GSEB Class 8 Maths Solutions Chapter 11 Mensuration Exercise 11.3

Question 1. There are two cuboidal boxes as shown in the adjoining figure. Which box requires the lesser amount of material to make?

Answer:
To find out which box needs less material, we calculate the total surface area for each cuboid.
For cuboid (a):
Length (l) = 60 cm
Breadth (b) = 40 cm
Height (h) = 50 cm
Total surface area = \( 2[lb + bh + hl] \)
= \( 2[(60 \times 40) + (40 \times 50) + (50 \times 60)] \) cm\(^2\)
= \( 2[2400 + 2000 + 3000] \) cm\(^2\)
= \( 2[7400] \) cm\(^2\)
= \( 14800 \) cm\(^2\)

For cuboid (b):
Length (l) = 50 cm
Breadth (b) = 50 cm
Height (h) = 50 cm
Total surface area = \( 2[lb + bh + hl] \)
= \( 2[(50 \times 50) + (50 \times 50) + (50 \times 50)] \) cm\(^2\)
= \( 2[2500 + 2500 + 2500] \) cm\(^2\)
= \( 2[7500] \) cm\(^2\)
= \( 15000 \) cm\(^2\)

Comparing the two, the total surface area of cuboid (a) is \( 14800 \) cm\(^2\) and cuboid (b) is \( 15000 \) cm\(^2\). Since the surface area of cuboid (b) is higher, cuboid (a) will require a smaller amount of material.
In simple words: We calculate the total outside area for both boxes. Box (a) has an area of 14800 cm² and box (b) has an area of 15000 cm². Because box (a) has a smaller surface area, it will need less material to make.

Exam Tip: When comparing material usage, always calculate the total surface area of each object and then compare the final values.

Question 2. A suitcase with measures 80 cm × 48 cm × 24 cm is to be covered with a tarpaulin cloth. How many metres of tarpaulin of width 96 cm is required to cover 100 such suitcases?
Answer:
To begin, let's find the total surface area of one suitcase.
Given dimensions of the suitcase:
Length (l) = 80 cm
Breadth (b) = 48 cm
Height (h) = 24 cm
Total surface area of a suitcase = \( 2(lb + bh + hl) \)
= \( 2(80 \times 48 + 48 \times 24 + 24 \times 80) \)
= \( 2(3840 + 1152 + 1920) \) cm\(^2\)
= \( 2(6912) \) cm\(^2\)
= \( 13824 \) cm\(^2\)

Next, we calculate the total surface area for 100 suitcases.
Total S.A. of 100 suitcases = \( 100 \times 13824 \) cm\(^2\)
= \( 1382400 \) cm\(^2\)

Now, consider the tarpaulin cloth.
The tarpaulin's width is given as 96 cm.
Let the required length of the tarpaulin be L cm.
Surface area of tarpaulin = Length \( \times \) Width = L \( \times \) 96 cm\(^2\)

To cover 100 suitcases, the tarpaulin's area must equal the total surface area of the suitcases.
So, L \( \times \) 96 = \( 1382400 \)
\( L = \frac{1382400}{96} \)
\( L = 14400 \) cm

Finally, convert the length from centimetres to metres.
\( 1 \) metre = \( 100 \) cm
\( L = \frac{14400}{100} \) metres
\( L = 144 \) metres

Therefore, 144 metres of tarpaulin will be needed to cover 100 suitcases.
In simple words: First, we found the total outside area of one suitcase. Then, we multiplied it by 100 for all suitcases. Since the tarpaulin is 96 cm wide, we divided the total area by 96 cm to find the length needed. Finally, we changed that length from centimetres to metres.

Exam Tip: Remember to convert all units to be consistent (e.g., all to cm or all to m) before performing calculations, and always double-check the final unit requested in the question.

