NCERT Solutions for Chapter 13 Surface Areas and Volumes Class 9 Maths
Book Solutions1
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 m2 costs ₹ 20.
(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
(i) Here, l = 1.5 m,b = 1.25 m
h = 65 cm = (65/100) m = 0.65 m
∵ It is open from the top.
∵ Its surface area = [Lateral surface area] + [Base area]
= [2(l + b)h] + [l × b]
= [2(1.50 + 1.25)0.65 m2] + [1.50 x 1.25 m2]
= [2 x 2.75 x 0.65 m2] + [1. 875 m2]
= 3.575 m2 + 1.875 m2 = 5.45 m2
∵ The total surface area of the box = 5.45 m2
∴ Area of the sheet required for making the box = 5.45 m2
(ii) ∵ Rate of sheet = ₹ 20 per m2
∴ Cost of 5.45 m2 = ₹ 20 x 5.45
⇒ Cost of the required sheet = ₹ 109
Exercise 13.1
Page Number 213
2
The length, breadth and height of a room are 5m, 4m and 3m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of ₹ 7.50 per m2.
Answer
Length of the room (l)= 5m Breadth of the room (b) = 4 mHeight of the room (h) = 3 m
The room is like a cuboid whose four walls (lateral surface) and ceiling are to be white washed.
∴ Area for white washing = [Lateral surface area] + [Area of the ceiling]
= [2(l + b)h] + [l x b]
= [2(5 + 4) x 3 m2] + [5 x 4 m2]
= [54 m2] + [20 m2] = 74 m2
Cost of white washing: Cost of white washing for 1 m2 = ₹ 7.50
∴ Cost of white washing for 74 m2 = ₹ 7.50 x 74
= ₹ 555
The required cost of white washing is ₹555.
Exercise 13.1
Page Number 213
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 ₹15000, find the height of the hall.
Answer
Note: Area of four walls = Lateral surface area.A rectangular hall means a cubiod.
Let the length and breadth of the hall be ‘l’ and ‘b’ respectively.
∴ [Perimeter of the floor] = 2(l + b) ⇒ 2(l + b) = 250 m.
∵ Area of four walls = Lateral surface area = [2(l + b)] x h [where ‘h’ is the height of hall.]
∴ Cost of painting the four walls = ₹ 10 x 250 h = ₹ 2500 h
⇒₹2500 h = ₹ 15000
Exercise 13.1
Page Number 213
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
Total area that can be painted = 9.375 m2, since a brick is like a cuboid∴ Total surface area of a brick = 2[lb + bh + hl]
= 2[(22.5 x 10) + (10 x 7.5) + (7.5 x 22.5)] cm2
= 2[(225) + (75) + (168.75)] cm2
= 2[468.75] cm2
Exercise 13.1
Page Number 213
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?
(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 of the cubical box = 10 cm
∴ Lateral surface area = 4a2
= 4 × 102 cm2
= 4 × 100 cm2
= 400 cm2
Total surface area = 6a2
= 6 × 102 = 6 × 100 cm2 = 600 cm2
∵ For the cuboidal box,
l = 12.5 cm, b = 10 cm, h = 8 cm
∴ Lateral surface area = 2[(l + b)] x h
= 2[12.5 + 10] × 8 cm2
= 2[22.5 x 8] cm2 = 360 cm2
Total surface area = 2[lb + bh + hl]
= 2[(12.5 × 10) + (10 x 8) + (8×12.5)] cm2
= 2[125 + 80 + 100] cm2
= 2[305] cm2
= 610 cm2
Obviously,
(i) ∵ 400 cm2 > 360 cm2 and 400 – 360 = 40
∴ The cubical box, has greater lateral surface area by 40 m2
(ii) ∵ 610 cm2 > 600 cm2 and 610 – 600 = 10
∴ The cuboidal box has greater total surface area by 10 m2
Exercise 13.1
Page Number 213
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?
(i) What is the area of the glass?
(ii) How much of tape is needed for all the 12 edges?
Answer
The herbarium is like a cuboid Here, l = 30 cm, b = 25 cm, h = 25 cm(i) ∵ Area of a cuboid = 2[lb + bh + hl]
∴ Surface area of the herbarium (glass) = 2[(30 × 25) + (25 × 25) + (25 × 30)] cm2
= 2[750 + 625 + 750] cm2
= 2[2125] cm2
= 4250 cm2
Thus, the required area of glass is 4250 cm2.
