Solution for Exercise 13.1

Q1 )
A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is opened at the top. Ignoring the thickness of the plastic sheet, determine:

i) The area of the sheet required for making the box.

ii) The cost of sheet for it, if a sheet measuring 1

Answer :

Given :

length (l) of box = 1.5m

Breadth (b) of box = 1.25 m

Depth (h) of box = 0.65 m

i) Box is to be open at top,

So, Area of sheet required

= 2lh + 2bh + lb

= [1.95 + 1.625 + 1.875]

= 5.45

Hence, the area of the sheet required for making the box is 5.45

ii) Cost of sheet per 1

Therefore, Cost of sheet of 5.45

= Rs. (5.45 x 20)

= Rs. 109

Hence, The cost of sheet for it is Rs. 109

Q2 )
The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of Rs. 7.50 per

Answer :

Given :

Length (l) of room = 5 m

Breadth (b) of room = 4 m

Height (h) of room = 3 m

It can be observed that four walls and the ceiling of the room are to be white- washed. The floor of the room is not to be white-washed.

So, we get,

Area to be white-washed

= Area of walls + Area of ceiling of room

= 2lh + 2bh + lb

= [2 x 5 x 3 + 2 x 4 x 3 + 5 x 4]

= [30 + 24 + 20]

= 74

Now, we have,

Cost of white-washing per

Cost of white-washing 74

Therefore, the cost of white washing the walls of the room and the ceiling is Rs. 555.

Q3 )
The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of Rs. 10 per

[Hint : Area of the four walls = Lateral surface area.]

Answer :

Let the length, breadth, and height of the rectangular hall be l m, b m, and h m respectively.

Now, perimeter of the floor of hall

= 2(l + b) = 250 m ...(Given)

Also, Area of four walls = 2lh + 2bh

Now, Cost of painting per

However, it is given that the cost of painting the walls is Rs 15000.

Hence, 15000 = 2500h

Therefore, the height of the hall is 6 m.

Q4 )
The paint in a certain container is suffcient to paint an area equal to 9.375

Answer :

We know that,

Total surface area of one brick

= 2(lb + bh + lh)

= [2(22.5 X 10 + 10 x 7.5 + 22.5 x 7.5)]

= [2(225 + 75 + 168.75)]

= [2 x 468.75]

= 937.5

Now, Let m bricks can be painted out by the paint of the container.

Thus, we get,

Area of m bricks = (m x 937.5) = 937.5m

Therefore, area that can be painted by the paint of the container = 9.375

= 93750

Thus, equating (i) and (ii), we get,

93750 = 937.5m

Therefore, 100 bricks can be painted out by the paint of the container.

Q5 )
A cubical box has each edge 10 cm and another cuboidal box is 12.5 cm long, 10 cm wide and 8 cm high.

i) Which box has the greater lateral surface area and by how much?

ii) Which box has the smaller total surface area and by how much?

Answer :

Given:

Edge of cube = 10 cm

Length (l) of box = 12.5 cm

Breadth (b) of box = 10 cm

Height (h) of box = 8 cm

i) We know that, Lateral surface area of cubical box

=

=

= 400

Also, We have,

Lateral surface area of cuboidal box

= 2 [lh + bh]

= [2 (12.5 x 10 x 8)]

= [2 x 180]

= 360

Clearly, the lateral surface area of the cubical box is greater than the lateral surface area of the cuboidal box.

Now, Lateral surface area of cubical box - Lateral surface area of cuboidal box

= 400

Therefore, the lateral surface area of the cubical box is greater than the lateral surface area of the cuboidal box by 40

ii)Similarly, Total surface area of cubical box

=

=

= 600

Also, We have,

Total surface area of cuboidal box

= 2(lh + bh + lb)

=[2 (12.5 x 8 + 10 x 8 + 12.5 x 100)]

= 610

Clearly, the total surface area of the cubical box is smaller than that of the cuboidal box.

Now, Total surface area of cuboidal box - Total surface area of cubical box

= 610

= 10

Therefore, the total surface area of the cubical box is smaller than that of the cuboidal box by 10

Q6 )
A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30 cm long, 25 cm wide and 25 cm high.

i) What is the area of the glass?

ii) How much of tape is needed for all the 12 edges?

Answer :

Given :

Length (l) of greenhouse = 30 cm

Breadth (b) of greenhouse = 25 cm

Height (h) of greenhouse = 25 cm

We know that,

Total surface area of greenhouse

=2 [lb + lh + bh]

= [2 (30 x 25 + 30 x 25 + 25 x 25)]

= [2(750 + 750 + 625)]

= [2 x 2125]

= 4250

Therefore, the area of glass is 4250

It can be observed that tape is required alongside AD, OC, CD, DA, EF, FG, GH, HE, AH, BE, DG, and CF.

