Students can Download Maths Chapter 16 Mensuration Additional 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 State Syllabus Class 8 Maths Chapter 16 Mensuration Additional Questions

Question 1.

Three metal cubes whose edges measure 3 cm, 4 cm and 5 cm respectively are melted to form a single cube find (1) side – length (2) total surface area of the new cube. What is the difference between total surface area of the new cube and the sum of total surface areas of the original cubes?

Solution:

Volume of a cube = l^{3}

Volume of cube 1 = 3^{3} = 27 cm^{3}

Volume of cube 2 = 4^{3} = 64 cm^{3}

Volume of cube 3 = 5^{3} = 125 cm^{3}

The volume of 3 cubes = 216 cm^{3}

Volume of the New cube formed = 216 cm^{3}

l^{3} = 216

l = 3√216 =6 cm

∴ (i) side – length = 6 cm

(ii) T.S.A of New cube = 6 l^{2}

= 6 x 6^{2} = 6 x 36 = 216 cm^{2}

T.S.A of cube 1 = 6 × l^{2} = 6 × 3^{2} = 6 × 9 = 54 cm^{2}

T.S.A of cube 2 = 6 × l^{2} – 6 × 42 = 6 × 16 = 96 cm^{2}

T.S.A of cube 3 = 6 × l^{2} = 6 × 5^{2} = 6 × 25 = 150 cm^{2}

Total T.S.A of 3 cubes = 300 cm^{2}

(iii) Difference between the surface areas = 300 – 216 = 84 cm^{2}

Question 2.

Two cubes, each of volume 512 cm^{3} are joined end to end. Find the lateral and total surface areas of resulting cuboid.

Solution:

Volume of a cube = 512 cm^{3}

l^{3} = 512

l = 3√512 = 8 cm

When two cubes joined end to end the length of the resulting cuboid = 8 + 8 = 16 cm, breadth = 8 cm and height = 8 cm.

L.S.A of cuboid = 2h (l + b)

= 2 × 8 (16 + 8) = 16 (24)

= 384 cm^{2}

T.S.A of cuboid = 2(lb + bh + lh)

= 2[16 × 8 + 8 × 8 +16 × 8]

=2 [128 + 64 + 128]

= 2 [320] = 640 cm^{2}

Question 3.

The length, breadth and height of a cuboid,are in the ratio 6 : 5 : 3. If the total surface area is 504 cm^{2} find its dimensions. Also find the volume of the cuboid.

Solution:

l : b : h = 6 : 5 : 3

Let the length, breadth and height be 6x, 5x and 3x respectively.

T.S.A of cuboid = 504 cm^{2}

2(lb + bh + lh) = 504

2[6x × 5x + 5x × 3x + 6x × 3x] = 504

2 [30x^{2} + 15x^{2} + 18x^{2}] = 504

2[63x^{2}] = 504

126x^{2} = 504

x^{2} = \(\frac{504}{126}\) = 4

x = √4 = 2cm

l = 6x = 6 × 2 = 12 cm, b = 5x = 5 × 2 = 10 cm,

h = 3x = 3 × 2 – 6 cm

Volume = lbh = 12 × 10 × 6 = 720 cm^{3}

Question 4.

Find the area of four walls of a room having length, breadth, and heighnt as 8m, 5m and 3m respectively. Find the cost of white washing the walls at the rate of Rs. 15/m^{2}.

Solution:

l = 8m, b = 5m, h = 3m, L.S.A = ?

Area of the 4 walls = L.S.A of cuboid

= 2h (l + b) = 2 × 3. (8 + 5) = 6 (13)

Area of the 4 walls = 78 m^{2}

The cost of white washing 1 m^{2} = Rs.15

The cost of white washing 78 m^{2} = 78 × 15

= Rs. 1170

Question 5.

A room is 6m long, 4m broad and 3m high. Find the cost of laying tiles on its floor and four walls at the cost of Rs. 80/m3.