(ii) Total length of 12 edges = 4l + 4b + 4h = 4(l + b + h)
= 4(30 + 25 + 25) cm
= 4 × 80 cm = 320 cm
Thus, length of tape needed = 320 cm
Exercise 13.1
Page Number 213
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 x 20 cm x 5 cm and the smaller of dimensions 15 cm x 12 cm x 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 box:Length (l) = 25 cm,
Breadth (b) = 20 cm,
Height (h) = 5 cm
∵ The box is like a cuboid and total surface area of a cuboid
= 2(lb + bh + hl)
∴ Area of a box = 2([25 × 20) + (20 x 5) + (5 × 25)] cm2
= 2[500 + 100 + 125] cm2
= 2[725] cm2
= 1450 cm2
⇒ Total surface area of 250 boxes = 250 × 1450 cm2
= 362500 cm2
For smaller box:
l = 15 cm, b = 12 cm, h = 5 cm
Total surface area of a box = 2[lb + bh + hl]
= 2[(15 × 12) + (12 × 5) + (5 × 15)]cm2
= 2[180 + 60 + 75] cm2
= 2[315] cm2
= 630 cm2
⇒ Total surface area of 250 boxes = 250 x 630 cm2 = 157500 cm2
Now, total surface area of both kinds of boxes = 362500 cm2 + 157500 cm2
= 5,20,000 cm2
Area for overlaps = 5% of [total surface area]
= (5/100)x 520000 cm2 = 26000 cm2
∴ Total area of the cardboard required = [Total area of 250 boxes] + [5% of total surface area]
= 520000 cm2 + 26000 cm2
= 546000 cm2
Cost of cardboard:
∵ Cost of 1000 cm2 = ₹ 4
∴ Cost of 546000 cm2
= (4/546000)/ 1000
= ₹ 4×546
= ₹ 2184
Exercise 13.1
Page Number 213
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
Here, height (h) = 2.5 mBase dimension = 4 m × 3 m
⇒ Length (l) = 4 m and
Breadth (b) = 3 m
∵ The structure is like a cuboid.
∴ The surface area of the cuboid, excluding the base
= [Lateral surface area] + [Area of ceiling]
= [2 (l + b)h] + [lb]
= [2 (4 + 3) × 2.5] + [4 x 3] m2
= [35] + [12] m2
= 47 m2
Thus, 47 m2 tarpaulin would be required.
SURFACE AREA OF A RIGHT CIRCULAR CYLINDER
When axis of a cylinder is perpendicular to the base radius then it is called a right circular cylinder.
If ‘r’ and ‘h’ be the radius and height of a cylinder respectively, then (i) Base area (= top area) = πr2
(ii) Curved surface area (Lateral surface area) of a cylinder = 2πrh
(iii) Total surface area of a cylinder = 2πrh + 2πr2 = 2πr(h + r)
Note: Unless it is mentioned assume π = (22/7)
When axis of a cylinder is perpendicular to the base radius then it is called a right circular cylinder.
If ‘r’ and ‘h’ be the radius and height of a cylinder respectively, then (i) Base area (= top area) = πr2
(ii) Curved surface area (Lateral surface area) of a cylinder = 2πrh
(iii) Total surface area of a cylinder = 2πrh + 2πr2 = 2πr(h + r)
Note: Unless it is mentioned assume π = (22/7)
Exercise 13.1
Page Number 213
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
Let ‘r ’ be the radius of the cylinder.Here, height (h) = 14 cm and curved surface area = 88 cm2
∵ Curved surface area of a cylinder = 2πrh
∴ 2 πrh = 88
Exercise 13.2
Page Number 216
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
Here, height (h) = 1 m∵ Diameter of the base = 140 cm = 1.40 m
Exercise 13.2
Page Number 216
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 figure) Find its.
(i) inner curved surface area,
(ii) outer curved surface area,
(iii) total surface area.
(i) inner curved surface area,
(ii) outer curved surface area,
(iii) total surface area.
Answer
Length of the metal pipe = 77 cm
Exercise 13.2
Page Number 216
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.
