Students can Download Class 8 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 to clear all their doubts, score well in final exams.

## Karnataka State Syllabus 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 l^{2} = 6 x 12^{2} = 6 x 144 = 864 cm^{2}

Volume of cube = P = (12)^{3} = 1728 cm^{3}

Question 2.

Find the volume of a cube whose surface area is 486 cm^{2}.

Answer:

T.S.A of a cube = 486cm²

6^{12} = 486

l² = \(\frac { 486 }{ 6 }\)

l² = 81

∴ l = √81 = 9 cm

Volume=l^{3} = 9^{3}= 729 cm^{3}

Question 3.

A tank, which is cuboidal in shape has volume 6.4 m^{3}. The length and breadth of the base are 2 m and 1.6 m respectively. Find the depth of the tank.

Answer:

V = 6.4m^{3}

l = 2m

b = 1.6m

h = ?

Volume of cuboid = 6.4m^{3}

l × b × h = 6.4

2 × l.6 × h = 6.4

3.2h = 6.4

h = \(\frac { 6.4 }{ 3.2 }\)

h = 2m

h = 2m .

∴ The depth of the tank is 2m.

Question 4.

How many m^{3} of soil has to be excavated from a rectangular well 28 cm deep and whose base dimensions are 10cm and 8m. Also, find the cost of plastering its vertical walls at the rate of Rs. 15/m^{2}.

Answer:

l= 10m

b = 8

h = 28m

Volume of soil = volume of cuboid

= l × b × h = 10 × 8 × 28 = 2240 m^{3}

2240m^{3} of soil has to be excavated.

Area to be plastered = L.S.A of cuboid

= 2h(l + b)

= 2 ×28(10 + 8)

= 56 × 18

= 1008m²

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.

Answer:

The cost of Rs. 500 is for 1 m^{3}.

The cost of Rs. 256 is for

∴ Volume of the wood = 0.512 m^{3}

Volume = 0.512 m

l^{3} = 0.512

l = 3√0.512

l = 0.8 m²

= 0.8 × 100

Length of the side = 80cm, l = 80 cm