NCERT Class 8 - Exercise 9.2

1. There are two cuboidal boxes as shown in the adjoining figure. Which box requires the lesser amount of material to make?
For the first box (Cuboid A):

Dimensions:

Length \(l = 60\) cm, Breadth \(b = 40\) cm, Height \(h = 50\) cm.

Total Surface Area (TSA):

\( TSA_A = 2(lb + bh + lh) \)
\( TSA_A = 2((60 \times 40) + (40 \times 50) + (60 \times 50)) \)
\( TSA_A = 2(2400 + 2000 + 3000) = 2(7400) = 14800 \) cm\(^2\)
For the second box (Cuboid B):

Dimensions:

Length \(l = 50\) cm, Breadth \(b = 50\) cm, Height \(h = 50\) cm.

Total Surface Area (TSA):

This is a cube with side \(a = 50\) cm.
\( TSA_B = 6a^2 = 6(50^2) = 6(2500) = 15000 \) cm\(^2\)
The first box (Cuboid A) has a smaller surface area (14800 cm\(^2\)) compared to the second box (15000 cm\(^2\)), so it requires a lesser amount of material to make.
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?

Step 1: Find the surface area of one suitcase

Dimensions of one suitcase: \(l = 80\) cm, \(b = 48\) cm, \(h = 24\) cm.
\( \text{Area of one suitcase} = 2(lb + bh + lh) \)
\( = 2((80 \times 48) + (48 \times 24) + (80 \times 24)) \)
\( = 2(3840 + 1152 + 1920) = 2(6912) = 13824 \) cm\(^2\)

Step 2: Find the total area for 100 suitcases

\( \text{Total area} = 100 \times 13824 = 1,382,400 \) cm\(^2\)

Step 3: Find the required length of tarpaulin

Width of tarpaulin = 96 cm. Let the required length be \(x\) cm.
Area of tarpaulin = \( \text{Length} \times \text{Width} \)
\( 1,382,400 = x \times 96 \)
\( x = \frac{1,382,400}{96} = 14400 \) cm

Step 4: Convert the length to metres

Since 1 m = 100 cm, we divide by 100.
\( \text{Length in metres} = \frac{14400}{100} = 144 \) m
144 metres of tarpaulin are required to cover 100 suitcases.
3. Find the side of a cube whose surface area is 600 cm\(^2\).

Given:

Total Surface Area (TSA) = 600 cm\(^2\).

Formula:

The formula for the surface area of a cube is \( TSA = 6a^2 \), where \(a\) is the side length.

Calculation:

\( 600 = 6a^2 \)
\( a^2 = \frac{600}{6} = 100 \)
\( a = \sqrt{100} = 10 \) cm
The side of the cube is 10 cm.
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 the bottom of the cabinet.

Given:

Dimensions of the cabinet: \(l = 2\) m, \(b = 1\) m, \(h = 1.5\) m.

Formula:

Area to be painted = Total Surface Area of Cabinet - Area of Bottom
Total Surface Area (TSA) of cuboid = \( 2(lb + bh + lh) \)
= \( 2(2 \times 1 + 1 \times 1.5 + 2 \times 1.5) \)
= \( 2(2 + 1.5 + 3) = 2(6.5) = 13 \) m\(^2\)
Area of Bottom = \( l \times b \) (since the bottom is a rectangle with length and breadth)
= \( 2 \times 1 = 2 \) m\(^2\).

Thus, the area she painted

= \( \text{TSA of Cabinet} - \text{Area of Bottom} \)
\[= 13 - 2 = 11 m^2 \]
The surface area she covered is 11 m\(^2\).
5. Dawood 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 m\(^2\) of area is painted. How many cans of paint will he need to paint the room?

Given:

Hall dimensions: \(l = 15\) m, \(b = 10\) m, \(h = 7\) m.
Area covered by one can = 100 m\(^2\).

Step 1: Find the total area to be painted

= The area (TSA) of Room - Area of floor.
Area of Room = 2(lb + bh + lh)
\[= 2((15 \times 10) + (10 \times 7) + (15 \times 7)) \] \[= 2(150 + 70 + 105) = 2(325) = 650 m^2 \] Area of floor = \(l \times b\). \[= 15 \times 10 = 150 m^2 \].
Thus, Total Area to be painted = \( 650 - 150 = 500 m^2 \)

Step 2: Find number of cans needed for painting room:

Area covered by 1 can = \(100 m^2\)


Thus, Number of cans used for painting \(500 m^2\) = \( \frac{\text{Total Area}}{\text{Area per can}} \)
\( = \frac{500}{100} = 5 \)
He will need 5 cans of paint.
6. Describe how the two figures at the right are alike and how they are different. Which box has larger lateral surface area?