Question 3. Find the side of a cube whose surface area is 600 cm².
Answer:
Let the side length of the cube be \( x \) cm.
The formula for the total surface area of a cube is \( 6x^2 \).
We are given that the surface area of the cube is \( 600 \) cm\(^2\).
So, we can set up the equation:
\( 6x^2 = 600 \)

To find \( x^2 \), divide both sides by 6:
\( x^2 = \frac{600}{6} \)
\( x^2 = 100 \)

To find \( x \), take the square root of both sides:
\( x = \sqrt{100} \)
\( x = 10 \)

Therefore, the required side of the cube is \( 10 \) cm.
In simple words: We know the total outside area of a cube is 6 times one side multiplied by itself. If the total area is 600 cm², we can find that the side of the cube is 10 cm.

Exam Tip: Recall that all faces of a cube are identical squares. The total surface area is the sum of the areas of these six squares.

Question 4. Rukhsar painted the outside of the cabinet of measure 1 m × 2 m × 1.5 m. How much surface area did she cover if she painted all except bottom of the cabinet?
Answer:
The cabinet is in the shape of a cuboid with the following dimensions:
Length (l) = 2 m
Breadth (b) = 1 m
Height (h) = 1.5 m

Rukhsar painted the entire outside area except for the bottom. This means she painted the area of the four walls and the top.
The total surface area of a cuboid is \( 2(lb + bh + hl) \).
The area of the bottom is \( lb \).
So, the area painted = Total surface area - Area of the bottom
Area painted = \( 2(lb + bh + hl) - lb \)
Area painted = \( 2(bh + hl) + lb \)

Let's substitute the values:
Area painted = \( 2[(1 \times 1.5) + (1.5 \times 2)] + (2 \times 1) \) m\(^2\)
= \( 2[1.5 + 3] + 2 \) m\(^2\)
= \( 2[4.5] + 2 \) m\(^2\)
= \( 9 + 2 \) m\(^2\)
= \( 11 \) m\(^2\)

Therefore, Rukhsar covered \( 11 \) m\(^2\) of surface area.
In simple words: Rukhsar painted a cabinet shaped like a rectangular box. She painted everything except the bottom. So, we calculate the area of the two side walls, the front and back walls, and the top surface. The total area she painted was 11 square metres.

Exam Tip: When painting a specific part of a 3D figure, ensure you only include the areas specified, deducting any unpainted parts from the total surface area.

Question 5. Daniel is painting the walls and ceiling of a cuboidal hall with length, breadth and height of 15 m, 10 m and 7 m respectively. From each can of paint 100 m² of area is painted. How many cans of paint will she need to paint the room?
Answer:
The dimensions of the cuboidal hall are:
Length (l) = 15 m
Breadth (b) = 10 m
Height (h) = 7 m

Daniel needs to paint the four walls and the ceiling.
Area to be painted = Area of 4 walls + Area of ceiling
Area of 4 walls = \( 2 \times (l+b) \times h \)
Area of ceiling = \( l \times b \)

So, Area to be painted = \( 2(l+b)h + lb \)
= \( 2(15+10) \times 7 + (15 \times 10) \) m\(^2\)
= \( 2(25) \times 7 + 150 \) m\(^2\)
= \( 50 \times 7 + 150 \) m\(^2\)
= \( 350 + 150 \) m\(^2\)
= \( 500 \) m\(^2\)

Now, we know that one can of paint covers \( 100 \) m\(^2\) of area.
Number of cans required = \( \frac{\text{Total area to be painted}}{\text{Area covered by one can}} \)
= \( \frac{500}{100} \)
= \( 5 \)

Therefore, Daniel will need 5 cans of paint to paint the room.
In simple words: We added up the area of the hall's four walls and its ceiling. The total area to paint was 500 m². Since one can covers 100 m², Daniel will need 5 cans of paint.

Exam Tip: For painting questions, identify which surfaces are to be covered. Remember that a room's floor is usually not painted when calculating wall and ceiling area.

Question 6. Describe how the two figures at the right are alike and how they are different. Which box has larger surface area?

Answer:
**Similarity:** Both figures have the same height, which is 7 cm.

