# Ncert Solutions For Class 8 Maths Ex 11.3

## Ncert Solutions For Class 8 Maths Chapter 11 Ex 11.3

1. Two cuboidal boxes as shown in the given figure. Find out, which box would use up less material?

Solution:

As we know that,

Total surface area of the cuboid = 2 (lh + bh + lb)

Total surface area of the cube = 6 (l) 2$^{2}$

(a) From the figure-:

Total surface area of the cuboid = [2{(70) (30) + (30) (45) + (70) (45)}] cm2$^{2}$

= [2(2100 + 1350 + 3150)] cm2$^{2}$

= (2 x 6600) cm2$^{2}$

= 13200 cm2$^{2}$

(b) From the figure-:

Total surface area of the cube = 6 (70cm)2$\left ( 70 cm\right )^{2}$ = 29400                  cm2$^{2}$

Thus, the cuboidal box (a) would use less material.

2.  A suitcase with measures 70 cm x 36 cm x 14 cm need to be covered with a blanket cloth. How many meters of cloth having width of 100 cm is required to cover 150 such suitcases?

Solution:

Here,  Total surface area of suitcase = 2 (lh + bh + lb)

= 2[(70) (36) + (36) (14) + (14) (70)]

= 2[2520 + 504 + 980]

= 8008 cm2$^{2}$

Total surface area of 150 suitcases = (8008 x 150) cm2$^{2}$ = 1201200 cm2$^{2}$

therefore$\ therefore$ Required blanket cloth = Length × Breadth

1201200 cm2$^{2}$ = Length × 100 cm

Length = (1201200100)cm=12012cm=120.12m$\left ( \frac{1201200}{100} \right )cm=12012cm=120.12m$

Hence, 120.12m of blanket cloth is required to cover 150 suitcases.

3. A cube having surface area 1200 cm2$^{2}$. Calculate the side of the cube.

Solution:

Given, surface area of cube = 1200 cm2$^{2}$

Let us assume the length of each side of the cube be ‘m’.

Now, Surface area of cube = 6 (side)2$\left (side \right )^{2}$

1200 cm2$^{2}$ = 6m2$6m^{2}$

m2$m^{2}$ = 200 cm2$^{2}$

m = 14.142 cm

Hence, the side of the cube is 14.142 cm.

4. Tejus painted the exterior of the house of measure 3 m x 6 m x 2.5 m. How much surface area did he cover if he painted all except the bottom of the house?

Solution:

Length (l) of the house = 6 m

Breadth (b) of the house = 3 m

Height (h) of the house = 2.5 m

Area of the house that was painted = 2h (l + b) + lb

= [2 X 2.5 X (6 + 3) + (6) (3)] m2$m^{2}$

= [5 X 9 + 18] m2$m^{2}$

= 135 m2$m^{2}$

5. Raghav is painting the interior walls and the ceiling of a cuboidal hall having length (l), breadth (b) and height (h) of 30 m, 20 m and 10 m respectively. From each tin of paint 200 m2of area is covered. How many tins of paint will he need to paint the total interior of the room?

Solution:

Here,  Length (l) = 30 m, breadth (b) = 20 m, height (h) = 10 m

Area of the hall to be painted = Area of the wall + Area of the ceiling

= 2h (l + b) + lb

=  [2(10) (30 + 20) + 30 ×20] m2$m^{2}$

= [20 x 50 + 600] m2$m^{2}$

= 1600 m2$m^{2}$

Given that, 200 m2$m^{2}$ area can be painted from each tin.

Number of tins required to paint an area of 1600 m2$m^{2}$

= 1600200=8$\frac{1600}{200}=8$

Hence, 8 tins are required to paint the interior walls and the ceiling of the cuboidal hall.

6. Describe what are the similarities and the difference between these two figures. Which box has larger lateral surface area?

Solution:

From the given two figures, the similarity is that both of them have same heights.

And the difference between both the figures is that one is cylinder and the other one is a cube.

Lateral surface area of the cube = 4l2$4l^{2}$ = 4(10cm)2=400cm2$4\left ( 10cm \right )^{2}=400cm^{2}$

Lateral surface area of the cylinder = 2πrh$2\pi rh$ = (2×227×102×10)cm2$\left (2 \times \frac{22}{7} \times \frac{10}{2} \times 10 \right ) cm^{2}$

= 314.285 cm2

Therefore, the cube has larger lateral surface area.

