PSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes

 PSEB Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes

PSEB 9th Class Maths Solutions Chapter 13 Surface Areas and Volumes

Ex 13.1

Question 1.
A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is open 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 m2 costs ₹ 20.
Answer:
The plastic box to be made is open at the top. Hence, the plastic sheet is required for the lateral surfaces and the base.
Here, for the box to be made,
length l = 1.5 m;
breadth b = 1.25 m and
height h = 65 cm = 0.65 m.
Area of the plastic sheet required for open box = Lateral surface area + Area of base
= 2 h(l + b) + l × b
= 2 × 0.65 (1.5 + 1.25) + 1.5 × 1.25 m2
= 1.3 × 2.75 + 1.875 m2
= 3.575 + 1.875 m2
= 5.45 m2
Cost of 1 m2 sheet = ₹ 20
∴ Cost of 5.45 m2 sheet = ₹ (5.45 × ₹ 20)
= ₹ 109

Question 2.
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 ₹ 7.50 per m2.
Answer:
For the given room, length 1 = 5 m; breadth b = 4 m and height h = 3 m.
Area of the region to be white washed
= Area of four walls + Area of ceiling
= 2 h(l + b) + l × b
= 2 × 3 (5 + 4) + 5 × 4 m2
= 54 + 20 m2
= 74 m2
Cost of white washing 1 m2 region = ₹ 7.5
∴ Cost of white washing 74 m2 region
= ₹ (74 × 7.5)
= ₹ 555

Question 3.
The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of ₹ 10 per m2 is ₹ 15,000, find the height of the hall. [Hint: Area of the four walls = Lateral surface area.]
Answer:
Area painted at the cost of ₹ 10 = 1 m2
∴ Area painted at the cost of ₹ 15,000
= 15000/10
= 1500 m2
∴ Area of the four walls = 1500m2
∴ Lateral surface area = 1500 m2
∴ Perimeter Of the floor × Height = 1500 m2
∴ 250 m × Height = 1500 m2
∴ Height = 15000/250
∴ Height = 6 m

Question 4.
The paint in a certain container is sufficient to paint an area equal to 9.375 m2. How many bricks of dimensions 22.5 cm × 10 cm× 7.5 cm can be painted out of this container?
Answer:
For each brick, length l = 22.5 cm; breadth b = 10 cm and height h = 7.5 cm.
Total surface area of one brick
= 2 (lb + bh + hl)
= 2 (22.5 × 10 + 10 × 7.5 + 7.5 × 22.5) cm2
= 2 (225 + 75 + 168.75) cm2
= 2 (468.75) cm2
= 937.5 cm2
= 937.5/10000 m2 = 0.09375 m2
No. of bricks that can be painted with paint sufficient to paint 0.09375 m2 area = 1
∴ No. of bricks that can be painted with paint sufficient to paint 9.375 m2 area 9.375
= 9.375/0.09375 = 100

Question 5.
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:
For the cubical box, edge a = 10 cm and for the cuboidal box, length l = 12.5 cm; breadth b = 10 cm and height h = 8 cm
(i) Lateral surface area of cubical box
= 4a2
= 4 (10)2 cm2
= 400 cm
Lateral surface area of cuboidal box
= 2h(l + b)
= 2 × 8(12.5 + 10) cm2
= 16 × 22.5 cm2
= 360 cm2
Thus, the lateral surface area of cubical box is greater by 40 cm2 (400 – 360).

(ii) Total surface area of cubical box = 6a2
= 6 (10)2 cm2
= 600 cm2
Total surface area of cuboidal box
= 2 (lb + bh + hl)
= 2(12.5 × 10 + 10 × 8 + 8 × 12.5) cm2
= 2 (125 + 80 + 100) cm2
= 2 (305) cm2
= 610 cm2
Thus, the total surface area of cubical box is smaller by 10 cm2 (610 – 600).

