Mathematics – Class 8 – Chapter 11 -Mensuration- Exercise 11.3 – NCERT Exercise Solution

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

Mathematics - Class 8 - Chapter 11 -Mensuration- Exercise 11.3 - NCERT Exercise Solution

Solution:

(a)

Given,

Length of cuboidal box (l) = 60 cm

Breadth of cuboidal box (b) = 40 cm

Height of cuboidal box (h) = 50 cm

According to question:

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

                                                       = 2×(60×40 +40×50 + 50×60)

                                                       = 2×(2400+2000+3000)

                                                        = 14800 cm2

(b)  Given,

Length of cubical box (l) = 50 cm

Breadth of cubical box (b) = 50 cm

Height of cubical box (h) = 50 cm

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

                                                     = 6(50×50)

                                                      = 6×2500

                                                      = 15000 cm2

Hence, according to the result of (a) and (b), we get that 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:

Given,

Length of suitcase (l) = 80 cm,

Breadth of suitcase (b)= 48 cm

Height of cuboidal (h) = 24 cm

Then,

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

                                                      = 2(80×48+48×24+24×80)

                                                      = 2 (3840+1152+1920)

                                                      = 2×6912

                                                      = 13824 cm2

Now,

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

3. Find the side of a cube whose surface area is 600cm2.

Solution:

Given, surface area of cube = 600 cm2

From formula we know that,

Surface area of a cube = 6(side)2

So,

6(side)2 = 600

(side)2 = 100

side = ±10

Hence, side cannot be negative. So, 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?

Mathematics - Class 8 - Chapter 11 -Mensuration- Exercise 11.3 - NCERT Exercise Solution

Solution:

Given,

Length of cabinet (l) = 2 m

Breadth of cabinet (b) = 1 m

Height of cabinet (h) = 1.5 m

Then,

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

= 2×1+2(1×1.5+1.5×2)

= 2+2(1.5+3.0)

= 2+9.0

= 11 m2

Hence, 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 m2 of area is painted. How many cans of paint will she need to paint the room?

Solution:

Given,

Length of wall (l) = 15 m

Breadth of wall (b) = 10 m  

Height of wall (h) = 7 m

Then,

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

= 15×10+2(10×7+7×15)

= 150+2(70+105)

= 150+350

= 500 m2

Now,

Total number of required cans = Area of classroom/Area of one can

= 500/100 = 5

Hence, 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?

Mathematics - Class 8 - Chapter 11 -Mensuration- Exercise 11.3 - NCERT Exercise Solution

Solution:

Given,

Diameter of cylinder (d) = 7 cm

So, radius of cylinder (r) = 7/2 cm [ r = d/2]

Height of cylinder (h) = 7 cm

Now,

Lateral surface area of cylinder = 2πrh

= 2 × (22/7) × (7/2) × 7

= 154 cm2

Now,

Lateral surface area of cube = 4 (side)2

= 4×72

= 4×49

= 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:

Given,

Radius of cylindrical tank (r) = 7 m

Height of cylindrical tank (h) = 3 m

Then,

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

= 2×(22/7)×7(3+7)

= 44×10 = 440 m2

Hence, 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:

Given, lateral surface area of hollow cylinder = 4224 cm2

Width of rectangular sheet (b) = 33 cm

Let l be the length of rectangular sheet.

Now,

Lateral surface area of cylinder = Area of rectangular sheet

4224 = b × l

4224 = 33 × l

l = 4224/33 = 128 cm

Hence, the length of the rectangular sheet is 128 cm.

Now,

Perimeter of rectangular sheet =  2(l+b)

= 2(128+33)

= 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.

Mathematics - Class 8 - Chapter 11 -Mensuration- Exercise 11.3 - NCERT Exercise Solution

Solution:

Given,

Diameter of road roller (d) = 84 cm

So, radius of road roller (r) = 84/2 = 42 cm [r = d/2]

Length of road roller (h) = 1 m = 100 cm

Now,

Curved surface area of road roller = 2πrh

= 2×(22/7)×42×100

= 26400 cm2

Now,

 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?

Mathematics - Class 8 - Chapter 11 -Mensuration- Exercise 11.3 - NCERT Exercise Solution

Solution:

Given,

Diameter of cylindrical container (d) = 14 cm

So, radius of cylindrical container (r) = 14/2  = 7 cm [r = d/2]

Height of cylindrical container (H) = 20 cm

According to the figure,

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

= 16 cm

Now,

Curved surface area of label = 2πrh

=  2×(22/7)×7×16

= 704

Hence, the area of the label is 704 cm2.

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