DAV Public school
Class 8 chapter - 14
Mensuration Worksheet -3
1.
P(i) Total surface of area of the cuboid = 2[lb + bh + lh]
= 2[50 × 40 + 40 × 30 + 50 × 30]
= 2[2000 + 1200 + 1500]
= 2[4700]
= 9400 cm2
(ii) Total surface area of the cube = 6l2
= 6 × (40)2
= 6 × 40 × 40
= 9600 cm2
Hence, (ii) box will need more paper to cover
2. Find the length of the side of a cube whose total surface area measures 600 cm².
Answer 2
Given - total surface area is 600 cm²
Formula for -
The total surface area of the cube = 6l2
⇒ 600 = 6l2
⇒ l2 = 600/6
⇒ l2 = 100
⇒ l = √100
⇒ l = 10 cm
The length of the side of a cube whose total surface area measures 600 cm² is 10 cm.
3. The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the inner of the room and ceiling at the rate of 50 per square metre.
Answer 3
Given- length = 5 m, breadth = 4 m and height = 3 m
∴ Area of the walls and the ceiling to be whitewashed = area of the 4 walls + area of the ceiling
= 2[bh + lh] + lb
= 2[4 × 3 + 5 × 3] + 5 × 4
= 2[12 + 15] + 20
= 54 + 20
= 74 m²
Ceiling at the rate of 50 per square meter.
So, we multiply with area of the walls and the ceiling to be whitewashed.
The cost of whitewashing = 74 × 50 = ₹ 3700.
4. Find the total surface area of a closed cardboard box of length 0.5 m, breadth 25 cm and height 15 cm.
Answer 4
Given - length = 0.5 m = 50 cm, breadth = 25 cm and height = 15 cm
∴ Total surface area = 2[lb + bh + lh]
= 2[50 × 25 + 25 × 15 + 50 × 15]
= 2[1250 + 375 + 750]
= 2[2375]
= 4750 cm²
Hence, T.S.A. = 4750 cm²
5. The dimensions of an oil tin are 26 cm x 26 cm x 45 cm. Find the area of tin sheet required to make 20 such tins.
Answer 5
Given - length= 26 cm, breadth = 26 cm and height = 45 cm
∴ T.S.A. of the tin sheet = 2[lb + bh + lh]
= 2126 × 26 + 26 × 45 + 26 × 45]
= 2[676 + 1170 + 1170]
= 2[3016]
= 6032 cm2
∴ The area of the tin sheet to make 20 such tins
= 6032 × 20
= 120640 cm2 = 12.064 m2.
6. A swimming pool is 20 m in length, 15 m in breadth, 4 m in depth. Find the cost of cementing its floor and walls at 35 per m².
Answer 6
Given - Length - 20m, breadth -15m, depth - 4m
Area of the walls and floor = 2 [lh + bh] + lb
= 2[20 × 4 + 15 × 4] + 20 × 15
= 2[80 + 60] + 300
= 2 × 140 + 300
= 280 + 300
= 580 m2
∴ The cost of cementing = 580 × 35 = ₹ 20300.
7. A cubical box with lid has a length of 30 cm. Find the cost of painting the inside and outside of the box at 2 per m².
Answer 7
Total surface area of the cubical box outside and inner side = 2 × 6l2
= 2 × 6 × 30 × 30
= 10800 cm2
= 1.08 m2
∴ The cost of painting the box on both sides = ₹ 5.50 × 1.08 = ₹ 5.94.
8. Two cubes of side 4 cm are fixed together. Find the total surface area of the new solid formed.
Answer 8
When the two cubes are fixed together the new solid formed will be a cuboid whose length
= 4 + 4 = 8 cm
breadth = 4 cm
height = 4 cm
∴ T.S.A. of the cuboid = 2[lb + bh + lh]
= 2[8 × 4 + 4 × 4 + 8 × 4]
= 2[32 + 16 + 32]
= 2 × 80
= 160 cm2