Question 6.
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:
(i) For the cuboidal greenhouse, length l = 30 cm; breadth fa = 25 cm and height h = 25 cm.
Area of glass used
= Total surface area of cuboid
= 2 (lb + bh + hl)
= 2 (30 × 25 + 25 × 25 + 25 × 30) cm2
= 2 (750 + 625 + 750) cm2
= 2 (2125) cm2
= 4250 cm2

(ii) 12 edges of the cuboidal greenhouse is made-up of 4 lengths, 4 breadths and 4 heights.
∴ Length of tape needed for 12 edges
= 4l + 4b + 4h
= 4 (l + b + h)
= 4 (30 + 25 + 25) cm
= 4 (80) cm
= 320 cm

Question 7.
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 ₹ 4 for 1000 cm2, find the cost of cardboard required for supplying 250 boxes of each kind.
Answer:
For bigger cuboidal boxes, length l = 25 cm;
breadth b = 20 cm and height h = 5 cm.
Total surface area of a bigger box
= 2 (lb + bh + hl)
= 2 (25 × 20 + 20 × 5 + 5 × 25) cm2
= 2 (500 + 100 + 125) cm2
= 1450 cm2
Area of cardboard required for overlap
= 5 % of 1450 cm2
= 72.5 cm2
Thus, the total area of cardboard required for 1 bigger box = 1450 + 72.5 cm2
= 1522.5 cm2
∴ The total area of cardboard required for 250 bigger boxes = (1522.5 × 250) cm2
For smaller cuboidal boxes, length l = 15 cm; breadth b = 12 cm and height h = 5 cm.
Total surface area of a smaller box
= 2 (lb + bh + hl)
= 2 (15 × 12 + 12 × 5 + 5 × 15) cm2
= 2(180 +60 + 75) cm2
= 2 (315) cm2
= 630 cm2
Area of cardboard required for overlap
= 5% of 630 cm2
= 31.5 cm2
Thus, the total area of cardboard required for 1 smaller box = 630 + 31.5 cm2 = 661.5 cm2
∴ The total area of cardboard required for 250 smaller boxes = (661.5 × 250) cm2
Now, the total area of cardboard required for all the boxes
= (1522.5 × 250) + (661.5 × 250) cm2
= 250(1522.5 + 661.5) cm2
= 250 × 2184 cm2
Cost of 1000 cm2 cardboard = ₹ 4
∴ Cost of 250 × 2184 cm2 cardboard
= ₹ (4×250×2184/1000)
= ₹ 2184

Question 8.
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 4 m × 3 m ?
Answer:
For the box-like structure without base, length
l = 4m; breadth b = 3m and height h = 2.5m.
Area of tarpaulin required
= Area of lateral surfaces + Area of top
= 2 h(l + b) + l × b
= 2 × 2.5 (4 + 3) + 4 × 3 m2
= 35 + 12 m2
= 47 m2

Ex 13.2

Note: Assume π = 22/7, unless stated otherwise.

Question 1.
The curved surface area of a right circular cylinder of height 14 cm is 88 cm2. Find the diameter of the base of the cylinder.
Answer:
Height of cylinder h = 14 cm.
Curved surface area of a cylinder = 2 πrh
∴ 88 cm2 = 2 × r × 14cm
∴ 88×7/2×22×14 cm = r
∴ r = 1 cm
Now, diameter of the cylinder = 2r = 2 × 1 cm
= 2 cm
Thus, the diameter of the base of the cylinder is 2 cm.

Question 2.
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:
Height of cylindrical tank h = 1 m
Diameter of the cylinder =140 cm
∴ Radius of the cylinder r =  diameter/2
= 140/2 cm
= 70 cm
= 0.7 m
Total surface area of the closed cylindrical tank
= 2πr (r + h)
= 2 × 227 × 0.7 (0.7 + 1) m2
= 4.4 × 1.7 m2
= 7.48 m2
Thus, 7.48 m2 sheet is required to make the closed cylindrical tank.

Question 3.
A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4 cm (see the given figure). Find its

(i) inner curved surface area,
Answer:
For inner cylinder, diameter = 4 cm
∴ For inner cylinder,
radius r =  diameter/2 = 2 cm
and height (length) h = 77 cm.
Inner curved surface area of the pipe
= 2πrh
= 2 × 22/7 × 2 × 77 cm2
= 968 cm2
Thus, the inner curved surface area is 968 cm2.

