glass required is the total surface area of the cuboid, as it includes the base.
Area of the glass \( = 2(lb + bh + hl) \)
\( = 2[(30)(25) + (25)(25) + (25)(30)] \)
\( = 2[750 + 625 + 750] \)
\( = 2[2125] \)
\( = 4250 \) cm²
So, \( 4250 \) cm² of glass is required.

2. A cuboid has 12 edges: 4 lengths, 4 breadths, and 4 heights. The total length of tape needed is the sum of all edge lengths.
Total length of tape \( = 4(l + b + h) \)
\( = 4(30 + 25 + 25) \)
\( = 4(80) \)
\( = 320 \) cm
So, \( 320 \) cm of tape is needed for all 12 edges.
In simple words: First, find the total glass area by adding up the areas of all six sides of the box. Second, find the total length of tape needed by adding up all the lengths of the edges of the box.

Exam Tip: For the area of glass, use the full total surface area formula since the base is included. For the tape needed, visualize a cuboid's edges and remember there are four of each dimension (length, breadth, height) to sum up.

Question 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 x 20 cm x 5 cm and the smaller of dimensions 15 cm x 12 cm x 5 cm. 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:
Let's calculate the cardboard needed for each box size.

For the bigger box:
Length \( l = 25 \) cm
Breadth \( b = 20 \) cm
Height \( h = 5 \) cm
Total surface area (TSA) of one bigger box \( = 2(lb + bh + hl) \)
\( = 2[(25)(20) + (20)(5) + (5)(25)] \)
\( = 2[500 + 100 + 125] \)
\( = 2[725] \)
\( = 1450 \) cm²
Extra cardboard for overlaps (5% of TSA) \( = 1450 \times \frac {5}{100} = 72.5 \) cm²
Net surface area per bigger box (including overlaps) \( = 1450 + 72.5 = 1522.5 \) cm²
Total cardboard needed for 250 bigger boxes \( = 1522.5 \times 250 = 380625 \) cm²
Cost of cardboard for bigger boxes (at Rs. 4 per 1000 cm²) \( = \frac {4}{1000} \times 380625 = Rs. 1522.50 \)

For the smaller box:
Length \( l = 15 \) cm
Breadth \( b = 12 \) cm
Height \( h = 5 \) cm
Total surface area (TSA) of one smaller box \( = 2(lb + bh + hl) \)
\( = 2[(15)(12) + (12)(5) + (5)(15)] \)
\( = 2[180 + 60 + 75] \)
\( = 2[315] \)
\( = 630 \) cm²
Extra cardboard for overlaps (5% of TSA) \( = 630 \times \frac {5}{100} = 31.5 \) cm²
Net surface area per smaller box (including overlaps) \( = 630 + 31.5 = 661.5 \) cm²
Total cardboard needed for 250 smaller boxes \( = 661.5 \times 250 = 165375 \) cm²
Cost of cardboard for smaller boxes (at Rs. 4 per 1000 cm²) \( = \frac {4}{1000} \times 165375 = Rs. 661.50 \)

Total cost of cardboard required for supplying 250 boxes of each kind:
Total cost \( = \) Cost for bigger boxes \( + \) Cost for smaller boxes
Total cost \( = Rs. 1522.50 + Rs. 661.50 = Rs. 2184 \)
Therefore, the total cost for the cardboard needed is Rs. 2184.
In simple words: First, calculate the total surface area for one big box and one small box. Add 5% extra to each for overlaps. Multiply these new areas by 250 to find the total cardboard for each type of box. Calculate the cost for each type of box. Finally, add these two costs together to get the grand total.

Exam Tip: Break down complex problems into smaller, manageable steps. Calculate the area and cost for each box size separately before summing them up. Remember to account for the extra material needed for overlaps when specified.

Question 8. Parveen wanted to make a temporary shelter for her car, by making a box-like structure with a 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 many tarpaulins would be required to make the shelter of height 2.5 m, with base dimensions 4 m x 3 m?
Answer:
The car shelter needs to cover all four sides and the top. The front face acts as a flap, implying it's also covered. This means we need to find the area of five faces (top + four sides).
Given dimensions for the shelter:
Length \( l = 4 \) m
Breadth \( b = 3 \) m
Height \( h = 2.5 \) m
The total surface area of the shelter (top and four walls) is calculated as:
Area \( = lb + 2(bh + hl) \)
\( = (4)(3) + 2[(3)(2.5) + (2.5)(4)] \)
\( = 12 + 2[7.5 + 10] \)
\( = 12 + 2[17.5] \)
\( = 12 + 35 \)
\( = 47 \) m²
Therefore, \( 47 \) m² of tarpaulin will be required to make the shelter.
In simple words: The car shelter covers the top and all four sides, including the front. So, you need to calculate the area of the top and the four side walls. Add these areas together to find the total amount of tarpaulin required.

Exam Tip: Read the question carefully to identify which surfaces need to be covered. If a 'box-like structure' needs 'four sides and the top' covered, it means five faces in total, not six. Use the appropriate surface area formula for these five faces.