Solution ‘

l = 6m, b = 4m, h = 3m

Area on. which tiles are to be laid

= Area of the base + LSA = lb + 2h (l + b) ‘

= 6 × 4 + 2 × 3 (6 + 4)

= 24 + 6(10)

= 24 + 60 = 84 m^{2}

The cost for laying tiles for 1 m^{2} = Rs.80

The cost for laying tiles for 84 m^{2}

= 84 × 80 = Rs. 6720

Question 6.

The length, breadth and height of a cuboid are in the ratio 5:3: 2. If its volume is 1078.110 mm^{3} find its dimensions. Also find the total surface area of the cuboid.

Solution:

l : b : h = 5 : 3 :2, Volume = 1078.110 mm^{3}

Volume = 1078.110 mm^{3}

l × b × h = 1078.110

5x × 3x × 2x = 1078.110

x^{3} = \(\frac{1078.110}{30}\) = 35.937

x = 3√35.937

x = 3.3 m

5x = 5 × 3.3 = 16.5 m

3x = 3 × 3.3 = 9.9 m

2x = 2 × 3.3 = 6,6 m

The dimensions are 16.5 m, 9.9m, 6.6 m.

T.S.A =2(lb + bh + lh)

= 2[16.5 × 9.9 + 9.9 × 6.6 +16.5 × 6.6]

= 2[163.35 + 65.34 + 108.9]

= 2[337.59] = 675.18 m^{3}

Question 7.

Suppose the perimeter of one face of a cube is 24 cm what is its volume?

Solution:

Perimeter = 24cm

4l = 24; l = \(\frac{24}{4}\) = 6

l = 6

Volume = l^{3} – 6^{3}

Volume = 216 cm^{3}

Question 8.

A wooden box has inner dimension l = 6m, b = 8m and h = 9m and it has uniform thickness of 10cm. The lateral surface of the outerside has to be painted at the rate of 50/m^{2}. What is the cost of painting?

Solution:

Inner length of 6m. The thickness is 10 cm

∴ the outer length is 6m + 20 cm = 6.2 m

Similarly the outer breadth is 8 m + 20 cm = 8.2 m

The height is 9 m + 20 m = 9.2 m of the box is 9m + 20 cm = 9.2 m

L.S.A of outer surface = 2h (l + b)

= 2 × 9.2 [6.2 + 8.2] = 18.4 (14.4)

= 264.96 m^{2}

Area to be painted = 264.96 m^{2}

The cost of painting 1 sq m = Rs. 50

The cost of painting 264.96 m^{2} = 264.96 × 50 = Rs. 13,248

Question 9.

Each edge of a cube is increased by 20%. What is the percentage increase in the volume of the cube?

Solution:

Let the length of each side by 1 cm

∴ Volume of cube = l^{3} = l^{3} = 1 cm^{3}

If the length is increased by 20% the length of each side is 1.2 cm.

∴ Volume of the new cube is = (1.2)^{3}

= 1.728 cm^{3}

Increase in the volume

= 1.728 – 1 = 0.728

∴ The percentage increase is

= 0.728 × 100 = 72.8%

Question 10.

Suppose the length of a cube is increased by 10% and its breadth is decreased by 10% will the volume of the new cuboid be the same as that of the cube? What about the total surface areas? If they change what would be the percentage change in both the cases?

Solution:

Let the length, breadth and height of the cube be 1 cm.

Length is increased by 10%

∴ length = 1.1 cm Breadth is decreased by 10%

∴ breadth = 0.9cm height remains the same.

Volume of cube = l^{3} – l^{3} = 1 cm^{3}

Volume of cuboid = lbh =1.1 × 0.9 × 1 = 0.99 cm^{3}

Difference in volume = 1 – 0.99 = 0.01 cm^{3}

Percentage in decrease = 0.01 × 100 = 1 %

T.S.A of cube = 6 l^{2} = 6 × 12 = 6 cm^{2}

T.S.A of cuboid = 2 (lb + bh + lh)

= 2 [1.1 × 0:9 + 0.9 × 1 + 1.1 × 1]

= 2 [0.99 + 0.9 + 1.1] = 2 [2.99]

= 5.98 cm^{2}

Difference in T.S.A = 6 – 5.98 = 0.02

Percentage in decrease = 0.02 × 100

= 42%