7. A closed cylindrical container of radius and height 14m and 4m is made of steel sheet, Calculate the quantity of sheet required to make the container.

Solution:

Total surface area of cylinder =  2πr(r+h)$2\pi r(r + h)$

= 2×227×14(14+4)$2\times \frac{22}{7}\times 14 \left ( 14 + 4 \right  )$

= 1584 m2

Therefore, the required metal sheet would be 1584 m2.

8. Given a hollow cylinder of lateral surface area as 5000 cm2. When we will cut it along the height then it will form a rectangular sheet of width 50 cm. Calculate the perimeter of the new formed sheet.

Solution:

When a rectangle sheet will form by cutting a hollow cylinder, then the area for both will remain the same.

=> Area of Cylinder = Area of Rectangular sheet

=> 5000 cm2 = 50 cm X Length

=> Length=500050$Length = \frac{5000}{50}$ = 100 cm

Therefore, the length of the rectangle sheet will be: 100 cm

Now, the perimeter will be = 2 ( length + width)

= [2 ( 100 + 50 )]cm

= ( 2 X 150 ) cm

= 300 cm

9. It takes a road roller 700 complete revolution to move once over a labeled road. Calculate the total area of road while the diameter of the road roller is 90 cm and its length is 15 m.

Solution:

In one revolution, the roller will cover an area equal to its lateral surface area.

Thus, in 1 revolution, area of the road covered = =  2πrh$2\pi rh$

= 2×227×45cm×15m$2 \times \frac{22}{7} \times 45 cm \times 15m$

= 2×227×45100m×15m$2 \times \frac{22}{7} \times \frac{45}{100} m \times 15m$

= 29700700$\frac{29700}{700}$ m2

In 700 revolutions, area of the road covered

= (700×29700700)$\left (700 \times \frac{29700}{700} \right )$ m2

= 29700 m2

10. A manufacturer packages its chocolate powder in a cylindrical box, which is has a diameter of 22 cm and height 30 cm. The manufacturer places its sticker around the body of the box (Refer to the figure). If the sticker placed is 3 cm from top and bottom, calculate the area of the sticker.

Solution:

Height of the sticker = 30cm – 3cm – 3cm = 24cm

Radius of the Sticker = (222)$\left ( \frac{22}{2} \right )$ = 11cm

Sticker is in the form of a cylinder with radius 11cm and height 24cm.

Area of label = 2π(Radius)(Height)$2\pi \left ( Radius \right ) \left ( Height \right )$

= (2×227×11×24)cm2$\left (2 \times \frac{22}{7} \times 11 \times 24 \right ) cm^{2}$

= 1659.428 cm2

Exercise 11.4

1. You are given two cylinders 1 and 2. The diameter and height of cylinder 2 is 14cm and 21cm respectively. For Cylinder 1, diameter and height is 21 cm and 14 cm respectively. Without calculating, can you suggest which one of the cylinder will have greater volume? Also, verify the volumes by doing the calculations. Check whether the cylinder with greater volume have greater surface area?

Solution:

The heights and diameters of these cylinders A and B are interchanged.

We know that,

Volume of cylinder = πr2h$\pi r^{2}h$

If measures of radius and height are same, then the cylinder with greater radius will have greater area.

Radius of Cylinder 1 = 142$\frac{14}{2}$ = 7cm

Radius of Cylinder 2 = 212$\frac{21}{2}$ cm

Here the radius of cylinder 2 is larger, then, the cylinder 2 will have larger volume.

Now, verifying by calculating.

Volume of cylinder 1 = πr2h$\pi r^{2}h$

= (227×142×142×21)cm3$\left (\frac{22}{7} \times \frac{14}{2} \times \frac{14}{2} \times 21 \right )cm^{3}$

= 3234cm3

Volume of cylinder 2 = πr2h$\pi r^{2}h$

= (227×212×212×14)cm3$\left (\frac{22}{7}\times \frac{21}{2} \times \frac{21}{2}\times 14 \right ) cm^{3}$

= 4851 cm3

Volume of Cylinder 2 is greater.

