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SEBA Class 9 Mathematics Chapter 13 Surface Areas and Volumes
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Surface Areas and Volumes
Chapter – 13
| Exercise 13.1 |
1. A plastic 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is to be open at the top. Ignoring the thickness of the plastic sheet, determine.
(i) The area of the sheet, required for making the box.
Ans: l = 1.5m.
b = 1.25 m.
h = 65cm = 0.65m.
∴ The area of the sheet required for making the box
(ii) The cost of a sheet for it, if a sheet measuring 1 m² costs R 20.
Ans: The cost of sheet for it = Rs. 5.45 x 20 = Rs.109.
2. The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of whitewashing the walls of the room and the ceiling at the rate of Rs. 7.50 per m²
Ans: l = 5m
b = 4m
h = 3m
Area of the walls of the room
= 2(l + b) h = 2(5 + 4) = 54m²
Area of the ceiling = lb = (5)(4) = 20m²
∴ Total area of the walls of the room and the ceiling
= 54m² + 20m² = 74m²
∴ Cost of white washing the walls of the room and the ceiling
= 74 x 7.5 = Rs. 0.555
3. 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 m² is Rs. 15000, find the height of the hall.
[Hint. Area of the four walls = Lateral surface area.]
Ans: Let the length, breadth and height of the rectangular hall be I m, b m and h m respectively.
Perimeter = 250 m
⇒ 2(l + b) = 250 ⇒ l + b = 125…(1)
⇒ 2(l+b)h = 1500 ⇒ (l + b) h = 750
⇒ 125h = 750 | Using (1)
Hence the height of the hall is 6 m.
4. 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?
Ans: Each edge of the cubical box (a) = 10cm
∴ Lateral surface area of the cubical box = 4a²
= 4 (10)² = 400cm²
For cuboidal box
l = 12.5cm
b = 10 cm
h = 8cm
∴ Lateral surface area of the cuboidal box = 2(1+b) h
= 2(12.5 + 10)(8) = 360cm²
∴ Cubical box has the greater lateral surface area than the cuboidal box by (400-360) cm²
(ii) Which box has the smaller total surface area and by how much?
Ans: Total surface area of the cubical box = 6a²
= 6 (10)² = 600cm²
Total surface area of the cuboidal box
= 2(lb + bh + hl)
= 2[ (12.5(10) + (10)(8) + (8)(12.5)]
= 2[125 + 80 + 100] = 610cm²
∴ Cubical box has a smaller total surface area than the cuboidal box by (610-600) cm², i.e., 10 cm².
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?
Ans: For herbarium
l = 30cm
b = 25 cm
h = cm
∴ Area of the lass = 2(lb + bh + hl)
= 2[(30)(25) + (25)(25) + (25)(30)]
= 2[750 + 625 + 750] = 4250cm²
(ii) How much tape is needed for all the 12 edges?
Ans: The tape needed for all the 12 edges
= 4(l + b + h) = 4(30 + 25 + 25) = 320cm
7. Shanti Sweets Stall was placing an order for making cardboard boxes for packing their sweets. Two sizes were required. The bigger of dimensions 25cm x 20cm x 5 cm and the smaller of dimensions 15 cm x 12cm x 5 cm. For all the overlaps, 5% of the total surface area is required extra, for all the overlaps. If the cost of the cardboard is Rs. 4 for 1000 cm², find the cost of cardboard required for supplying 250 boxes of each kind.
Ans: For bigger box
l = 25cm
b = 20cm
h = 5cm
∴ Total surface area of the bigger box
= 2(lb + bh + hl)
= 2[(25)(20) + (20)(5) + (5)(25)]
= 2[500 + 100 + 125] = 1450cm²
Cardboard required for all the overlaps
∴ Net surface area of the bigger box
= 1450cm² + 72.5cm² = 1522.5cm²
∴ Net surface area of 250 bigger boxes
= 1522.5 x 250 = 280625cm²
For smaller box:
l = 15cm
b = 12cm
h = 5cm
∴ Total surface area of the box = 2(lb + bh + hl)
= 2[(12)(12) + (12)(5) + (5)(15)]
= 2[180 + 60 + 75] = 630cm²
Cardboard required for all the overlaps
∴ Net surface area of the smaller box
= 630cm² + 31.5cm² = 661.5cm²
∴ Net surface area of 250 smaller boxes
= 661.5 x 250 = 165375cm²
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 4m x 3m?