We have, Total length of tape

= 4(l+ b + h)

= [4(30 + 25 + 25)] cm

= 320 cm

Therefore, 320 cm tape is required for all the 12 edges.

Q7 )
Shanti Sweets Stall was placing an order for making cardboard boxes for packing their sweets.Two sizes of boxes were required. The bigger of dimensions 25 cm × 20 cm × 5 cm and the smaller of dimensions 15 cm × 12 cm × 5 cm. For all the overlaps, 5% of the total surface area is required extra. If the cost of the cardboard is Rs. 4 for 1000

Answer :

Given:

Length (l) of bigger box = 25 cm

Breadth (b) of bigger box = 20 cm

Height (h) of bigger box = 5 cm

we know that,

Total surface area of bigger box

= 2(lb + lh + bh)

= [2 (25 x 20 + 25 x 5 + 20 x 5)]

= [2(500 + 125 + 100)]

= 1450

Therefore, Extra area required for overlapping

=

While considering all overlaps, total surface area of 1 bigger box

= (1450 + 72.5)

So, Area of cardboard sheet required for 250 such bigger boxes = (1522.5 x 250)

Similarly, total surface area of smaller box = [2 (15 + 15 x 5 + 12 x 5)]

= [2 (180 + 75 + 60)]

= (2 x 315)

= 630

Therefore, extra area required for overlapping

=

So, Total surface area of 1 smaller box while considering all overlaps

= (630 + 31.5)

= 661.5

So, Area of cardboard sheet required for 250 smaller boxes

= (250 x 661.5)

Thus, Total cardboard sheet required

= (380625 + 165375)

= 546000

Cost of 1000

Therefore, Cost of 546000

Therefore, the cost of cardboard sheet required for 250 such boxes of each kind will be Rs 2184.

Q8 ) Parveen wanted to make a temporary shelter for her car, by making a box-like structure with tarpaulin that covers all the four sides and the top of the car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small, and therefore negligible, how much tarpaulin would be required to make the shelter of height 2.5 m, with base dimensions 4m × 3 m?

Answer :

Given:

Length (l) of shelter = 4 m

Breadth (b) of shelter = 3 m

Height (h) of shelter = 2.5 m

Tarpaulin will be required for the top and four wall sides of the shelter.

So, Area of Tarpaulin required

= 2(lh + bh) + lb

= [2(4 x 2.5 + 3 x 2.5) + 4 x 3]

= [2(10 + 7.5) + 12]

= 47

Therefore, 47

Solution for Exercise 13.2

Q1 )
The curved surface area of a right circular cylinder of height 14 cm is 88

Answer :

Given :

Height (h) of cylinder = 14 cm

Curved surface area of cylinder = 88

Let the diameter of the cylinder be d.

So, (

(where, r is the radius of the base of the cylinder)

Therefore, the diameter of the base of the cylinder is 2 cm.

Q2 ) It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. How many square metres of the sheet are required for the same?

Answer :

Given :

height(h) of the cylindrical tank = 1m

Base radius(r) of cylindrical tank

=(

Now, Area of sheet required = total surface of tank

But, we know that,

Area of sheet required

=

=

=

= 7.48

Therefore , it will required 7.48

Q3 )
A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4 cm. Find its

i) inner curved surface area

ii) outer curved surface area

iii) total surface area.

Answer :

Given :

Inner radius (r) of cylindrical pipe =

outer radius (R) of cylindrical pipe =

Height (h) Of cylindrical pipe = Length Of cylindrical pipe = 77 cm

i) CSA of inner surface of pipe

=

=

=

= 968

ii) CSA of inner surface of pipe

=

=

=

= 1064.8

iii) Total surface area of pipe

= CSA of inner surface + CSA of outer surface + Area of both circular ends of pipe

=

=

=

= (2032.8 + 5.28)

= 2038.08

Therefore, the total surface area of the cylindrical pipe is 2038.08

Q4 )
The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in

Answer :

Given :

Height (h) of cylindrical roller = Length of roller = 120 cm

And radius (r) of the circular end roller =

It can be observed that a roller is cylindrical.

Now, we know that,

CSA of rollar

=

=

=

= 31680

Therefore, Area of field

= 500 x CSA of roller

=

= 1584

Q5 )
A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of Rs. 12.50 per

Answer :

Given :

Height (h) cylindrical pillar = 3.5 m

Radius (r) of the circular end of pillar =

Now, we know that,

CSA of cylindrical pillar =

=

=

= 5.5

Now, Cost of painting 1

Cost of painting 5.5

Therefore, the cost of painting the CSA of the pillar is Rs. 68.75

Q6 )
Curved surface area of a right circular cylinder is 4.4

Answer :

Given :

Radius (r) of the base of cylinder = 0.7 m

CSA Of cylinder = 4.4

Let the height of the circular cylinder be h.