  

Question 5.
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 ₹ 12.50 per m2.
Answer:
For the cylindrical pillar, diameter d = 50 cm = 0.5 m and height h = 3.5 m.
Curved surface area of the cylindrical pillar
= πdh
= 22/7 × 0.5 × 3.5 m2
= 5.5 m2
Cost of painting 1 m2 area = ₹ 12.50
∴ Cost of painting 5.5 m2 area = ₹ (12.50 x 5.5)
= ₹ 68.75
Thus, the cost of painting the curved surface of the pillar is ₹ 68.75.

Question 6.
Curved surface area of a right circular cylinder is 4.4 m2. If the radius of the base of the cylinder is 0.7 m, find its height.
Answer:
For the given cylinder, radius r = 0.7 m and
curved surface area = 4.4 m2.
Curved surface area of a cylinder = 2πrh
∴ 4.4 m2 = 2 × 22/7 × 0.7m × h
∴ h = 4.4×7/2×22×0.7m
∴ h = 1 m
Thus, the height of the cylinder is 1 m.

Question 7.
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 ₹ 40 per m2.
Answer:
A circular well means a cylindrical well. For the cylindrical well, diameter d = 3.5 m and height (depth) h = 10 m.
(i) Curved surface area of the well
= πdh
= 22/7 × 3.5 × 10 m2
= 110 m2

(ii) Cost of plastering 1 m2 region = ₹ 40
∴ Cost of plastering 110 m2 region
= ₹ (40 × 110)
= ₹ 4400

Question 8.
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:
For the cylindrical pipe, diameter d = 5 cm = 0.05 m and height (length) h = 28 m.
The radiation surface in the system is the •curved surface of the pipe.
Hence, we find the curved surface area of the cylindrical pipe.
Curved surface area of the cylindrical pipe
= πdh
= 22/7 × 0.05 × 28 m2
= 4.4 m2
Thus, the total radiating surface in the system is 4.4 m2.

Question 9.
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 1/12 of the steel actually used was wasted in making the tank.
Answer:
For the closed cylindrical tank, diameter d = 4.2 m, hence radius
r = 4.2/2 = 2.1 m and height h = 4.5 m.

Question 10.
In the given figure, 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:
The shape of the decorative cloth will be cylindrical.
For the cylinder of cloth, diameter d = 20 cm and height h = 30 cm + 2.5 cm + 2.5 cm = 35 cm.
Curved surface area of the cylinder of cloth
= πdh
= 22/7 × 20 × 35 cm2
Thus, 2200 cm2 cloth is required for covering the lampshade.

Question 11.
The students of 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 cardboards. If there were 35 competitors, how much cardboard was required to be bought for the competition ?
Answer:
The cylindrical penholders to be made have base but open at the top. Thus, to prepare a penholder, the area of the cardboard required will be given by the curved surface area of the cylinder and the area of base.

Ex 13.3

Note: Assume π = 22/7, unless stated otherwise.

Question 1.
Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.
Answer:

    

Question 5.
What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm.
(Use π = 3.14)
Answer:
For the conical tent to be made, radius r = 6 m and height h = 8 m.
l2 = h2 + r2 = 82 + 62 = 64 + 36 = 100
∴ l = √100 = 10 m
Area of the tarpaulin used in making tent
= Curved surface area of conical tent
= πrl
= 3.14 × 6 × 10 m2
= 188.4 m2
Now, the width of the tarpaulin is 3 m.
∴ Length of tarpaulin required = 188.4/3 m
= 62.8 m
But, 20 cm, i.e., 0.2 m of tarpaulin is required more for margins and wastage.
∴ Total length of the tarpaulin required = 62.8 + 0.2 m = 63 m
Thus, total length of tarpaulin required is 63 m.

Question 6.
The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of whitewashing its curved surface at the rate of ₹ 210 per 100 m2.
Answer:

Question 7.
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:
For the conical cap, radius r = 7 cm and height h = 24 cm.