Ans: For shelter:
l = 4m
b = 3m
h = 2.5m
∴ Total surface area of the shelter = lb + 2(bh + hl)
= (4)(3) + 2[(3)(2.5) + (2.5)(4)]
= 12 + 2[7.5 + 10] = 47m²
Hence 47 m² of tarpaulin will be required.
∴ Cost of cardboard required for supplying 250 boxes of each kind
= Rs.1522 .50 + Rs.661.50 = Rs. 2184
9. The total surface area of a right circular cone-
(a) π (r + h)
(b) πrl
(c) πr²h
(d) πr(r + 1)
Ans: (d) πr(r + 1)
10. The surface area of a sphere of diameter 14 cm is
(a) 661cm²
(c) 166cm²
(c) 576cm²
(d) 616cm²
Ans: (d) 616cm²
| Exercise 13.2 |
1. The curved surface area of a right circular cylinder of height 14 cm is 88cm². Find the diameter of the base of the cylinder.
Ans: Let the radius of the base of the cylinder be r cm.
h = 14cm
Curved surface area = 88cm² | given
⇒ 2r = 2
Hence the diameter of the base of the cylinder is 2 cm.
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?
Ans: h = 1m = 100cm
2r = 140cm
∴ Total surface area of the closed cylindrical tank
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. Find its
(i) inner curved surface area.
(ii) outer curved surface area.
(iii) total surface area.
Ans: h = 77cm, 2r = 4 cm
⇒ r = 2cm, 2R = 4.4cm
⇒ R = 2.2cm
(i) Inner curved surface area = 2πrh
(ii) Outer curved surface area = 2πrh
(iii) Total surface area
= 1064.8 + 968 + 5.28 = 2038.08cm²
4. 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 m².
Ans: 2r = 84 cm, ⇒ r = 42cm, h = 120cm
∴ Area of the playground levelled in taking 1 complete revolution
= 2πrh
∴ Area of the playground = 31680 x 500
= 1584m²
Hence the area of the playground is 1584m²
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 Rs. 12.50 per m².
Ans: 2r = 50cm
∴ r = 25cm = 0.25m
h = 3.5m
∴ Curved surface area of the pillar = 2πrh
∴ Cost of painting the curved surface of the pillar at the rate of Rs. 12.50 per m²
= Rs. 5.5 x 12.50 = Rs.68.75
6. Curved surface area of a right circular cylinder is 4.4m². If the radius of the base of the cylinder is 0.7 m, find its height.
Ans: Let the height of the right circular cylinder be h m.
r = 0.7m
Curved surface area = 4.4m²
⇒ 4.4h = 4.4 ⇒ h = 1m
Hence the height of the right circular cylinder is 1 m.
7. The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find
(i) its inner curved surface area.
Ans:
h = 10m
∴ Inner curved surface area of the circular well
= 2πrh
(ii) the cost of plastering this curved surface at the rate of Rs. 40 per m².
Ans: Cost of plastering the curved surface at the rate of Rs. 40 per m²
= Rs.11 x 40 = Rs. 4400
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.
Ans: h = 28m,
2r = 5cm
∴ Total radiating surface in the system
= 2πrh
Example 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.
Ans: 2r = 4.2m
h = 4.5m
∴ Lateral or curved surface area = 2πrh
(ii) how much steel was actually used if it was wasted in making the closed tank? 1/12 of the steel actually used.
Ans: Total surface area = 2πr(h + r)
Let the actual area of steel used be x m²
Since 1/12 of the actual steel used was wasted, the area of the steel
∴ Steel actually used = 95.04m².
10. In the 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.
Ans: 2r = 20 cm ⇒ r = 10 cm
h = 30cm
= 2200cm²
11. The students of a Vidyalaya were asked to participate in a competition for making and decorating pen holders 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 com-petitors with cardboard. If there were 35 competitors, how much card-board was required to be bought for the competition?
Ans: r = 3cm,
h = 10.5cm
∴ Cardboard required for 1 competitor = 2πrh + πr²
∴ Cardboard required for 35 competitors
Hence 7920cm² of cardboard was required to be bought for the competition.
| Exercise 13.3 |
1. Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.
