Class 8 | NCERT Solution Maths Chapter 11 | Mensuration | Exercise 11.3

Exercise 11.3 Page No: 186

1. There are two cuboidal boxes as shown in the adjoining figure. Which box requires the lesser amount of material to make?

Solution:

(a) Given: Length of cuboidal box  (l) = 60 cm

Breadth of cuboidal box  (b)  = 40 cm

Height of cuboidal box  (h)  = 50 cm

Total surface area of cuboidal box =  2×(lb+bh+hl)

= 2×(60×40+40×50+50×60)

= 2×(2400+2000+3000)

= 14800 cm2

(b)  Length of cubical box (l) = 50 cm

Breadth of cubicalbox (b) = 50 cm

Height of cubicalbox (h) = 50 cm

Total surface area of cubical box = 6(side)2

= 6(50×50)

= 6×2500

= 15000

Surface area of the cubical box is 15000 cm2

From the result of (a) and (b), cuboidal box requires the lesser amount of material to make.

2. A suitcase with measures 80 cm x 48 cm x 24 cm is to be covered with a tarpaulin cloth. How many meters of tarpaulin of width 96 cm is required to cover 100 such suitcases?

Solution:

Length of suitcase box, l = 80 cm,

Breadth of suitcase box, b= 48 cm

And Height of cuboidal box , h = 24 cm

Total surface area of suitcase box = 2(lb+bh+hl)

= 2(80×48+48×24+24×80)

= 2 (3840+1152+1920)

= 2×6912

= 13824

Total surface area of suitcase box is 13824 cm2

Area of Tarpaulin cloth = Surface area of suitcase

l×b = 13824

l ×96 = 13824

l = 144

Required tarpaulin for 100 suitcases = 144×100 = 14400 cm = 144 m

Hence tarpaulin cloth required to cover 100 suitcases is 144 m.

3. Find the side of a cube whose surface area is 600cm^2 .

Solution:

Surface area of cube = 600 cm2 (Given)

Formula for surface area of a cube = 6(side)2

Substituting the values, we get

6(side)2 = 600

(side)2 = 100

Or side = ±10

Since side cannot be negative, the measure of each side of a cube is 10 cm

4. Rukshar painted the outside of the cabinet of measure 1 m ×2 m ×1.5 m. How much surface area did she cover if she painted all except the bottom of the cabinet?

Solution:

Length of cabinet, l = 2 m, Breadth of cabinet, b = 1 m and Height of cabinet, h  = 1.5 m

Surface area of cabinet =  lb+2(bh+hl )

= 2×1+2(1×1.5+1.5×2)

= 2+2(1.5+3.0)

= 2+9.0

= 11

Required surface area of cabinet is 11m2.

5. Daniel is paining the walls and ceiling of a cuboidal hall with length, breadth and height of 15 m, 10 m and 7 m respectively. From each can of paint 100 m^2of area is painted. How many cans of paint will she need to paint the room?

Solution:

Length of wall, l  = 15 m, Breadth of wall, b = 10 m and Height of wall, h  = 7 m

Total Surface area of classroom = lb+2(bh+hl )

= 15×10+2(10×7+7×15)

= 150+2(70+105)

= 150+350

= 500

Now, Required number of cans =  Area of hall/Area of one can

= 500/100 = 5

Therefore, 5 cans are required to paint the room.

6. Describe how the two figures below are alike and how they are different. Which box has larger lateral surface areas?

Solution:

Diameter of cylinder = 7 cm (Given)

Radius of cylinder, r  =  7/2  cm

Height of cylinder, h = 7 cm

Lateral surface area of cylinder = 2πrh

=  2×(22/7)×(7/2)×7 = 154

So, Lateral surface area of cylinder is154 cm2

Now, lateral surface area of cube =  4 (side)2=4×72 = 4×49 = 196

Lateral surface area of cube is 196 cm2

Hence, the cube has larger lateral surface area.

7. A closed cylindrical tank of radius 7 m and height 3 m is made from a sheet of metal. How much sheet of metal is required?

Solution:

Radius of cylindrical tank, r  = 7 m

Height of cylindrical tank , h = 3 m

Total surface area of cylindrical tank = 2πr(h+r)

=  2×(22/7)×7(3+7)

= 44×10 = 440

Therefore, 440  m2 metal sheet is required.

8. The lateral surface area of a hollow cylinder is 4224cm2. It is cut along its height and formed a rectangular sheet of width 33 cm. Find the perimeter of rectangular sheet?

Solution:

Lateral surface area of hollow cylinder = 4224  cm2

Height of hollow cylinder, h = 33 cm and say r be the radius of the hollow cylinder

Curved surface area of hollow cylinder = 2πrh

4224 =  2×π×r×33

r = (4224)/(2π×33)

r = 64/π

Now, Length of rectangular sheet, l =  2πr

l = 2 π×(64/π) = 128 (using value of r)

So the length of the rectangular sheet is 128 cm.

Also, Perimeter of rectangular sheet =  2(l+b)

= 2(128+33)

= 322

The perimeter of rectangular sheet is 322 cm.

9. A road roller takes 750 complete revolutions to move once over to level a road. Find the area of the road if the diameter of a road roller is 84 cm and length 1 m.

Solution:

Diameter of road roller, d = 84 cm

Radius of road roller, r = d/2 = 84/2 = 42 cm

Length of road roller, h = 1 m = 100 cm

Formula for Curved surface area of road roller = 2πrh

= 2×(22/7)×42×100 = 26400

Curved surface area of road roller is 26400 cm2

Again, Area covered by road roller in 750 revolutions = 26400×750cm2

= 1,98,00,000cm2

= 1980 m2  [∵ 1 m2= 10,000 cm2]

Hence the area of the road is 1980 m2.

10. A company packages its milk powder in cylindrical container whose base has a diameter of 14 cm and height 20 cm. Company places a label around the surface of the container (as shown in figure). If the label is placed 2 cm from top and bottom, what is the area of the label?

Solution:

Diameter of cylindrical container , d = 14 cm

Radius of cylindrical container, r = d/2 = 14/2  = 7 cm

Height of cylindrical container = 20 cm

Height of the label, say h = 20–2–2 (from the figure)

= 16 cm

Curved surface area of label = 2πrh

=  2×(22/7)×7×16

= 704

Hence, the area of the label is 704 cm2.