Therefore, we have,

Therefore, the height of the cylinder is 1 m.

Q7 )
The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find

i) its inner curved surface area

ii) the cost of plastering this curved surface at the rate of Rs. 40 per

Answer :

Given :

Inner radius (r) of circular well =

Depth (h) of circular well = 10 m

i) We know that,

Inner curved surface area

=

=

= (44 × 0.25 × 10)

= 110

Therefore, the inner curved surface area of the circular well is 110

ii) We have, Cost of plastering 1

Cost of plastering 100

Therefore, the cost of plastering the CSA of this well is Rs. 4400.

Q8 ) In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.

Answer :

Given :

Height (h) of cylindrical pipe Length of cylindrical pipe = 28 m

Radius (r) of circular end of pipe

=

We know that,

CSA of cylindrical pipe

=

=

= 4.4

Therefore, the area of the radiating surface of the system is 4.4

Q9 )
Find

i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5 m high.

ii) how much steel was actually used, if

Answer :

Given :

Height (h) of the cylindrical tank = 4.5 m

radius(r) of the circular end of cylindrical tank =

i) lateral or curved surface area of tank

=

=

= (44 × 0.3 × 4.5)

= 59.4

Therefore , CSA of tank is 59.4

ii) Now, so as to find steel used,

total surface area

=

=

=

= 87.12

Now, Let actual area of x

Since,

This means that the actual area used

=

Therefore, 95.04

Q10 )
In figure below, you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade.

Answer :

Given :

Height (h) of the frame of lampshade = (2.5 + 30 + 2.5) cm = 35cm

Radius (r) of the circular end of the frame of lampshade =

Now, Cloth required for covering the lampshade

=

=

= (44 × 10 × 5)

= 2200

Hence, for covering the lampshade, 2200

Q11 ) The students of a Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition?

Answer :

Given :

Radius (r) of the circular end of cylindrical penholder = 3 cm

Height (h) Of penholder = 10.5 cm

We have,

Surface area of 1 penholder

= CSA of penholder + Area of base of penholder

=

= [

= [

=

=

So, we get,

Area Of cardboard sheet used by 1 competitor =

Thus, Area of cardboard sheet used by 35 competitors

=

Solution for Exercise 13.3

Q1 ) Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.

Answer :

Given :

Radius (r) of the base of cone =

Slant height (l) of cone = 10 cm

We know that, CSA of cone

=

= [

= [

= 165

Therefore, the curved surface area of the cone is 165

Q2 ) Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.

Answer :

Given :

Radius (r) of the base of cone =

Slant height (l) of cone = 21 cm

We know that, TSA of cone

=

= [

= [

= 1244.57

Therefore, the total surface area of the cone is 1244.57

Q3 )
Curved surface area of a cone is 308

i) radius of the base and

ii) total surface area of the cone.

Answer :

Given :

Slant height (l) of cone = 14 cm

Curved surface area of a cone is 308

i) Let the radius Of the circular end of the cone be r.

We know that,

CSA of cone

=

= [

= [

= 44r cm

Therefore, the radius Of the circular end Of the cone is = 7 cm.

ii) Total surface area of cone

= CSA of cone + Area of base

= [

= [

= [308 + 154]

= 462

Therefore, the total surface area of the cone is 462

Q4 )
A conical tent is 10 m high and the radius of its base is 24 m. Find

i) slant height of the tent.

ii) cost of the canvas required to make the tent, if the cost of 1

Answer :

i) Let ABC be a conical tent.

Given :

Height (h) of conical tent = 10 m

Radius (r) of conical tent = 24 m

Let the slant height of the tent be l.

In

Therefore, the slant height Of the tent is 26 m.

ii) We know, CSA of tent

=

= [

=

Cost of 1

Cost of [

= Rs. [

= Rs. 137280.

Therefore, the cost of the canvas required to make such a tent is Rs. 137280.

Q5 )
What length of tarpaulin 3m wide will be required to make conical tent of height 8 m and base radius 6 m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm (Use

Answer :

Given :

Height (h) of conical tent = 8 cm

Radius (r) of base of tent = 6 m

Slant height (l) of tent =

Therefore, the slant height Of the tent is 10 m.

We know, CSA of tent

=

= [

=

Now, let the length of tarpaulin sheet required be l.

As, 20 cm will be wasted, therefore, the effective length will be (l- 0.2) m.

Breadth of tarpaulin = 3m ...(Given)

We have, Area of sheet = CSA of tent

Q6 )
The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of Rs. 210 per 100

Answer :

Given :

Slant height (l) of conical tomb = 25 m

Base radius (r) of tomb =

We know, CSA of tent =

= [

= [

=

Now, we know that,

Cost of white-washing 100

So, Cost of white-washing 550

= Rs. [

= Rs. 1155

Therefore, it will cost of Rs. 1155 while white-washing such a conical tomb.