**Differences:**
* **Cylinder:** It possesses curved and circular surfaces. Its base and top are circles.
* **Cube:** All of its faces are identical squares. It has flat surfaces and straight edges.

Now, let's calculate the lateral surface area for both to determine which has a larger surface area.

**For the Cylinder:**
Height (h) = 7 cm
Diameter = 7 cm, so Radius (r) = \( \frac{7}{2} \) cm = 3.5 cm
Lateral surface area of the cylinder = \( 2\pi rh \)
= \( 2 \times \frac{22}{7} \times \frac{7}{2} \times 7 \) cm\(^2\)
= \( 22 \times 7 \) cm\(^2\)
= \( 154 \) cm\(^2\)

**For the Cube:**
Side length (l) = 7 cm
The cube has 4 lateral faces (excluding top and bottom).
Lateral surface area of the cube = \( 4l^2 \)
= \( 4 \times (7 \times 7) \) cm\(^2\)
= \( 4 \times 49 \) cm\(^2\)
= \( 196 \) cm\(^2\)

Comparing the lateral surface areas:
Cylinder's lateral surface area = \( 154 \) cm\(^2\)
Cube's lateral surface area = \( 196 \) cm\(^2\)

Clearly, the cubical box has a larger lateral surface area.
In simple words: Both shapes are 7 cm tall. The cylinder is round with curved sides, while the cube has flat square sides. When we compare only their side areas, the cube has a bigger side area than the cylinder.

Exam Tip: Ensure you understand the difference between total surface area and lateral/curved surface area. Lateral surface area excludes the top and bottom faces.

Question 7. A closed cylindrical tank of radius 7 m and height 3 m is made from a sheet of metal. How much sheet of metal is required?
Answer:
Given the dimensions of the closed cylindrical tank:
Radius (r) = 7 m
Height (h) = 3 m

Since the tank is closed, the amount of sheet metal needed is equal to its total surface area.
The formula for the total surface area of a closed cylinder is \( 2\pi r(r + h) \).

Let's substitute the given values:
Total surface area = \( 2 \times \frac{22}{7} \times 7 \times (7 + 3) \) m\(^2\)
= \( 2 \times \frac{22}{7} \times 7 \times 10 \) m\(^2\)
= \( 2 \times 22 \times 10 \) m\(^2\)
= \( 44 \times 10 \) m\(^2\)
= \( 440 \) m\(^2\)

Thus, \( 440 \) m\(^2\) of metal sheet is required to make the tank.
In simple words: We need to find the total outside area of the cylindrical tank. Using the radius of 7m and height of 3m, we calculated that 440 square metres of metal sheet are required.

Exam Tip: For a closed cylinder, both the top and bottom circular bases are included in the total surface area calculation, in addition to the curved side surface.

Question 8. The lateral surface area of a hollow cylinder is 4224 cm². It is cut along its height and formed a rectangular sheet of width 33 cm. Find the perimeter of rectangular sheet?
Answer:
When a hollow cylinder is cut along its height and flattened, it forms a rectangular sheet. In this case, the lateral surface area of the cylinder becomes the area of the rectangular sheet.
Given: Lateral surface area of the cylinder = \( 4224 \) cm\(^2\)
This means, Area of the rectangular sheet = \( 4224 \) cm\(^2\)

The width of the rectangular sheet is given as 33 cm.
Let the length of the rectangular sheet be \( l \) cm.

The formula for the area of a rectangle is Length \( \times \) Width.
So, \( l \times 33 = 4224 \)
To find \( l \), divide the area by the width:
\( l = \frac{4224}{33} \)
\( l = 128 \) cm

Now we have the length and width of the rectangular sheet:
Length = 128 cm
Width = 33 cm

The perimeter of a rectangular sheet = \( 2 \times (\text{Length} + \text{Width}) \)
= \( 2 \times (128 + 33) \)
= \( 2 \times 161 \)
= \( 322 \) cm

Thus, the perimeter of the rectangular sheet is \( 322 \) cm.
In simple words: When a hollow cylinder is unrolled, it makes a rectangle. We know its area and width. We used these to find its length. Then, we used the length and width to calculate the perimeter of that rectangular sheet.