Ex 13.4

Note: Assume π = 22/7, unless stated otherwise.

Question 1.
Find the surface area of a sphere of radius:
(i) 10.5 cm
Answer:

   

Note: We can also use the formula “Surface area of a sphere = πd2” as
4πr2 = π × 4r2 = π × (2r)2 = πd2, where r and d are radius and diameter of the sphere respectively.

Question 3.
Find the total surface area of a hemisphere of radius 10 cm. (Use π =3.14)
Answer:
For the given hemisphere, radius r = 10 cm.
Total surface area of a hemisphere
= 3πr2
= 3 × 3.14 × 10 × 10 cm2
= 942 cm2

Question 4.
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:

  

Question 8.
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:
For the given hemispherical bowl, the inner radius is 5 cm and the thickness of steel is 0.25 cm.
∴ Outer radius r of the given hemispherical bowl = 5 + 0.25 cm = 5.25 cm.
Curved surface area of a hemisphere

Question 9.
A right circular cylinder just encloses a sphere of radius r (see the given figure). 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:
Here,
radius of the cylinder = radius of the sphere = r and height of the cylinder h
= 2 × radius of the sphere = 2r
(i) Surface area of the sphere = 4πr2

(ii) Curved surface area of the cylinder
= 2 πrh
= 2 × 1 × r × 2r
= 4 πr2

Ex 13.5

Question 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:
For the cuboidal matchbox, length l = 4 cm; breadth b = 2.5 cm and height h = 1.5 cm.
Volume of a cuboidal matchbox
= l × b × h
= 4 × 2.5 × 1.5 cm3
= 15 cm3
Then, volume of 12 matchboxes = 12 × 15 cm3 = 180 cm3
Thus, the volume of a packet containing 12 matchboxes is 180 cm3.

Question 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 m3 = 1000 l)
Answer:
For the cuboidal water tank, length l = 6m; breadth b = 5 m and height h = 4.5 m.
Capacity of the cuboidal tank = l × b × h
= 6 × 5 × 4.5 m3
= 135 m3
1 m3 = 1000 litres
∴ 135 m3 = 135000 litres
Thus, the given cuboidal water tank can hold 1,35,000 litres of water.

Question 3.
A cuboidal vessel is 10 m long and 8 m wide. How high must it be made to hold 380 cubic metres of a liquid?
Answer:
For the cuboidal vessel, length l = 10 m;
breadth b = 8 m and capacity = 380 m3.
Capacity of a cuboidal vessel = l × b × h
∴ 380 m3 = 10 m × 8 m × h m
∴ h = 380/10×8 m
∴ h = 4.75 m
The height of the cuboidal vessel must be made 4.75 m.

Question 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 m3.
Answer:
For the cuboidal pit, length l = 8m; breadth b = 6 m and height (depth) h = 3 m.
Volume of the earth to be dugout to make the cuboidal pit = Volume of a cuboid
= l × b × h
= 8 × 6 × 3 m3
= 144 m3
Cost of digging out 1 m3 of earth = ₹ 30
∴ Cost of digging out 144 m3 of earth
= ₹ (30 × 144)
= ₹ 4320
Thus, the cost of digging the cuboidal pit is ₹ 4320.

Question 5.
The capacity of a cuboidal tank is 50,000 litres of water. Find the breadth of the tank, if its length and depth are respectively 2.5 m and 10 m.
Answer:
For the cuboidal tank, length l = 2.5 m;
height (depth) h = 10 m and
capacity = 50,000 litres.
1000 litres = 1 m3
∴ 50,000 litres = 50,000/1000 m3 = 50 m3
Capacity of cuboidal tank = l × b × h
∴ 50 m3 = 2.5 m × b m × 10 m
∴ b = 50/2.5×10 m
∴ b = 2 m
Thus, the breadth of the cuboidal tank is 2 m.