Q7 ) A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.

Answer :

Given :

Radius (r) of conical cap = 7 cm

Height (h) of conical cap = 24 cm

Slant height (l) of the conical cap

Therefore, the slant height Of the tent is 25 cm.

We know, CSA of tent

=

= [

= [

=

Thus, CSA of 10 such caps

= (10 x 550)

= 5500

Q8 )
A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is Rs. 12 per

Answer :

Given :

Radius (r) of conical cap =

Height (h) of conical cap = 1 m

Slant height (l) of the conical cap

We know, CSA of tent

=

= [

=

So, CSA of 50 such cones

=

= 32.028

Now, we have,

Cost of painting 1

Cost of painting 32.02

= Rs. 384.336

= Rs 384.34 (approximately)

Therefore, it Will cost Rs. 384.34 in painting 50 such hollow cones.

Solution for Exercise 13.4

Q1 )
Find the surface area of a sphere of radius:

i) 10.5 cm

ii) 5.6 cm

iii) 14 cm

Answer :

i) Radius (r) of sphere = 10.5 cm

Surface area of sphere

=

= [

= [

= (88 × 1.5 × 10.5)

= 1386

Therefore, the surface area of a sphere having radius 10.5cm is 1386

ii) Radius (r) of sphere = 5.6 cm

Surface area of sphere

=

= [

= [

= (88 × 0.8 × 5.6)

= 394.24

Therefore, the surface area of a sphere having radius 5.6 cm is 394.24

iii) Radius (r) of sphere = 14 cm

Surface area of sphere

=

= [

= [

= (4 × 44 × 14)

= 2464

Therefore, the surface area of a sphere having radius 14 cm is 2464

Q2 )
Find the surface area of a sphere of diameter:

i) 14 cm

ii) 21 cm

iii) 3.5 m

Answer :

i) Radius (r) of sphere =

Surface area of sphere

=

= [

= [

= (88 × 7)

= 616

Therefore, the surface area of a sphere having radius 14 cm is 616

ii) Radius (r) of sphere =

Surface area of sphere

=

= [

= [

= (88 × 1.5 × 10.5)

= 1386

Therefore, the surface area of a sphere having radius 10.5cm is 1386

iii) Radius (r) of sphere =

Surface area of sphere

=

= [

= [

= ((88 × 1.75)

= 38.5

Therefore, the surface area of a sphere having radius 3.5 m is 38.5

Q3 )
Find the total surface area of a hemisphere of radius 10 cm. (Use

Answer :

We know that,

total surface area Of hemisphere

= CSA of hemisphere + Area of circular end of hemisphere

=

=

= [

= 942

Therefore, the total surface area of such a hemisphere is 942

Q4 ) The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.

Answer :

Given :

Radius (r) of spherical balloon = 7 cm

Radius (R) of spherical balloon, when air is pumped into it = 14 cm

=

=

=

Therefore, the ratio between the surface areas in these two cases is 1 : 4.

Q5 )
A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of Rs. 16 per 100

Answer :

Inner Radius (r) of sphere =

Surface area of sphere

=

= [

= [

= ((88 × 5.25)

= 173.25

Therefore, the surface area of a sphere having radius 5.25 cm is 173.25

Cost of tin-plating 100

Cost of tin-plating 173.25

Therefore, the cost of tin-plating the inner side of the hemispherical bowl is Rs. 27.72.

Q6 )
Find the radius of a sphere whose surface area is 154

Answer :

Let the radius of the sphere be r.

Given :

Surface area of sphere = 154

Q7 ) The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.

Answer :

Let the diameter of earth be d.

Therefore, the diameter of moon will be

Thus, Radius of earth =

So, Radius of moon =

Hence, Surface area of moon =

And Surface area of earth =

So, Required ratio =

Therefore, the ratio between their surface areas will be 1 : 16.

Q8 ) A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.

Answer :

Given :

Inner radius Of hemispherical bowl = 5cm

Thickness of the bowl = 0.25 cm

Outer radius (r) Of hemispherical bowl = (5 + 0.25) cm = 5.25 cm

We also know that,

Outer CSA of hemispherical bowl

=

=

=

Therefore, the outer curved surface area Of the bowl is 173.25

Q9 )
A right circular cylinder just encloses a sphere of radius r (see Fig). Find

i) surface area of the sphere

ii) curved surface area of the cylinder

iii) ratio of the areas obtained in i) and ii).

Answer :

i) Surface area of sphere =

ii) Height of cylinder = r + r = 2r

Radius of cylinder = r

We know that,

CSA of cylinder

=

=

=

iii) ratio of the areas obtained

=

=

=

Therefore, the ratio between these two surface areas is 1:1.