Ans: ∵ Diameter of the base = 10.5cm
Slant height (l) = 10cm
∴ Curved surface area of the cone = πrl
2. Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
Ans: Slant height (l) = 21m
Diameter of base = 24m
∴ Total curved surface area of the cone = πr(l + r)
Q. 3 Curved surface area of a cone is 308cm² and its slant height is 14 cm. Find
(i) radius of the base.
Ans: Slant height (l) = 14cm
Curved surface area = 308cm²
Hence the radius of the base is 7 cm.
(ii) total surface area of the cone.
Ans: Total surface area of the cone = πr(l + r)
Hence the total surface area of the cone is 462cm²
4. A conical tent is 10 m high and the radius of its base is 24 m. Find
(i) slant height of the tent.
Ans: h = 10m
r = 24m
Hence the slant height of the tent is 26 m.
(ii) cost of the canvas required.
Ans: Curved surface area of the tent = πrl
∴ Cost of the canvas required to make the tent, if the cost of 1 m² canvas is Rs. 70
Hence the cost of the canvas is Rs. 137280
5. What length of tarpaulin 3 m 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 wast-age in cutting is approximately 20 cm (Use π = 3.14)
Ans: For conical tent
∴ Curved surface area = πrl
= 3.14 x 6 x 10 = 188.4m²
Width of tarpaulin = 3m
Extra length of the material required = 20cm = 0.2m
∴ Actual length of tarpaulin required = 62.8m + 0.2m = 63m
6. 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 100m².
Ans: Slant height (l) = 25 m
Base diameter = 14m
∴ Cost of white-washing the curved surface of the tomb at the rate of Rs. 210 per 100 m².
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.
Ans: Base radius (r) = 7cm
Height (h) = 24cm
∴ Curved surface area of a cap
∴ Curved surface area of 10 caps = 550 x 10 = 5500 cm²
Hence the area of the sheet required to make 10 such caps is 5500cm².
8. 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 m² what will be the cost of painting all these cones? (Use π = 3.14 and take √(1.04) = 1.02 )
Ans: Base diameter = 40cm
Height (h) = 1m
∴ Curved surface area = πrl
= 3.14 x 0.2 x 1.02 = 0.64056m²
∴ Curved surface area of 50 cones
= 0.64056 x 50m² = 32.028m²
∴ Cost of painting all these cones = 32.028 x 12
= 384.336 = Rs. 384.34 (approximately).
| Exercise 13.4 |
1. Find the surface area of a sphere of radius:
(i) 10.5 cm
Ans: r = 10.5cm
∴ Surface area = 4πr²
(ii) 5.6 cm
Ans: r = 5.6cm
(iii) 14 cm
Ans: r = 14cm
2. Find the surface area of a sphere of radius:
(i) 14 cm
Ans: Diameter = 14cm
(ii) 21 cm
Ans: Diameter = 21 cm
(iii) 3.5 cm
Ans: Diameter = 3.5 cm
3. Find the total surface area of a hemisphere of radius 10 cm. (Use π = 3.14)
Ans: r = 10cm
∴ Total surface area of the hemisphere
= 3πr² = 3 x 3.14 x (10)² = 942cm²
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.
Ans: Case I. r = 7cm
Case II. r = 14cm
∴ Ratio of surface areas of the balloon
5. A Hemisphere bowl made of brass has an inner diameter 10.5 cm. Find the cost of tin-plating it on the insident the rate of Rs. 16 per 100 cm².
Ans: Inner diameter = 10.5cm
∴ Inner surface area
∴ Cost of tin-plating at the rate of Rs. 16 per 100 cm²
6. Find the radius of a sphere whose surface area is 154 cm².
Ans: Let the radius of the sphere be r cm.
Surface area = 154 cm²
Hence the radius of the sphere is 3.5 cm.
7. The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.
Ans: Let the diameter of the earth be 2r.
∴ Surface area of the earth = 4πr²
∴ Ratio of their surface area
8. A Hemisphere 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.
Ans: Inner radius of the bowl = 5 cm
Thickness of steel = 0.25 cm
∴ Outer radius of the bowl = 5 + 0.25 = 5.25cm
∴ Outer curved surface of the bowl
9. A right circular cylinder just encloses a sphere of radius r. Find
(i) surface area of the sphere.