Question 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:
Total requirement of water per day
= No. of people × daily requirement per person
= 4000 × 150 litres
= 6,00,000 litres
= 6,00,000/1000 m3
= 600 m3
For the cuboidal tank, length l = 20 m;
breadth b = 15 m and height h = 6 m
Capacity of the cuboidal tank = l × b × h
= 20 × 15 × 6 m3
= 1800 m3
600 m3 of water can last for 1 day in the village.
∴ 1800 m3 of water can last for 1800/600 = 3 days in the village.

Question 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:
For the cuboidal godown, length l = 40 m;
breadth b = 25 m and height h = 15 m.
Capacity of cuboidal godown = l × b × h
= 40 × 25 × 15 m3
For the wooden cuboidal crate, length l = 1.5 m; breadth b = 1.25 m and height h = 0,5 m.
Volume of 1 cuboidal crate
= l × b × h
= 1.5 × 1.25 × 0.5 m3
∴The no. of crates that can be stored in the godown =

Question 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, find the ratio between their surface areas.
Answer:
For the original cube, edge a =12 cm.
Volume of original cube = a3 = 123 cm3
= 1728 cm3
8 cubes of equal volume are made from the original cube.

Thus, the required ratio of the total surface area of the original cube and the total surface area of a new cube is 4:1.
Note: If the ratio of TSA of the original ‘ cube and TSA of all the new cubes is required, then it will be 1 : 2.

Question 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:
2 km = 2000 m and 1 hour = 60 minutes
Rate of flow of water in the river

Ex 13.6

Question 1.
The circumference of the base of a 7 cylindrical vessel is 132 cm and its height is 25 cm. How many litres of water can it hold? (1000 cm3 = 1l)

 

Question 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:
(i) For the cuboidal container with rectangular base, length l = 5 cm; breadth b = 4 cm and height h = 15 cm.
Capacity of the cuboidal container
= Volume of cuboid
= l × b × h
= 5 × 4 × 15 cm3
= 300 cm3

(ii) Volume of a cylinder
= πr2h
= 3.14 × 3 × 3 × 5 cm3
= 141.3 cm3
Thus, the volume of the cylinder is 141.3 cm3.

Question 5.
It costs ₹ 2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of painting is at the rate of ₹ 20 per m2, find
(i) inner curved surface area of the vessel,
(ii) radius of the base and
(iii) capacity of the vessel.

 

Question 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 ?

Question 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, find the volume of the wood and that of the graphite.

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

Ex 13.7

Question 1.
Find the volume of the right circular cone with
(i) radius 6 cm, height 7 cm

Question 2.
Find the capacity in litres of a conical vessel with
(i) radius 7 cm, slant height 25 cm

 

Question 3.
The height of a cone is 15 cm. If its volume is 1570 cm3, find the radius of the base. (Use π = 3.14.)

Question 4.
If the volume of a right circular cone of height 9 cm is 48 πcm3, find the diameter of its base.

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

Question 6.
The volume of a right circular cone is 9856 cm3. If the diameter of the base is 28 cm, find
(i) height of the cone,
(ii) slant height of the cone and
(iii) curved surface area of the cone.

(iii) Curved surface area of a cone
= πrl
= 22/7 × 14 × 50 cm2
= 2200 cm2

Question 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:
A right circular cone is received when ∆ ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm.
For the cone so obtained, radius r = 5 cm, height h = 12 cm and slant height l = 13 cm.
Volume of the cone obtained
= 1/3 πr2h
= 1/3 × π × 5 × 5 × 12 cm3
= 100π cm3

Question 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:
Now, if ∆ ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 5 cm, again a right circular cone is received.
For the cone so obtained, radiqs r = 12cm; height h = 5 cm and slant height l = 13 cm.
Volume of the cone obtained

Question 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.

Ex 13.8

Question 1.
Find the volume of a sphere whose radius is
(i) 7 cm

       

Question 8.
A dome of a building is in the form of a hemisphere. From inside, it was white washed at the cost of ₹ 4989.60. If the s cost of whitewashing is ₹ 20 per square metre, find the
(i) inside surface area of s the dome.

  

Ex 13.9

Question 1.
A wooden bookshelf has external dimensions as follows : Height =110 cm, Depth = 25 cm, Breadth = 85 cm (see the given 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 20paise per cm2 and the rate of painting is 10 paise per cm2, find the total expenses required for polishing and painting the surface of the bookshelf.