Solution for Exercise 13.5

1. A matchbox measures 4 cm × 2.5 cm × 1.5 cm. What will be the volume of a packet containing 12 such boxes?

Answer :

Given :

Matchbox is a cuboid having its length (l), breadth (b), height (h) as 4 cm, 2.5cm,and 1.5 cm.

We know that,

Volume of 1 match box = l x b x h

= (4 x 2.5 x 1.5)

= 15

So, the volume Of 12 match boxes is = Volume of 1 match box x 12

i.e., 15

= 180

Therefore, the volume Of 12 match boxes is 180

2. A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold? (1 (\{m}^{2}\) = 1000 L)

Answer :

Given :

The cuboidal water tank has its length (l) as 6 m, breadth (b) as 5 m, and
height (h) as 4.5 m.

We know that,

Volume of tank = l x b x h

= (6 x 5 x 4.5)

= 135

But, Amount of water that 1

Thus, Amount of water that 135 volume can hold = (135 x 1000) litres
= 135000 litres

Therefore, such tank can hold up to 135000 litres of Water.

3. A cuboidal vessel is 10 m long and 8 m wide. How high must it be made to hold 380

Answer :

Given :

Length (l) of vessel = 10 m

Width (b) Of vessel = 8 m

We know that,

Volume of tank = l x b x h

But, Volume of vessel = 380

Thus, l x b x h = 380

(10 x 8 x h)

(80 x h)

h =

h = 4.75 m

Therefore, the height Of the vessel should be 4.75 m.

4. Find the cost of digging a cuboidal pit 8 m long, 6 m broad and 3 m deep at the rate of ` 30 per

Answer :

Given: The cuboidal pit has its length (l) as 8 m, width (b) as 6 m, and depth (h)as 3 m.

We know that,

Volume Of pit = (8 x 6 x 3)

Now, Cost of digging per

Therfore, Cost of digging 144

5. The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank, if its length and depth are respectively 2.5 m and 10 m.

Answer :

Given :

Length (l) and depth (h) of tank is 2.5 m and 10 m respectively.

Let the breadth of the tank be b m.

Volume of tank = l x b x h

= (2.5 x b x 10)

= 25b

But, Capacity of tank = 50000 litres of water

Thus, 25b

i.e., 25000 b = 50000

i.e., b = 20 m.

Therefore, the breadth of the tank is 2 m.

6. A village, having a population of 4000, requires 150 litres of water per head per day. It has a tank measuring 20 m × 15 m × 6 m. For how many days will the water of this tank last?

Answer :

Given :

The tank is cuboidal in shape having its length (l) as 20 m, breadth (b) as 15 m, and height (h) as 6 m.

Capacity of tank = l x b x h

= (20 x 15 x 6)

= 1800

Thus, Water consumed by the people Of the village in 1 day = (4000 x 150) litres

i.e., = 600000 litres

Now, Let water in this tank last for n days.

So, we get that,

Water consumed by all people of village in n days = Capacity of tank

i.e., n x 600000 = 1800000

i.e., n = 3

Therefore, the water of this tank will last for 3 days.

7. A godown measures 40 m × 25 m × 15 m. Find the maximum number of wooden crates each measuring 1.5 m × 1.25 m × 0.5 m that can be stored in the godown.

Answer :

The godown has its length

while the wooden crate has its length

We get that, volume of godown = (40 x 25 x 10)

= 10000

Also, Volume of 1 wooden crate

= (1.5 x 1.25 x 0.5)

= 0.9375

Now, Let n wooden crates can be stored in the godown.

Therefore, volume of n wooden crates = Volume of godown

= 0.9375 x n = 10000

i.e., n = 10666.66

Therefore, 10666 Wooden crates can be stored in the godown.

8. A solid cube of side 12 cm is cut into eight cubes of equal volume. What will be the side of the new cube? Also, and the ratio between their surface areas.

Answer :

Given :

Side (a) of cube = 12 cm

We know that,

Volume of 8 cubes =

=

= 1728

Now, Let the side of the smaller cube be

Thus, volume of 1 smaller cube =

= 216

i.e.,

i.e.,

Therefore, the side of the smaller cubes will be 6 cm.

Ratio between surface areas of cube =

=

=

Therefore, the ratio between the surface areas of these cubes is 4:1.

9. A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute?

Answer :

Given :

Rate of water flow = 2 km per hour

=

=

Depth (h) of river = 3 m

Width (b) of = 40 m

So, we get,

Volume of water flowed in 1 min =

Therefore, in 1 minute, 4000

Solution for Exercise 13.6

1. The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. How many litres of water can it hold? (1000

Answer :

Given :

Height (h) of vessel = 25 cm

Circumference of vessel = 132 cm

Let the radius Of the cylindrical vessel be r.

i.e.,

i.e.,

i.e.,

We know that,

Volume of cylindrical vessel

= 34650

i.e.,

Therefore, such vessel can hold 34.65 litres Of water.