Answer
The roller is in the form of a cylinder diameter of the roller = 84 cm⇒ Radius of the cylinder =(84/2) = 42 cm
Length of the roller = 120 cm
⇒ Height of the cylinder (h) = 120 cm
∴ Curved surface area of the roller (cylinder) = 2 πrh
= 2 × 22 × 6 × 120 cm2
= 31680 cm2
∴ Area of the playground levelled in one revolution of the roller = 31680 cm2
Exercise 13.2
Page Number 217
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
Exercise 13.2
Page Number 217
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
Radius (r) = 0.7 m Let height of the cylinder be ‘h’ metres.∴ Curved surface area = 2πrh
Exercise 13.2
Page Number 217
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.
(i) its inner curved surface area.
(ii) the cost of plastering this curved surface at the rate of 40 per m2.
Answer
Inner diameter of the well = 3.5 m
Exercise 13.2
Page Number 217
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
Length of the cylindrical pipe = 28 m⇒ h = 28 m
Diameter of the pipe = 5 cm
Exercise 13.2
Page Number 217
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.
(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
The storage tank is in the form of a cylinder, and diameter of the tank = 4.2 m
Thus, the required area of the steel that was actually used is 95.04 m2
Exercise 13.2
Page Number 217
10
In the figure you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has 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 lampshade is in the form of a cylinder, where
Height = 30 cm.
∵ A margin of 2.5 cm is to be added to top and bottom.
∴ Total height of the cylinder h = 30 cm + 2.5 cm + 2.5 cm = 35 cm
Now, curved surface area = 2 πrh
= 2 × 22 × 10 × 5 cm2
= 2200 cm2
Thus, the required area of the cloth = 2200 cm2
Exercise 13.2
Page Number 217
11
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
Here, the penholders are in the form of cylinders Radius of a cylinder (r) = 3 cmHeight of a cylinder (h) = 10.5 cm
Since, a penholder must be open from the top.
∴ Surface area of a penholder (cylinder) = [Lateral surface area] + [Base area]
= 5 × 1584 cm2
= 7920 cm2 [∵ There are 35 competitors]
Thus, 7920 cm2 of cardboard was required to be bought.
Exercise 13.2
Page Number 217
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
Here, diameter of the base = 10.5 cm
Slant height (l) = 10 cm
∴ Curvered surface area of the cone = πrl
Exercise 13.3
Page Number 221
2
Find the total surfce area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
Answer
Here, diameter = 24 m
Exercise 13.3
Page Number 221
3
Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find:
(i) radius of the base and
(ii) total surface area of the cone.
(i) radius of the base and
(ii) total surface area of the cone.
Answer
Here, curved surface area = 308 cm2Slant height (l) = 14 cm
(i) Let the radius of the base be ‘r’ cm.
∴ πrl = 308
and curved surface area = 308 cm2 [Given]
∴ Total surface area
= [Curved surface area] + [Base area]
= 308 cm2 + 154 cm2
= 462 cm2
Exercise 13.3
Page Number 221
4
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 m2 canvas is ₹ 70.
(i) slant height of the tent.
(ii) cost of the canvas required to make the tent, if the cost of 1 m2 canvas is ₹ 70.
Answer
Here, height of the tent (h) = 10 mRadius of the base (r) = 24 m
Exercise 13.3
Page Number 221
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 wastage in cutting is apπroximately 20 cm (Use π = 3.14).
Answer
Here, Base radius (r) = 6 mHeight (h) = 8 m
∴ Area of the canvas (tarpaulin) required to make the tent = 188.4 m2
Let the length of the tarpaulin = ‘L’ m
∴ Length x Breadth = 188.4
⇒ L x 3 = 188.4
⇒ L= (188.4/3) = 62.8 m
Extra length of tarpaulin for margins = 20 cm = (20/100) m = 0.2 m
Thus, total length of tarpaulin required = 62.8 m + 0.2 m = 63.0 m
Exercise 13.3
Page Number 221
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 ₹ 210 per 100 m2.
Find the cost of white washing its curved surface at the rate of ₹ 210 per 100 m2.