Ans: Surface area of the sphere = 4πr²
(ii) curved surface area of the cylinder.
Ans: For cylinder
Radius of the base = r
Height = 2r
∴ Curved surface area of the cylinder = 2π(r)(2r) = 4πr²
(iii) ratio of the areas obtained in (i) and (ii).
Ans: Ratio of the areas obtained in (i) and (ii)
| Exercise 13.5 |
1. A matchbox measures 4cm x 2.5cm x 1.5cm. What will be the volume of a packet containing 12 such boxes?
Ans: Volume of a matchbox = 4 x 2.5 x 1.5cm³ = 15cm³
∴ Volume of packet containing 12 such
boxes = 15×12 cm² = 180 cm³
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? (1m³ = 1000l)
Ans: Capacity of the tank = 6 × 5 × 4.5 m³ = 135 m³
∴ Volume of water it can hold = 135m³ = 135 x 1000l = 135000l.
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?
Ans: Let the height of the cuboidal vessel be h m.
l = 10m
b = 8m
Capacity of the cuboidal vessel = 380m³.
Hence the cuboidal vessel must be made 4.75 m high.
4. Find the cost of digging a cuboidal pit 8 m long, 6 m broad and 3 m deep at the rate of Rs. 30 per m³.
Ans: l = 8m
b = 6m
h = 3m
∴ Volume of the cuboidal pit = lbh = 8 × 6 × 3 m³ = 144 m³
∴ Cost of digging the cuboidal pit @ Rs. 30 per m³
= Rs.144 x 30 = Rs. 4320
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.
Ans: Let the breadth of the cuboidal tank be b m.
l = 2.5m
h = 10m
Capacity of the cuboidal tank
Hence the breadth of the cuboidal 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 20m x 15m x 6m.
For how many days will the water of this tank last?
Ans: Requirement of water per head per day = 150 litres
∴ Requirement of water for the total population of the village per day
= 150 x 4000 litres
For tank
l = 20m
b = 15m
h = 6m
∴ Capacity of the tank = 20 x 15 x 6m³ = 1800m³
∴ Number of days for which the water of this tank last
7 A godown measures 40m x 25m x 10m. Find the maximum number of wooden crates each measuring 1.5m x 1.25m x 0.5m that can be stored in the godown.
Ans: For godown
l = 40m
b = 25m
h = 10m
∴ Capacity of the godown
= lbh = 40 x 25 x 10m³ = 10000m³
For a wooden crate
l = 1.5m
b = 1.25m
h = 0.5m
∴ Capacity of a wooden crate = lbh
= 1.5 x 1.25 x 0.5m³ = 0.9375m³
Hence the maximum number of wooden crates that can be stored in the godown is 10666.
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.
Ans: Side of the solid cube (a) = 12cm
∴ Volume of the solid cube
= a³ = (12)³ = 12 x 12 x12cm³ = 1728cm³
∴ It is cut into eight cubes of equal volume.
Let the side of the new cube be x cm.
Then, volume of the new cube = x³cm³
According to the question, x³ = 216
⇒ x = 6cm
Hence the side of the new cube will be 6 cm.
Surface area of the original cube = 6a² = 6 (12)² cm²
Surface area of the new cube = 6x² = 6 (6)² cm²
∴ Ration between their surface areas
Hence the ratio between their surface areas 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?
Ans: In one hour
l = 2 km = 2 x 1000m = 2000m
b = 40m
h = 3m
∴ Water fell into the sea in one hour
= lbh = 2000 x 40 x 3m³
∴ Water fell into the sea in a minute
Hence 4000m³ of water will fall into the sea in a minute.
| 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? (1000cm³= 11)
Ans: Let the base radius of the vessel ber cm.
Then, circumference of the base of the cylindrical vessel = 2πr cm
According to the question, 2πr = 132
∴ Capacity of the cylindrical vessel
Hence the cylindrical vessel can hold 34.65 I 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 1cm³ of wood has a mass of 0.6 g.
Ans: ∵ Inner diameter = 24 cm
∴ Outer diameter = 28 cm
Length of the pipe (h) = 35cm
Outer volume = πR² h
Inner volume = πr² h
∴ Volume of the wood used = Outer volume – Inner volume
= 21560cm³- 15840cm³ = 5720cm³
∴ Mass of the pipe = 5720 x 0.6g = 3432g = 3.432kg.