2. The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1

Answer :

Given :

Inner radius of cylindrical pipe (r) =

Outer radius of cylindrical pipe (R) =

Height (h) of pipe = Length of pipe = 35 cm

We know that,

Volume of pipe =

=

= 110 x 52

= 5720

Now, Mass of 1

So, Mass of 5720

= 3432 g

= 3.432 kg

Therefore, mass of the pipe is 3.432 kg.

3. A soft drink is available in two packs –

i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and

ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much?

Answer :

The tin can will be cuboidal in shape while the plastic cylinder Hill be cylindrical in shape.

i)

Given :

Length (l) of tin can = 5 cm

Breadth (b) of tin can = 4 cm

Height (h) Of tin can = 15 cm

Therefore, Capacity of tin can = l x b x h

= (5 x 4 x 15)

= 300

ii)

Given :

Radius (r) of circular end of plastic cylinder =

Height (h) of plastic cylinder = 10 cm

Therefore, Capacity of plastic cylinder =

=

=

Therefore, plastic cylinder has the greater capacity.

So, we get,

Difference in capacity = (385 - 300)

4. If the lateral surface of a cylinder is 94.2

i) radius of its base

ii) its volume.

(Use

Answer :

i) Given :

Height (h) of cylinder = 5 cm

Let radius of cylinder be r.

Also, CSA of cylinder = 94.2

Thus, we get that,

(2 × 3.14 × r × 5) cm = 94.2

i.e., r = 3 cm

ii) We know that,

Volume of cylinder =

=

= 141.3

5. It costs Rs. 2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of painting is at the rate of Rs. 20 per

i) inner curved surface area of the vessel,

ii) radius of the base,

iii) capacity of the vessel.

Answer :

i)Given :

Rs. 20 is the cost Of painting 1

So, when Rs. 2200 is the cost of painting, area is =

= 110

Therefore, the inner curved surface area of the vessel is 110

ii)Height (h) of vessel = 10 m

Surface area = 110

Let the radius of the base of the vessel be r.

We know that,

Surface area =

But, CSA = 110

i.e., 110

Therefore, r =

iii) Now, volume of vessel =

=

= 96.25

Therefore, the capacity of the vessel is 96.25

i) inner curved surface area of the vessel,

ii) radius of the base,

iii) capacity of the vessel.

Answer :

i)Given :

Rs. 20 is the cost Of painting 1

So, when Rs. 2200 is the cost of painting, area is =

= 110

Therefore, the inner curved surface area of the vessel is 110

ii)Height (h) of vessel = 10 m

Surface area = 110

Let the radius of the base of the vessel be r.

We know that,

Surface area =

But, CSA = 110

i.e., 110

Therefore, r =

iii) Now, volume of vessel =

=

= 96.25

Therefore, the capacity of the vessel is 96.25

6. The capacity of a closed cylindrical vessel of height 1 m is 15.4 litres. How many square metres of metal sheet would be needed to make it?

Answer :

Given :

Height (h) Of cylindrical vessel = 1 m

Volume of cylindrical vessel = 15.4 litres = 0.0154

Let the radius of the circular end be r.

But we know that,

Volume of cylinder =

Thus, we get that,

i.e., r = 0.07 m.

Also, we have,

Total surface area of vessel =

=

= 0.44 × 1.07

= 0.4708

Therefore, 0.4708

7. A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length of the pencil is 14 cm, and the volume of the wood and that of the graphite.

Answer :

Given :

Radius (R) of pencil =

Radius (r) of graphite =

Height (h) of pencil = 14 cm

We know that,

Volume of wood in pencil =

=

=

= 44 × 0.12

= 5.28

Similarly, Volume of graphite in pencil =

=

= 44 × 0.0025

= 0.11

8. A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients?

Answer :

Given :

Radius (r) of cylindrical bowl =

Height (h) of bowl, up to which bowl is filled with soup = 4 cm

We know that,

Volume Of soup in bowl =

=

= 11 × 3.5 × 4

= 154

Therefore, Volume of soup given to 250 patients = (250 × 154)

Solution for Exercise 13.7

1. Find the volume of the right circular cone with

i) radius 6 cm, height 7 cm

ii) radius 3.5 cm, height 12 cm.

Answer :

i)Given :

Radius (r) of cone = 6 cm

Height (h) of cone = 7 cm

We know that,

Volume of cone =

=

= 12 × 22

= 264

Therefore, the volume of the cone is 264

ii)Given :

Radius (r) of cone = 3.5 cm

Height (h) of cone = 12 cm

Similarly,

Volume of cone =

=

= 12.833 × 12

= 154

Therefore, the volume of the cone is 154

2. Find the capacity in litres of a conical vessel with

i) radius 7 cm, slant height 25 cm

ii) height 12 cm, slant height 13 cm .

