Question 1.

Find the total surface area of cubes having the following sides:

i. 3 cm

ii. 5 cm

iii. 7.2 m

iv. 6.8 m

v. 5.5 m

Solution:

i. Total surface area of cube = 6l²

= 6 × (3)²

= 6 × 9

= 54 sq. cm.

ii. Total surface area of cube = 6l²

= 6 × 5²

= 6 × 25

= 150 sq. cm.

iii. Total surface area of cube = 6l²

= 6 × (7.2)²

= 6 × 51.84

= 311.04 sq. m.

iv. Total surface area of cube = 6l²

= 6 × (6.8)²

= 6 × 46.24

= 277.44 sq. m.

v. Total surface area of cube = 6l²

= 6 × (5.5)²

= 6 × 30.25

= 181.5 sq. m.

Question 2.

Find the total surface area of the cuboids of length, breadth and height as given below:

i. 12 cm, 10 cm, 5 cm

ii. 5 cm, 3.5 cm, 1.4 cm

iii. 2.5 m, 2 m, 2.4 m

iv. 8 m, 5 m, 3.5 m

Solution:

i. Total surface area of cuboid

= 2 (lb + bh + lh)

= 2 (12 × 10 + 10 × 5 + 12 × 5)

= 2 (120 + 50 + 60)

= 2 × 230

= 460 sq. cm.

ii. Total surface area of cuboid

= 2 (lb + bh + lh)

= 2 (5 × 3.5 + 3.5 × 1.4 + 5 × 1.4)

= 2(17.5 + 4.9 + 7)

= 2 × 29.4

= 58.8 sq. cm.

iii. Total surface area of cuboid = 2 (lb + bh + lh)

= 2(2.5 × 2 + 2 × 2.4 + 2.5 × 2.4)

= 2 (5 + 4.8 + 6)

= 2 × 15.8

= 31.6 sq. m.

iv. Total surface area of cuboid = 2 (lb + bh + lh)

= 2 (8 × 5+ 5 × 3.5 + 8 × 3.5)

= 2(40 + 17.5 + 28)

= 2 × 85.5

= 171 sq. m.

Question 3.

A matchbox is 4 cm long, 2.5 cm broad and 1.5 cm in height. Its outer sides are to be covered exactly with craft paper. How much paper will be required to do so?

Solution:

Length of the matchbox (l) = 4 cm, breadth (b) = 2.5 cm, height (h) = 1.5 cm

∴ Total surface area of the matchbox = 2 (lb + bh + lh)

= 2 (4 × 2.5 + 2.5 × 1.5 + 4 × 1.5)

= 2 (10 + 3.75 + 6)

= 2 × 19.75

= 39.5 sq. cm.

∴ 39.5 sq. cm paper will be required.

Question 4.

An open box of length 1.5 m, breadth 1 m, and height 1 m is to be made for use on a trolley for carrying garden waste. How much sheet metal will be required to make this box? The inside and outside surface of the box is to be painted with rust proof paint. At a rate of Rs 150 per sq. m, how much will it cost to paint the box?

Solution:

Length of the box (l) = 1.5 m, breadth (b) = 1 m, height (h) = 1 m

Since, the box is open at top,

∴ Sheet required to make the box = total surface area of the box – area of the top

= 2 (lb + bh + lh) – lb

= 2lb + 2bh + 2lh – lb

= lb + 2bh + 2lh

= 1.5 × 1 + 2 × 1 × 1 + 2 × 1.5 × 1

= 1.5 + 2 + 3

= 6.5 sq. m.

Since, the inside and outside surface of the box are to be painted.

∴ Area to be painted = 2 × Area of the box = 2 × 6.5 = 13 sq. m.

Total cost of painting = area to be painted × rate per sq. m.

= 13 × 150

= Rs 1950

∴ 6.5 sq. m. sheet of metal will be required and the cost of painting the box will be Rs 1950.

**Intext Questions and Activities**

Question 1.

Measure the length and breadth of the courts laid out for games such as kho-kho, kabaddi, tennis, badminton, etc. Find out their perimeters and areas. (Textbook pg. no. 81)

Solution:

(Students should attempt the above activities on their own)

Question 2.

Take mobile handsets of different sizes and find the area of their screens. (Textbook pg. no. 82)

Solution:

(Students should attempt the above activities on their own)