Exam Tip: Visualize the transformation: the cylinder's height becomes the rectangle's width (or length), and the cylinder's circumference becomes the rectangle's length (or width).

Question 9. A road roller takes 750 complete revolutions to move once over to level a road. Find the area of the road if the diameter of a road roller is 84 cm and length is 1 m?
Answer:
A road roller is shaped like a cylinder. The area it levels in one revolution is equal to its lateral surface area.

Given dimensions of the road roller:
Diameter = 84 cm
Radius (r) = \( \frac{\text{Diameter}}{2} = \frac{84}{2} = 42 \) cm
Length (which is the height of the cylinder, h) = 1 m. Convert to cm: \( 1 \) m = \( 100 \) cm.
So, h = 100 cm.

First, calculate the lateral surface area of the cylinder:
Lateral surface area = \( 2\pi rh \)
= \( 2 \times \frac{22}{7} \times 42 \times 100 \) cm\(^2\)
= \( 2 \times 22 \times 6 \times 100 \) cm\(^2\)
= \( 44 \times 600 \) cm\(^2\)
= \( 26400 \) cm\(^2\)

This is the area leveled by the roller in one revolution.
The roller takes 750 complete revolutions.

Total area of the road = Area leveled in 1 revolution \( \times \) Number of revolutions
= \( 26400 \times 750 \) cm\(^2\)
= \( 19800000 \) cm\(^2\)

Now, convert the area from cm\(^2\) to m\(^2\).
Since \( 1 \) m = \( 100 \) cm, then \( 1 \) m\(^2\) = \( (100 \times 100) \) cm\(^2\) = \( 10000 \) cm\(^2\).
Total area of the road in m\(^2\) = \( \frac{19800000}{10000} \) m\(^2\)
= \( 1980 \) m\(^2\)

Thus, the required area of the road is \( 1980 \) m\(^2\).
In simple words: The road roller is a cylinder. We first found the area it covers in one spin using its diameter and length. Then, we multiplied that area by 750 because it makes 750 spins. Finally, we changed the total area from square centimetres to square metres.

Exam Tip: Remember to always match units. If the length is in metres and diameter in cm, convert one to match the other before starting calculations to avoid errors.

Question 10. A company packages its milk powder in cylindrical container whose base has a diameter of 14 cm and height 20 cm. Company places a label around the surface of the container as shown in the figure. If the label is placed 2 cm from top and bottom, what is the area of the label?

Answer:
The label covers the lateral surface of a cylinder, but not the entire height of the container.

Given dimensions of the cylindrical container:
Diameter = 14 cm
Radius (r) = \( \frac{14}{2} \) cm = 7 cm
Total Height of the container = 20 cm

The label is placed 2 cm from the top and 2 cm from the bottom of the container.
So, the height of the label (h_label) = Total height - (Space from top + Space from bottom)
h_label = \( 20 - (2 + 2) \) cm
h_label = \( 20 - 4 \) cm
h_label = \( 16 \) cm

The area of the label is the curved surface area of a cylinder with radius \( r \) and height \( h_{label} \).
Area of the label = \( 2\pi r h_{label} \)
= \( 2 \times \frac{22}{7} \times 7 \times 16 \) cm\(^2\)
= \( 2 \times 22 \times 16 \) cm\(^2\)
= \( 44 \times 16 \) cm\(^2\)
= \( 704 \) cm\(^2\)

Therefore, the area of the label is \( 704 \) cm\(^2\).
In simple words: The label wraps around the middle of the cylindrical milk container. We found its height by subtracting the empty space at the top and bottom from the total height. Then, we used this new height and the container's radius to calculate the label's curved area.

Exam Tip: For problems involving labels or partial coverings, always calculate the relevant dimensions (like the effective height for a label) before applying area formulas.