# KSEEB Solutions for Class 8 Maths Chapter 16 Mensuration Ex 16.2

Students can Download Maths Chapter 16 Mensuration Ex 16.2 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

## 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.
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.
T.S.A of a cube = 486 cm²
612 = 486
l² = $$\frac{486}{6}$$
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.
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 = $$\frac{6.4}{3.2}$$
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.
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

Question 5.
A solid cubical box of fine wood costs Rs. 256 at the rate of Rs. 500/m3. Find its volume and length of each side.
$$\frac{256}{500}=\frac{128}{250}=\frac{64}{125} \mathrm{m}^{3}$$
∴ Volume of the wood = $$\frac{64}{125} \mathrm{m}^{3}$$ = 0.512 m3
$$l=\sqrt{0.512}$$