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The length, breadth and height of Kavita's bedroom are 6 m, 4 m and 3 m respectively. It has two equal windows, each of dimensions 1 m × 0.5 m. It also has a door of dimensions 2 m × 1 m.

The length, breadth and height of Kavita's bedroom are 6 m, 4 m and 3 m respectively. It has two equal windows, each of dimensions 1 m × 0.5 m. It also has a door of dimensions 2 m × 1 m. Volume and Surface Area of Solids, R.S. Aggarwal Mathematics Solutions ICSE Class 9.

Based on the above information, answer the following questions:

  1. Area occupied by the door and the two windows is :
    (a) 1 m2
    (b) 2 m2
    (c) 2.5 m2
    (d) 3 m2

  2. Kavita wants to whitewash the four walls of the room. Area to be whitewashed is :
    (a) 60 m2
    (b) 57 m2
    (c) 55 m2
    (d) 50 m2

  3. Square tiles each of side 50 cm are laid on the floor of the room. The number of such tiles laid is :
    (a) 100
    (b) 98
    (c) 96
    (d) 72

  4. Volume of air contained in the room is :
    (a) 72 m3
    (b) 70 m3
    (c) 60 m3
    (d) 52 m3

  5. The length of the longest rod (to the nearest m) that can be placed in the room is :
    (a) 4 m
    (b) 5 m
    (c) 8 m
    (d) 7 m

Mensuration

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Answer

Given,

Length (l) = 6 m

Breadth (b) = 4 m

Height (h) = 3 m

Dimension of each window = 1 m × 0.5 m

Dimension of door = 2 m × 1 m

1. Area of one window = 1 × 0.5 = 0.5 m2

∴ Area of two windows = 2 (1 × 0.5) = 1 m2

Area of one door = 2 × 1 = 2 m2.

Total area = 2 + 1 = 3 m2.

Hence, option (d) is the correct option.

2. Area to be whitewashed = Area of four walls - Area of two windows and a door.

Calculating the area of four walls,

We know that,

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

= 2 × 3 × (6 + 4)

= 6 × 10

= 60 m2.

Area of two windows and a door = 3 m2.

Area to be whitewashed = 60 - 3 = 57 m2.

Hence, option (b) is the correct option.

3. Given,

Tile side = 50 cm = 0.5 m

Calculating the floor area,

Area of floor = Length × Breadth

= 6 × 4

= 24 m2.

Area of square tile = (side)2

= (0.5)2

= 0.25 m2

Number of tiles = Area of floorArea of each tile\dfrac{\text{Area of floor}}{\text{Area of each tile}}

= 240.25\dfrac{24}{0.25}

= 96.

Hence, option (c) is the correct option.

4. Volume of air contained in the room = l × b × h

= 6 × 4 × 3

= 72 m3.

Hence, option (a) is the correct option.

5. Length of the longest rod that can be placed in the room = Diagonal of room

Diagonal (d)=l2+b2+h2=62+42+32=36+16+9=61=7.818 m.\text{Diagonal (d)} = \sqrt{l^2 + b^2 + h^2} \\[1em] = \sqrt{6^2 + 4^2 + 3^2} \\[1em] = \sqrt{36 + 16 + 9} \\[1em] = \sqrt{61} \\[1em] = 7.81 \approx 8 \text{ m}.

Hence, option (c) is the correct option.

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