KSEEB Solutions for Class 8 Maths Chapter 16 Mensuration Ex 16.2

KSEEB Solutions for Class 8 Maths Chapter 16 Mensuration Ex 16.2

Karnataka Board Class 8 Maths Chapter 16 Mensuration Ex 16.2

Question 1.
Find the total surface area and volume of a cube whose length is 12 cm.
Answer:
l = 12 cm
T.S.A of cube = 6 l² = 6 x 12² = 6 x 144 = 864 cm²
Volume of cube = P = (12)³ = 1728 cm³

Question 2.
Find the volume of a cube whose surface area is 486 cm².
Answer:
T.S.A of a cube = 486 cm²
6¹² = 486
l² = 486/6
l² = 81
∴ l = √81 = 9 cm
Volume = l³ = 9³= 729 cm³

Question 3.
A tank, which is cuboidal in shape has a volume 6.4m³. The length and breadth of the base are 2m and 1.6m respectively. Find the depth of the tank.
Answer:
V = 6.4 m³
l = 2 m
b = 1.6 m
h = ?
Volume of cuboid = 6.4 m³
l × b × h = 6.4
2 × 1.6 × h = 6.4
3.2h = 6.4
h = 6.4/3.2
h = 2m
h = 2m .
∴ The depth of the tank is 2 m.

Question 4.
How many m³ of soil has to be excavated from a rectangular well 28cm deep and whose base dimensions are 10cm and 8m. Also, find the cost of plastering its vertical walls at the rate of Rs. 15/m2.
Answer:
l = 10 m
b = 8
h = 28 m
Volume of soil = volume of cuboid
= l × b × h = 10 × 8 × 28 = 2240 m³
2240 m³ of soil has to be excavated.
Area to be plastered = L.S.A of cuboid
= 2h(l + b)
= 2 × 28(10 + 8)
= 56 × 18
= 1008 m²
The cost of plastering 1m² = Rs. 15
∴ The cost of plastering = 15 × 1008 = Rs. 15,120