3. A soft drink is available in two packs – (i) a tin can with a rectan- gular 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?
Ans: (i) For tin can
l = 5cm
b = 4cm
h = 15cm
∴ Capacity = l x b x h = 5 x 4 x 15cm³ = 300cm³
(ii) For plastic cylinder
Height (h) = 10cm
Clearly the second container i.e., a plastic cylinder has greater capacс- ity than the first container i.e., a tin can by 385 – 380 = 85cm³.
4. If the lateral surface of a cylinder 94.2 cm² height is 5 cm, then find:
(i) radius of its base.
Ans: Let the radius of the base of the cylinder be r cm.
h = 5cm
Lateral surface = 94.2cm²
⇒ πrh = 94.2 ⇒ 2 x 3.14r x 5 = 94.2
⇒ r = 3 cm
Hence the radius of the base is 3 cm.
(ii) its volume. (Use π =3.14)
Ans: r = 3 cm
h = 4cm
∴ Volume of the cylinder = πr² h
= 3.14 × (3)² × 5 = 141.3cm³
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 m² find:
(i) inner curved surface area of the vessel.
Ans: Inner curved surface area of the vessel
(ii) radius of the base.
Ans: Let the radius of the base be r m
h = 10m
Inner curved surface area = 110m²
Hence the radius of the base is 1.75 m.
(iii) capacity of the vessel.
Ans: r = 1.75m
h = 10m
∴ Capacity of the vessel = πr²h
Hence the capacity of the vessel is 96.25m³ (or 96.25 kl).
6. The capacity of a closed cylindrical of height 1 m is 15.4 litres. How many square metres of metal sheet would be needed to make it?
Ans: h = 1m
Let the radius of the base be r m.
Capacity = 0.0154m³
∴ Curved surface area = 2πrh + 2πr²
= 0.44 + 0.0308 = 0.4708m²
Hence 0.47m² of metal sheet should be needed.
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.
Ans: For solid cylinder of graphite
Diameter = 1mm
Length of the pencil (h) = 14cm = 140mm
∴ Volume of the graphite = πr²h
For cylinder of wood
Diameter 7mm
Length of the pencil (h) = 14cm = 140mm
∴ Volume of the wood
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 does the hospital has to prepare daily to serve 250 patients?
Ans: Diameter = 7cm
Height (h) = 4 cm
∴ Volume of soup in the cylindrical bowl = πr²h
∴ Volume of soup to be prepared daily to serve 250 patients
= 154 × 250 cm³ = 38500 cm³ (or 38.51)
Hence the hospital has to prepare 38500 cm³ (or 38.5 l) of soup daily to serve 250 patient
| Exercise 13.7 |
1. Find the volume of the right circular cone with:
(i) radius 6 cm, height 7 cm.
Ans: r = 6 cm
h = 7cm
(ii) radius 3.5 cm, height 12 cm.
Ans: r = 3.5cm
h = 12cm
2. Find the capacity in litres of a conical vessel with:
(i) radius 7 cm, slant height 25 cm.
Ans: r = 7cm
l = 25cm
r² + h² = l²
⇒ (7)² + h² = (25)² ⇒ h² = (25)² – (7)²
⇒ h² = 625-49 ⇒ h² = 576
= 1232 cm³ = 1.232l.
(ii) height 12 cm, slant height 13 cm.
Ans: h = 12cm
l = 13cm
r² + h² = l²
⇒ r ² + (12)² = (13)² ⇒ r² + 144 = 169
⇒ r² = 169-144 ⇒ r² = 25
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)
Ans: Let the radius of the base of the cone be r cm.
h = 15 cm
Volume 1570 cm³
Hence the radius of the base of the cone is 10 cm.
4. If the volume of a right circular cone of height 9 cm is 48π cm³, find the diameter of its base.
Ans: Let the radius of the base of the circular cone be cm.
h = 9 cm
Volume = 48π cm³
⇒ 2r = 2(4) = 8 cm
Hence the diameter of the base of the right circular cone is 8 cm.
5. A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?
Ans: For conical pit
Diameter = 3.5 cm
Depth (h) = 12 m
= 38.5m³ = 38.5 × 100l = 38.5kl.