Answer :

i)Given :

Radius (r) of cone = 7 cm

Slant height (l) of cone = 25 cm

Height of cone =

=

=

We know that,

Volume of cone =

=

= 154 × 8

= 1232

Therefore, capacity of the conical vessel = 1232

ii) Height (h) of cone = 12 cm

Slant height (l) of cone = 13 cm

Radius of cone =

=

=

We know that,

Volume of cone =

=

= 300 × 1.047

= 314.28

Therefore, capacity of the conical vessel = 314.28

3. The height of a cone is 15 cm. If its volume is 1570

Answer :

Given :

Height (h) of cone = 9 cm

Also, Volume of cone = 1570

Let the radius of the cone be r.

But, Volume of cone =

i.e., 1570

i.e.,

Thus, r = 10 cm

Therefore, the radius of the base of cone is 10 cm.

4. If the volume of a right circular cone of height 9 cm is

Answer :

Given :

Height (h) of cone = 9 cm

Also, Volume of cone =

Let the radius of the cone be r.

But, Volume of cone =

i.e.,

i.e.,

i.e.,

i.e., r = 4cm

Thus, Diameter of base = 2r = 8 cm

Therefore, the diameter of the base of cone is 8 cm.

5. A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?

Answer :

Given :

Radius (r) of pit =

Height (h) of pit = depth of pit = 12 m

We know that,

Volume of pit =

=

= 38.5

Thus, capacity of the pit = (38.5 × 1) kilolitres = 38.5 kilolitres

6. The volume of a right circular cone is 9856

i) height of the cone

ii) slant height of the cone

iii) curved surface area of the cone.

Answer :

Given :

Radius of the base =

Also, Volume of a right circular cone = 9856

i) Let the height of the cone be h.

But, we know that,

Volume of right circular cone =

i.e., 9856

i.e., h = 48 cm.

Therefore, height of the cone is 48 cm.

ii)We know, slant height of cone =

=

l =

Therefore, slant height of the cone is 50 cm.

iii) We also know that,

Curved surface area of the cone =

=

= 2200

Therefore, the curved surface area of the cone is 2200

7. A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find the volume of the solid so obtained.

Answer :

When right-angled

Now, thus, we know that,

Volume of cone =

=

=

Therefore, the volume of the cone so formed is

8. If the triangle ABC in the Question 7 above is revolved about the side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in Questions 7 and 8.

Answer :

When right-angled

Now, thus, we know that,

Volume of cone =

=

=

=

Therefore, the volume of the cone so formed is

Thus, the ratio of the volumes of the two solids obtained in Questions 7 and 8

=

=

i.e., 5 : 12.

9. A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.

Answer :

Given :

Radius (r) of heap =

Height (h) of heap = 3 m

As we know that,

Volume of heap =

=

= 86.625

Therefore, the volume of the heap of wheat is 86.625

But, Area of canvas required = CSA of cone

So, we get,

=

=

=

= 99.825

Therefore, 99.825

Solution for Exercise 13.8

1. Find the volume of a sphere whose radius is

i) 7 cm

ii) 0.63 m

Answer :

i)Given :

Radius of sphere = 7 cm

We know that,

Volume of sphere =

=

=

=

Therefore, the volume of the sphere is

ii) Given :

Radius of sphere = 0.63 m

We know that,

Volume of sphere =

=

=

Therefore, the volume of the sphere is

2. Find the amount of water displaced by a solid spherical ball of diameter

i) 28 cm

ii) 0.21 m.

Answer :

i)Given :

Radius of sphere =

We know that,

Volume of sphere =

=

=

Therefore, the volume of the sphere is

ii)Given :

Radius of sphere =

We know that,

Volume of sphere =

=

=

Therefore, the volume of the sphere is

3. The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the density of the metal is 8.9 g per

Answer :

Given :

Radius of sphere =

We know that,

Volume of sphere =

=

=

Now, we know that,

i.e., Mass = Density × Volume

i.e, = 8.9 × 38.808 g

i.e., = 345.3912 g

Hence, the mass of the ball is 345.39 g (approximately).

4. The diameter of the moon is approximately one-fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon?

Answer :

Let the diameter of earth be d.

Therefore, the diameter of moon will be

Thus, Radius of earth =

So, Radius of moon =

Now, Volume of moon =

Similarly,

Volume of earth =

So,

Therefore, the volume of moon =

5. How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?

Answer :

Given :

Radius (r) of hemisphere bowl =

We know that,

Volume of hemisphere =

=

=

=

= 0.303 litre (approximately)

Therefore ,the volume hemisphere bowl is 0.303 litres.