Answer:
Outer faces to be polished:

• One face on back side of the bookshelf, measuring 110 cm × 85 cm.
• Two faces on the sides, each of those measuring 110 cm × 25 cm.
• The top and the base, each of those measuring 85 cm × 25 cm.
• Two vertical strips on the front side, each of those measuring 110 cm × 5 cm.
• Four horizontal strips on the front side, each of those measuring 75 cm × 5 cm.

Thus, total area of region to be polished
= [(110 × 85) + 2(110 × 25) + 2 (85 × 25) + 2(110 × 5) + 4(75 × 5)] cm2
= (9350 + 5500 + 4250 + 1100+ 1500) cm2
= 21700 cm2
20 paise per cm2 = ₹ 0.20 per cm2
Cost of polishing 1 cm2 region = ₹ 0.20
∴ Cost of polishing 21700 cm2 region
= ₹ (21700 × 0.20)
= ₹ 4340

Inner faces to be painted:

• Two faces on the sides each of those measuring 90 cm × 20 cm.
• Two faces each of two shelves, the top face and the bottom face, in all six face, each of those measuring 75 cm × 20 cm.
• Face on the back side, measuring 90 cm × 75 cm.

Thus, total area of the region to be painted
= [2 (90 × 20) + 6 (75 × 20) + (90 × 75)] cm2
= (3600 + 9000 + 6750) cm2
= 19350 cm2
10 paise per cm2 = ₹0.10 per cm2
Cost of painting 1 cm2 region = ₹ 0.10
∴ Cost of painting 19350 cm2 region = ₹ (19350 × 0.10) = ₹ 1935
Then, the total expense of polishing and painting = ₹ 4340 + ₹ 1935 = ₹ 6275

Question 2.
The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in the given 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 cm2 and black paint costs 5 paise per cm2.

Answer:
For each wooden sphere,
radius r =  diameter/2 = 21/2 cm
Curved surface area of 1 sphere
= 4πr2
= 4 × 22/7 × 21/2 × 21/2 cm2
= 1386 cm2
For each cylindrical support, radius r = 1.5 cm and height h = 7 cm.
Area of top of cylindrical support
= πr2
= 22/7 × 1.5 × 1.5 cm2
= 7.07 cm2 (approx.)
Hence, the area of each sphere to be painted silver = 1386 cm2 – 7.07 cm2 = 1378.93 cm2
∴ Total area of eight spheres to be painted silver = 1378.93 cm2 × 8 = 11031.44 cm2
25 paise per cm2 = ₹ 0.25 per cm2
Cost of painting silver in 1 cm2 region = ₹ 0.25
∴ Cost of painting silver in 11031.44 cm2 region
= ₹ (11031.44 x 0.25)
= ₹ 2757.86 (approx.)
Curved surface area of 1 cylindrical support
= 2πrh
= 2 × 22/7 × 1.5 × 7 cm
= 66 cm2
∴ Total area of eight cylindrical supports to be painted black = 66 cm2 × 8 = 528 cm2
5 paise per cm2 = ₹ 0.05 per cm2
Cost of painting black in 1 cm2 region = ₹ 0.05
∴ Cost of painting black in 528 cm2 region = ₹ (528 × 0.05)
= ₹ 26.40
Thus, the total cost of painting = ₹ 2757.86 + ₹ 26.40
= ₹ 2784.26 (approx.)

Question 3.
The diameter of a sphere is decreased by 25 %. By what per cent does its curved surface area decrease?
Answer:
Suppose, the initial diameter of the sphere is d units and radius is r units.
∴ d = 2r
Original curved surface area of the sphere
= 4πr2
= π (4r2)
= π (2r)2
= πd2 unit2
Now, the diameter of the sphere is reduced by 25 %. Hence, the new diameter of the sphere is 0.75d units.
New curved surface area of the sphere
= π (diameter)
= π (0.75d)2 unit2
= 0.5625 πd2 unit2
∴ The decrease in the curved surface area of the sphere = πd2 – 0.5625 πd2
= 0.4375 πd2 unit2
∴Percentage decrease in the curved surface area of the sphere = 0.4375πd2/πd2 × 100 = 43.75 %
Thus, when the diameter of a sphere is decreased by 25 %, its curved surface area decreases by 43.75 %.