6. The volume of a right circular cone is 9856cm³. If the diameter of the base is 28 cm, find
(i) height of the cone.
Ans: Diameter of the base = 28cm
Let the height of the cone be h cm.
Volume 9856 cm³
Hence the height of the cone is 48 cm.
(ii) slant height of the cone.
Ans: r = 14cm
h = 48cm
Hence the slant height of the cone is 50 cm.
(iii) curved surface area of the cone.
Ans: r = 14cm
l = 50cm
Hence the curved surface area of the cone is 2200 cm²
7. A right triangle ABC with sides 5 cm, 12 cm and 13 cm revolved about the side 12 cm. Find the volume of the solid so obtained.
Ans: The solid obtained will be a right circular cone whose radius of the base is 5 cm and height is 12 cm.
∴ r = 5 cm
h = 12cm
Hence the volume of the solid obtained is 100π cm³.
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 Question 7 and 8.
Ans: The solid obtained will be a right circular cone whose radius of the base is 12 cm and height is 5 cm.
∴ r = 12 cm
h = 5cm
= 240π cm³
Ratio of the volumes of the two solids obtained = 100π:240π = 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.
Ans: For heap of wheat
Diameter 10.5 m
Height (h) = 3m
Hence the area of the canvas required is 99.825 m².
| Exercise 13.8 |
1. Find the volume of a sphere whose radius is:
(i) r = 7 cm.
Ans: r = 7
(ii) 0.63 m.
Ans: r = 0.63
2. Find the amount of water displaced by a solid spherical ball of diameter
(i) 28 cm.
Ans: Diameter = 28 cm
(ii) 0.21 m.
Ans: Diameter = 0.21 m
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 cm³?
Ans: Diameter = 4.2 cm
Density = 8.9 g per cm³
∴ Mass of the ball = Volume X Density
= 38.808 × 8.9 = 345.39 g (approx.)
4. The diameter of the moon is approximately one-fourth the di-ameter of the earth. What fraction of the volume of the earth is the volume of the moon?
Ans: Let the radius of the earth be r.
Then, diameter of the earth = 2r
(volume of the earth)
Hence the volume of the moon is 1/64 fraction of the volume of the earth.
5. How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?
Ans: Diameter = 10.5 cm
(approx.) = 0.303 l (approx.)
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.
Ans: Inner radius (r) = 1m
Thickness of iron sheet = lcm = 0.101m
∴ Outer radius (R) = Inner radius (r) + Thickness of iron sheet
= 1m + 0.01m = 1.01m
7. Find the volume of a sphere whose surface area is 154c m2.
Ans: Let the radius of the sphere be r cm.
Surface area = 154cm²
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 Rs. 2.00 per square metre, find the
(i) inside the surface area of the dome.
Ans: Inside surface area of the dome
(ii) volume of the air inside the dome.
Ans: Let the radius of the hemisphere be r m.
Inside surface area = 249.48m²
∴ Volume of the air inside
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.
Ans:
∴ Volume of 27 solid iron spheres
∴ Volume of the new sphere = 36 πr³
Let the radius of the new sphere be r’.
Hence the radius r’ of the new-sphere is 3r.
(ii) ratio of S and S’.
Ans: S = 4πr²
Hence the ratio of S and S’ is 1:9
10. A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm³) is needed to fill this capsule?
Ans: ∵ Diameter of the capsule = 3.5mm
Hence 22.46 mm² (approx.) of medicine is needed to fill this capsule.
11. The outer diameter of a spherical shell is 10 cm and the inner diameter is 9 cm. Find the volume of the metal contained in the shell.
Ans: ∵ Outer diameter = 10 cm
Volume of the metal contained in the shell
12. If the number of square centimetres on the surface of a sphere is equal to the number of cubic centimetres in its volume, what is the diameter of the sphere?
Ans: Let the radius of the sphere be r cm. Then,
Surface area = 4πr² cm²
⇒ r = 3 ⇒ 2r = 6
Hence the diameter of the sphere is 6 cm.
13. A cone and a hemisphere have equal bases and equal volumes. Find the ratio of their heights.
Ans: Let the radius of base of hemisphere and cone, each ber cm. Let the height of the cone be h cm.
