**NCERT Solutions for Class 8 Maths Chapter 11 Exercise 11.3 – Mensuration, has been designed by the NCERT to test the knowledge of the student on the following topics:-**

- Solid Shapes
- Surface Area of Cube, Cuboid and Cylinder

– Cuboid

– Cube

– Cylinders

### NCERT Solutions for Class 8 Maths Chapter 11 Exercise 11.3 – Mensuration

**NCERT Solutions for Class 8 Maths Chapter 11 Exercise 11.3 – Mensuration**

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

**Sol.: **– To find which box requires the lesser amount of material to make, we need to find surface area of the boxes.

a). The box in figure is cuboidal in shape.

So, Surface area of box = 2 (lb + bh + hl)

Where, l = Length = 60 cm

b = Breadth = 50 cm

h= Height = 40 cm

So, Surface area of box = 2 (60 × 50 + 50 × 40 + 40 × 60)

= 2 (3000 + 2000 + 2400)

= 2 × 7400

= 14800 cm^{2}

b). The box in figure is cubical in shape.

So, Surface area of box = 6 a^{2 }Where, a = Side = 50 cm

So, Surface area of box = 6 × 50^{2 }= 6 × 2500

= 15000 cm^{2}

As we can see surface area of box (a) is less, so it will require less amount of material to make.

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

** ****Sol.: **– As we need to find the length of tarpaulin that is required to cover suitcase. First, we will find total surface area of suitcase.

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

Where, l = Length of suitcase = 80 cm

b = Breadth of suitcase = 48 cm

h= Height of suitcase = 24 cm

So, Surface area of suitcase = 2 (80×48 + 48×24 + 24×80)

= 2 (3840 + 1152 + 1920)

= 13824 cm^{2 }Now, Surface area of 100 suitcase = 100 × 13824

= 1382400 cm^{2}

To cover this area of suitcase, area of tarpaulin should be equal to this area.

1382400 cm^{2} = (Width × Length)of tarpaulin

1382400 cm^{2} = 96 × Length of tarpaulin

So, Length of tarpaulin = (1382400/96) = 14400 cm = 144 m

So, Length of Tarpaulin required to cover 100 suitcases is **144 m.**

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

** ****Sol.: **– We know that,

Surface area of cube = 6 a

^{2 }Where, a = Side of cube

600 = 6 a

^{2 }

Thus, side of cube whose surface area is 600 cm

^{2}is 10 cm.

**4. Rukhsar 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?**

**Sol.: **– Total surface area of the cabinet painted = Total surface area of the cabinet – Area of bottom

= 2 (lb + bh + hl) – (lb)

_{of bottom }

_{surface }Here, l = Length of cabinet = 2 m

b = Breadth of cabinet = 1 m

h= Height of cabinet = 1.5 m

So, Total surface area of the cabinet painted

= 2 (2×1 + 1×1.5 + 1.5×2) – 2×1

= 2 (2 + 1.5 + 3) – 2

= 2 × 7.5 – 2

= 11 m^{2}

Therefore, the total surface area of the cabinet painted by Rukhsar is **11 m ^{2}.**

**5. ****Daniel is painting 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 ^{2} of area is painted. How many cans of paint will she need to paint the room?**

**Sol.: **– As Daniel is only painting wall and ceiling of the cuboidal hall not floor.

So, Total area painted

= Total surface area of the hall – Area of floor

= 2 (lb + bh + hl) – (lb)_{of bottom }_{surface}

Here, l = Length of hall = 15 m

b = Breadth of hall = 10 m

h= Height of hall = 7 m

So, Total surface area of the hall painted

= 2 (15×10 + 10×7 + 7×15) – 15×10

= 2 (150 + 70 + 105) – 150

= 2 × 325 – 150

= 500 m^{2}

As we know that, 100 m^{2} of area is painted by 1 can.

500 m of area required = (500/100) = 5 cans

So, **5 cans** are required to paint the room.

**6. Describe how the two figures at the right are alike and how they are different. Which box has larger lateral surface area?**

**Sol.: **– As we can observer from figures that height of both figures is same i.e. 7 cm.

Also figure (a) is cylindrical in shape whereas (b) is cubical.

Lateral Surface Area of figure (a) = 2 ?rh

= 2 × 22/7 x 7/2 × 7

= 154 cm^{2}

And, Lateral Surface Area of figure (b) = 4a^{2 }= 4 × 7^{2 }= 196 cm^{2}

Hence, **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?**

** ****Sol.: **– Since Cylindrical tank is made from sheet of metal

So, Area of metal sheet = Area of Closed Cylindrical tank

= 2 ?r(h+r)

Here, r = Radius of Cylindrical tank = 7 m

h = Height of Cylindrical tank = 3 m

So, Area of metal sheet = 2 × (22/7) × 7 × (3 +7)

= 440 m

^{2 }Therefore, Area of metal sheet required to form closed cylindrical tank is

**440 m**.

^{2}**8. The lateral surface area of a hollow cylinder is 4224 cm ^{2}. It is cut along its height and formed a rectangular sheet of width 33 cm. Find the perimeter of rectangular sheet?**

**Sol.: **–

Since rectangular sheet is made from cylinder,

So, Area of sheet = Lateral surface Area of hollow Cylindrical

(Length × Breadth)_{of sheet} = 4224

Length × 33 = 4224

Length = (4224/33)

= 128 cm

Now perimeter of rectangular sheet = 2 (Length + Breadth)

= 2 (128 + 33)

= 322 cm

Thus, 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 is 1 m.**

**Sol.: **–

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

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

As, road roller takes 750 complete revolutions to move once over to level the road.

So, Area of road = 750 × Latent surface Area of road roller

= 750 × 2 ?rh

= 750 × 2 × (22/7) × 42 × 100

= 19800000 cm^{2 }= 1980 m^{2}

Thus, total area of road is **1980 m ^{2}.**

**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 the figure). If the label is placed 2 cm from top and bottom, what is the area of the label.**

**Sol.: **– As we can observe from above figure, the label is forming a shape hollow cylinder.

So, Area of label = 2 ?rh

Where, r = Radius of hollow cylinder formed by label = Diameter/2 = 14/2 = 7 cm

H = Height of hollow cylinder formed by label = 20 – (2+2) = 16 cm

So, Area of label = 2 × (22/7) × 7 × 16

= 704 cm^{2}

Thus, area of label on the milk powder container is **704 m ^{2}.**

**The next Exercise for** **NCERT Solutions for Class 8 Maths Chapter 11 Exercise 11.4 – Mensuration can be accessed by clicking here**

**Download NCERT Solutions for Class 8 Maths Chapter 11 Exercise 11.3 – Mensuration**

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