MCQ

Multiple Choice Questions and Answer

Answer each question by selecting the proper alternative from those given below each question to make the statement true:

Question 1.
The total surface area of a cuboid with length 20 cm, breadth 15 cm and height 10 cm is
A. 1300
B. 650
C. 3000
D. 1500
Answer:
A. 1300

Question 2.
The lateral surface area of a cuboid with length 15 cm, breadth 8 cm and height 5 cm is ………………. cm2.
A. 115
B. 230
C. 600
D. 300
Answer:
B. 230

Question 3.
The diameter of a cylinder is 7 cm and its curved surface area is 220 cm2. Then, its height is ……………….. cm.
A. 35
B. 10
C. 44
D. 20
Answer:
B. 10

Question 4.
The total surface area of a closed cylinder with radius 3.5 cm and height 6.5 cm is ………………… cm2.
A. 110
B. 220
C. 330
D. 440
Answer:
B. 220

Question 5.
The curved surface area of a cone is 880 cm2. If its slant height is 20 cm, then its diameter is …………………… cm.
A. 14
B. 7
C. 3.5
D. 28
Answer:
D. 28

Question 6.
The curved surface area of a cone with diameter 14 cm and slant height 10 cm is ………………. cm2.
A. 220
B. 1540
C. 110
D. 440
Answer:
A. 220

Question 7.
The height of a cone is 24 cm and its slant height is 25 cm. Then, its diameter is ………………. cm.
A. 14
B. 7
C. 4
D. 49
Answer:
A. 14

Question 8.
The circumference of the base of a cone is 44 cm and its slant height is 15 cm. Then, its curved surface area is ……………….. cm2.
A. 14
B. 154
C. 330
D. 115
Answer:
C. 330

Question 9.
The diameter of a cone is 7 cm and its slant ! height is 16.5 cm. Then, its total surface area is …………………. cm2.
A. 110
B. 220
C. 105
D. 154
Answer:
B. 220

Question 10.
Total surface area of a hemisphere with radius 7 cm is ………………….. cm2.
A. 231
B. 115.5
C. 462
D. 154
Answer:
C. 462

Question 11.
Total surface area of a hemisphere is 72 cm2.
Then, its curved surface area is ………………….. cm2.
A. 24
B. 36
C. 48
D. 72
Answer:
C. 48

Question 12.
The surface area of a sphere is 616 cm2.
Then, its radius is ……………. cm.
A. 6
B. 8
C. 7
D. 14
Answer:
C. 7

Question 13.
In a cuboid, the area of the face with sides length and breadth is 120 cm2. If the height of the cuboid is 5 cm, then its volume is …………………… cm3
A. 120
B. 240
C. 600
D. 300
Answer:
C. 600

Question 14.
The volume of a cylinder is 2200 cm3 and its height is 7 cm. Then, the radius of the cylinder is ……………….. cm.
A. 5
B. 15
C. 10
D. 20
Answer:
C. 10

Question 15.
The radius and height of a cone are 7 cm and 3 cm respectively. Then, the volume of the cone is ……………… cm3.
A. 154
B. 168
C. 148
D. 462
Answer:
A. 154

Question 16.
The volume of a sphere is 4.5 π cm3. Then, its diameter is …………………. cm.
A. 3
B. 2
C. 1.5
D. 4
Answer:
A. 3

Question 17.
The ratio of radii of two cones is 2 : 3 and the ratio of their heights is 9:4. Then, the ratio of their volumes is
A. 1 : 1
B. 3 : 2
C. 1 : 3
D. 2 : 3
Answer:
A. 1 : 1

Question 18.
The circumference of the base of a cone is 44 cm and its height is 6 cm. Then, its volume is ………………… cm3.
A. 49
B. 98
C. 308
D. 154
Answer:
C. 308