6. A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank.

Answer :

Given :

Inner radius of hemispherical tank (r) = 1 m

Thickness of hemispherical tank 1 cm = 0.01 m

Outer radius of hemispherical tank (R) = (1 + 0.01) m = 1.01 m

We know that,

Volume of iron is used to make such a tank =

=

=

= 0.06348

Therefore, the volume of the iron used to make the tank is 0.06348

7. Find the volume of a sphere whose surface area is 154

Answer :

Given :

Surface area of sphere = 154

Let radius of sphere be r.

But, we also know that,

Surface area of sphere =

i.e., 154

i.e.,

i.e., r =

i.e., r = 3.5 cm

Now, Volume of sphere =

=

=

Therefore , the volume of the sphere is =

8. A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of Rs. 498.96. If the cost of white-washing is 2.00 per square metre, find the

i) inside surface area of the dome,

ii) volume of the air inside the dome.

Answer :

i)Given :

Cost of white washing the dome from inside = Rs. 498.96

Also, Cost of white washing 1

Therefore, CSA of the inner side of dome = Rs.

ii) Let the inner radius of the hemispherical dome be r.

Now, CSA of inner side of dome = 249.48

But, CSA =

i.e., 249.48

i.e.,

Therefore, r = 6.3 m.

Also, we have,

Volume of air inside the dome = Volume of hemispherical dome

=

=

=

Therefore, volume of the air inside the dome is

9. Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S'. Find the

i) radius r' of the new sphere,

ii) ratio of S and S'.

Answer :

i) Radius of 1 solid iron sphere = r

Volume of 1 solid iron sphere =

Thus, Volume of 27 solid iron spheres = 27 ×

27 solid iron spheres are melted to form 1 iron sphere.

Therefore, the volume of this iron sphere will be equal to the volume of 27 solid iron spheres.

Now, Let the radius of this new sphere be r'.

Therefore, Volume of new solid iron sphere =

Thus,

i.e.,

i.e., r' = 3r.

ii) Now, we have,

surface area of 1 solid iron sphere of radius r =

So, Surface area of iron sphere of radius r' =

=

=

Therefore, ratio of S and S' =

=

10. A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in

Answer :

Given :

Radius (r) of capsule =

We know that,

Volume of spherical capsule =

=

=

=

Therefore ,the volume of the spherical capsule is

Solution for Exercise 13.9

Q.1 Height = 110cm, Depth = 25cm, Breadth = 85cm (see figure). The thickness of the plank is 5 cm everywhere. The external faces are to be polished and the inner faces are to be painted. If the rate of polishing-is 20 paise per

Answer :

External dimensions of book self,

Length, l = 85cm

Breadth, b = 25 cm

Height, h = 110 cm

External surface area of shelf while leaving out the front face of the shelf

= lh+2(lb+bh)

= [85×110+2(85×25+25×110)] = (9350+9750) = 19100

External surface area of shelf is 19100

Area of front face = [85×110-75×100+2(75×5)] = 1850+750 = 2600

So, area is 2600

Area to be polished = (19100+2600)

Cost of polishing 1

Cost of polishing 21700

Dimensions of row of the book shelf

Length(l) = 75 cm

Breadth (b) = 20 cm and

Height(h) = 30 cm

Area to be painted in one row= 2(l+h)b+lh = [2(75+30)× 20+75×30] = (4200+2250) = 6450

So, area is 6450

Area to be painted in 3 rows = (3×6450)

Cost of painting 1

Cost of painting 19350

Total expense required for polishing and painting = Rs. (4340+1935) = Rs. 6275

Q.2 The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in figure. Eight such spheres are-used for this purpose, and are to be painted silver. Each support is a cylinder of radius 1.5 cm and height 7 cm and is to be painted black. Find the cost of paint required if silver paint costs 25 paise per

Answer :

Diameter of wooden sphere = 21 cm

Radius of wooden sphere, r = diameter/ 2 = (

Surface area of a wooden sphere =

Radius of the circular end of cylindrical support = 1.5 cm

Height of cylindrical support = 7 cm

Curved surface area of one cylindrical support = 2

= 2×(

Now,

Area of the circular end of cylindrical support =

=

Again,

Area to be painted silver = [8 ×(1386-7.07)] = 8×1378.93 = 11031.44

Area to be painted is 11031.44

Cost for painting with silver colour = Rs(11031.44×0.25) =Rs 2757.86

Area to be painted black = (8×66)

Cost for painting with black colour =Rs (528×0.05) = Rs26.40

Therefore, the total painting cost is:

= Rs(2757.86 +26.40)

= Rs 2784.26

Q.3 The diameter of a sphere is decreased by 25%. By what per cent does its curved surface area decrease?

Answer :

Let the diameter of a sphere be d.

After decreasing, diameter of the sphere

= d –

= d –

Since, surface area of a sphere =

Surface area of a sphere, when diameter of the sphere is

Now, decrease percentage in curved surface area

=

= 43.75%