RBSE Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.3

Rajasthan Board RBSE Solutions for Class 8 Maths Chapter 11 Mensuration Ex 11.3 Textbook Exercise Questions and Answers.

RBSE Class 8 Maths Solutions Chapter 11 Mensuration Ex 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:
Total surface area of first box
= 2 (lb + bh + lh)
= 2(60 × 40 + 40 × 50 + 60 × 50) cm²
= 200(24 + 20 + 30) cm²
= 200 × 74 cm²
= 14,800 cm²
Total surface area of second box
= 6l² = 6 × 50 × 50 cm² = 15,000 cm²
Since the total surface area of first box is less than that of the second, therefore the first box, i.e. (a) requires the least amount of material to make.

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:
Total surface area of suitcase
= 2(80 × 48 + 48 × 24 + 80 × 24) cm²
= 2(3840 + 1152 + 1920) cm²
= 2 × 6912 cm²
= 13,824 cm²
Area of 1 metre tarpaulin = (100 × 96) cm²
= 9600 cm²
Surface area of 100 suitcases
= 100 × 13,824 cm²
Tarpaulin required to cover 100 suitcases
= \(\frac{100 \times 13824}{9600}\) metres
= 144 metres

Question 3.
Find the side of a cube whose surface area is 600 cm².
Answer:
Let x be the side of the cube having surface' area 600 cm².
∴ 6x² = 600 ⇒ x² = 100 ⇒ x = 10
Hence, the side of the cube = 10 cm

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 the bottom of the cabinet?

Answer:
Here, l = 2 m, b = 1 m and h = 1.5 m
Area to be painted
= 2bh + 2 lh + lb
= (2 × 1 × 1.5 + 2 × 2 × 1.5 + 2 × 1)m²
= (3 + 6 + 2) m²
= 11 m²

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:
Here, l = 15 m, b = 10 m and h = 7 m .
Area to be painted
= 2 bh + 2 lh + lb
= (2 × 10 × 7 + 2 × 15 × 7 + 15 × 10) m²
= (140 + 210 + 150) m²
= 500 m²
∵ Since each can of paint covers 100 m², therefore number of cans required
= \(\frac{500}{100}\) = 5

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

Answer:
Similarity: Both have same heights.
Difference: One is a cylinder, the other is a cube. Cylinder is a solid obtained by revolving a rectangular lamina about one of its sides. Whereas cube is solid bounded by six squares. Cylinder has two circular faces, whereas cube has six faces.
Lateral surface area of cylinder = 2πrh,
= \(\left(2 \times \frac{22}{7} \times \frac{7}{2} \times 7\right)\)cm² = 154cm²
and, lateral surface area of cube = 6l²
= 6 × (7)² = 6 × 7 × 7 = 6 × 49 = 294 cm²
Hence, cube has larger surface area.

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:
Here, r = 7 m and h = 3 m
Sheet of metal required to make a closed cylinder
= Total surface area of the cylinder
= (2πrh + 2πr²) sq. unit
= (2 × \(\frac{22}{7}\) × 7 × 3 + 2 × \(\frac{22}{7}\) × 7 x 7)m²
= (132 + 308) m² = 440 m²

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:
We have, lateral surface area of a cylinder = 4224 cm²
Let the length of rectangular sheet be ‘x’.
∴ It is cut along the height and a rectangular sheet is formed.
∴ Area of sheet = Lateral surface area of given cylinder
i. e. 33x = 4224
or x = \(\frac{4224}{33}\) = 128 cm
Now, perimeter of rectangular sheet
= 2(l + b) = 2 × (128 + 33) cm
= 2 × 161 cm = 322 cm

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:
Road roller is cylinder whose radius,
r = \(\left(\frac{84}{2}\right)\)cm = 42 cm = 0.42 m
and whose length,
h = 1 m

Curved surface area of the roller
= 2πrh
= (2 × \(\frac{22}{7}\) × 0.42 × 1)m²
= 2.64 m²
∴ Area levelled by the road roller is 750 revolutions = (750 × 2.64) m² = 1980 m²

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:
We know that the given container of milk powder is a cylinder.
Radius = \(\frac{14}{2}\)cm = 7 cm and Height of the level = 20 cm - (2 + 2) cm = 16 cm
[∵ It is given that the label is placed 2 cm from top and bottom]
∴ Area of the label = 2πrh
= 2 × \(\frac{22}{7}\) × 7 × 16 cm²
= 2 × 22 × 16 cm²
= 704 cm²