Surface area of cylinder 1 = 2πr(r+h)$2\pi r \left ( r + h \right )$

= [2×227×142×(142+21)]cm2$\left [ 2 \times \frac{22}{7} \times \frac{14}{2}\times \left ( \frac{14}{2} + 21 \right ) \right ]cm^{2}$

= (44 X 28) cm2

= 1232 cm2

Surface area for cylinder 2  = 2πr(r+h)$2\pi r \left ( r + h \right )$

= [2×227×212×(212+14)]cm2$\left [ 2 \times \frac{22}{7} \times \frac{21}{2}\times \left ( \frac{21}{2} + 14 \right ) \right ]cm^{2}$

= (33 X 49) cm2

= 1617 cm2

Thus the surface area of cylinder 2 is also greater.

2. A cuboid having dimensions 80 cm x 40 cm x 25 cm. Calculate the how many of small cubes with sides 8 cm can be placed inside the given cuboid.

Solution:

Volume of cuboid = 80 cm × 40 cm × 25 cm = 80000 cm3

Side of the cube = 8 cm

Volume of the cube = (8)3 cm3 = 512 cm3

Required number of cubes = VolumeofthecuboidVolumeofthecube$\frac{Volume\:of\:the\:cuboid}{Volume\:of\:the\:cube}$

= 80000512=156.25$\frac{80000}{512}=156.25$

Hence, 156 (approx.) cubes can be placed in the given cuboid.

3. Obtain the height of a cylinder whose volume is 2m3and diameter of the base is 200 cm?

Solution:

Diameter of the base = 200 cm

Radius (r) of the base = (2002)cm=100cm=100100m=1m$\left ( \frac{200}{2} \right )cm=100cm=\frac{100}{100}m=1m$

Volume of cylinder = πr2h$\pi r^{2}h$

2m3 = 227×1m×1m×h$\frac{22}{7}\times 1m\times 1m\times h$

h = 7×222=1422m=0.636m$\frac{7\times 2}{22}=\frac{14}{22}m=0.636m$

Hence, the height of the cylinder is 0.636m

4. A milk tank with radius and length of 2m and 8m respectively, is in the form of a cylinder. Calculate the quantity of milk (in liters) which can be stored in the tank.

Solution:

Given, Radius of cylinder(r) = 2m

Length of cylinder(h) = 8m

Volume of cylinder = πr2h$\pi r^{2}h$

= (227×2×2×8)m3$\left ( \frac{22}{7} \times 2\times 2\times 8\right )m^{3}$

= 100.57m3$m^{3}$

1m3=1000L$1m^{3}=1000L$

Quantity required = (100.57 x 1000)L = 100570 L

Hence therefore, 100570 L of milk can be stored in the tank.

5. If each edge of a cube is doubled,

(a) Find out the number of time its surface area will increase.

(b) Find out the number of times its volume will increase.

Solution:

(a) Let us assume the edge of the cube as l.

Initial surface area = 6l2$6l^{2}$

If each edge of the cube will be doubled, then it becomes 2l.

So, New surface area = 6(2l)2=24l2=4×6l2$6\left ( 2l \right )^{2}=24l^{2}=4\times 6l^{2}$

Hence, the surface area will be increased by 4 times.

(b) Let us assume the initial volume as l3$l^{3}$

If each edge of the cube will be doubled, then it becomes 2l.

So, New volume = (2l)3$\left ( 2l \right )^{3}$ = 8l3$l^{3}$ = 8×l3$8\times l^{3}$

Hence, the volume will increase 8 times.

6. Find the height of a cuboid whose base area is 200cm2$200cm^{2}$ and volume is 1200cm3$1200cm^{3}$?

Solution:

Given: Base area of cuboid = 200 cm2$cm^{2}$

Volume of cuboid = 1200 cm3$cm^{3}$

We know that,

Volume of cuboid = l b h              [ because$\ because$ Base area = l x b = 200(given)]

=> 1200 = 200 ´h

=> h = 1200200=6m$\frac{1200}{200}=6m$.

7. Water is pouring into a cuboidal tank at the rate of 90 liters per minute. If the volume of tank is 270 m3$m^{3}$, calculate the number of hours it will take to fill the tank.

Solution:

Volume of tank = 270 m3$m^{3}$ =(270 x 1000)L = 270000 L

Rate of pouring water into cuboidal tank = 90 liters/minute

=> (90 x 90)L = 8100 L per hour

∴ Required number of hours = 2700008100=33.33$\frac{270000}{8100} = 33.33$ hours = 33 hours (approx..)

It will take 33 hours to fill the tank.