Similarities:

Both figures are 3-dimensional solid shapes.

Differences:

- A cube has flat faces, while a cylinder has a curved surface.

- A cube has 6 faces, 12 edges, and 8 vertices.

A cylinder has 3 faces (two circular, one curved), 2 edges (circular), and no vertices.

Comparison of Lateral Surface Area (LSA):

Let's assume the dimensions are consistent (e.g., cube side = 7 cm, cylinder diameter = 7 cm, cylinder height = 7 cm).

Cube:

Side \(a = 7\) cm.
LSA = \(4a^2 = 4(7^2) = 4 \times 49 = 196\) cm\(^2\).
Cylinder:

Radius \(r = 7/2 = 3.5\) cm, Height \(h = 7\) cm.
LSA = \(2\pi rh = 2 \times \frac{22}{7} \times 3.5 \times 7 = 2 \times 22 \times 3.5 = 154\) cm\(^2\).
The cube has a larger lateral surface area (196 cm\(^2\)) than the cylinder (154 cm\(^2\)).
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?

Given:

Radius \( (r) = 7\) m, Height \((h) = 3\) m.

Formula:

The amount of sheet metal required for a closed tank is its Total Surface Area (TSA).
\( TSA = 2\pi r(r + h) \)

Calculation:

\( TSA = 2 \times \frac{22}{7} \times 7 \times (7 + 3) \)
\( = 2 \times 22 \times 10 = 440 \) m\(^2\)
440 m\(^2\) of sheet metal is required.
8. The lateral surface area of a hollow cylinder is 4224 cm\(^2\). It is cut along its height and formed a rectangular sheet of width 33 cm. Find the perimeter of rectangular sheet?

Analysis:

When a cylinder is cut and unrolled, it forms a rectangle.

The area of this rectangle is the Lateral Surface Area (LSA) of the cylinder.

The width of the rectangle is the height of the cylinder,

and the length of the rectangle is the circumference of the cylinder's base.

Given:

Area of rectangular sheet = 4224 cm\(^2\).
Width of rectangular sheet = 33 cm.

Step 1: Find the length of the rectangular sheet

Area = Length \(\times\) Width
\( 4224 = \text{Length} \times 33 \)
\( \text{Length} = \frac{4224}{33} = 128 \) cm

Step 2: Find the perimeter of the rectangular sheet

Perimeter = \(2(\text{Length} + \text{Width})\)
\( = 2(128 + 33) = 2(161) = 322 \) cm
The perimeter of the rectangular sheet is 322 cm.
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.

Analysis:

The area covered by the road roller in one revolution is its Lateral Surface Area (LSA).

Given:

Diameter = 84 cm, so Radius \(r = 84/2 = 42\) cm.
Length (height) \(h = 1\) m = 100 cm.

Step 1: Find the area covered in one revolution

Area per revolution = LSA of cylinder = \(2\pi rh\)
\( = 2 \times \frac{22}{7} \times 42 \times 100 \)
\( = 2 \times 22 \times 6 \times 100 = 26400 \) cm\(^2\)

Step 2: Find the total area of the road

Total area = Area per revolution \(\times\) Number of revolutions
\( = 26400 \times 750 = 19,800,000 \) cm\(^2\)

Step 3: Convert the total area to m\(^2\)

\( 1 \text{ m}^2 = (100 \times 100) \text{ cm}^2 = 10,000 \) cm\(^2\).
\( \text{Total Area in m}^2 = \frac{19,800,000}{10,000} = 1980 \) m\(^2\)
The area of the road is 1980 m\(^2\).
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.

Analysis:

The label is a rectangle with a length equal to the circumference of the cylinder's base and a height that is 4 cm less than the cylinder's height (2 cm from top and 2 cm from bottom).

Given:

Cylinder diameter = 14 cm, so Radius \(r = 14/2 = 7\) cm.
Cylinder height = 20 cm.

Step 1: Find the height of the label

Height of label \(h_{label} = 20 - 2 - 2 = 16\) cm.

Step 2: Find the area of the label

Area of label = (Circumference) \(\times\) (Height of label)
Area = \(2\pi r \times h_{label}\)
\( = 2 \times \frac{22}{7} \times 7 \times 16 \)
\( = 2 \times 22 \times 16 = 44 \times 16 = 704 \) cm\(^2\)
The area of the label is 704 cm\(^2\).