Answer
= 22 × 25 m2 = 550 m2
Cost of white washing:
Rate of whitewashing = ₹ 210 per 100 m2
∴ Cost of whitewashing for 550 m2
Exercise 13.3
Page Number 221
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
Here, Radius of the base (r) = 7 cmheight (h) = 24 cm
⇒ Lateral surface area of 10 caps = 10 x 550 cm2 = 5500 cm2
Thus, the required area of the sheet = 5500 cm2
Exercise 13.3
Page Number 221
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 ₹12 per m2, what will be the cost of painting all these cones? (Use π = 3.14 and take √1.04= 1.02)
Answer
Here, ∵ Diameter of the base = 40 cm
= 1.02 m [∵ √1.04= 1.02 (Given)]
Now, curved surface area = πrl
⇒ Curved surface area of 1 cone = 3.14 x 0.2 x 1.02 m2
Exercise 13.3
Page Number 221
1
Find the surface area of a sphere of radius:
(i) 10.5 cm
(ii) 5.6 cm
(iii) 14 cm
(i) 10.5 cm
(ii) 5.6 cm
(iii) 14 cm
Answer
(i) Here r = 10.5 cm∴ Surface area of the sphere = 4πr2
Exercise 13.4
Page Number 225
2
Find the surface area of a sphere of diameter:
(i) 14 cm
(ii) 21 cm
(iii) 3.5 m
(ii) 21 cm
(iii) 3.5 m
Answer
Exercise 13.4
Page Number 225
3
Find the total surface area of a hemisphere of radius 10 cm. (Use π = 3.14)
Answer
= 314 cm2
∴ Total surface area = 628 cm2 + 314 cm2
= 942 cm2
REMEMBER
(i) Curved surface area of a hemisphere = 2πr2
(ii) Total surface area of a hemisphere = 3πr2
Exercise 13.4
Page Number 225
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
Case I: When radius (r1) = 7 cm
Exercise 13.4
Page Number 225
5
A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tinplating it on the inside at the rate of ₹ 16 per 100 cm2.
Answer
Inner diameter of the hemisphere = 10.5 m
∵ Curved surface area of a hemisphere = 2πr2
∴ Inner curved surface area of hemispherical bowl
Exercise 13.4
Page Number 225
6
Find the radius of a sphere whose surface area is 154 cm2.
Answer
Let the radius of the sphere be ‘r’ cm.∴ Surface area = 4πr2
⇒ 4πr2 = 154
Exercise 13.4
Page Number 225
7
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 radius of the earth = r∴ Radius of the moon = (r/4)
∵ Surface area of a sphere = 4πr2
Since, the earth as well as the moon are considered to be spheres.
∴ Surface area of the earth = 4πr2
⇒ [Surface area of the moon] : [Surface area of the earth] = 1 : 16
Thus, the required ratio = 1 : 16
Exercise 13.4
Page Number 225
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
Inner radius (r) = 5cmThickness = 0.25 cm
∴ Outer radius (R) = [5.00 + 0.25] cm = 5.25 cm
∴ Outer surface area of the bowl = 2πr2
Exercise 13.4
Page Number 225
9
A right circular cylinder just encloses a sphere of radius r (see Fig. 13.22). Find
(i) surface area of the sphere,
(ii) curved surface area of the cylinder,
(iii) ratio of the areas obtained in (i) and (ii).
(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) For the sphere: Radius = r∴ Surface area of the sphere = 4 πr2
(ii) For the right circular cylinder:
∵ Radius of the cylinder = Radius of the sphere
∴ Radius of the cylinder = r
Height of the cylinder = Diameter of the sphere
⇒ Height of the cylinder (h) = 2r
Since, curved surface area of a cylinder = 2πrh
= 2πr(2r)
= 4πr2
Exercise 13.4
Page Number 225
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
Measures of matchbox (cuboid) is 4 cm × 2.5 cm × 1.5 cm⇒ l = 4 cm,
b = 2.5 cm
h = 1.5 cm
∴ Volume of matchbox = (l x b) × h
= [4 cm × 2.5 cm] × 1.5 cm3
= 15 cm3
⇒ Volume of 12 boxes = 12 × 15 cm3
= 180 cm3
Exercise 13.5
Page Number 228
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
Here, Length (l)= 6 mBreadth (b) = 5 m
Depth (h) = 4.5 m
∴ Capacity = l x b x h
= 6 x 5 x 4.5 m3
= 3 x 45 = 135 m3
∵ 1 m3 can hold 1000 l.
∴ 135 m3 can hold (135 x 1000 l = 135000 l) of water.
∴ The required amount of water in the tank = 135000 l.
Exercise 13.5
Page Number 228
3
A cuboidal vessel is 10 m long and 8 m wide. How high must it be made to hold 380 cubic metres of liquid?
Answer
Length (l) = 10 mBreadth (b) = 8 m
Volume (v) = 380 m3
Let height of the cuboid be ‘h’.
Since, volume of a cuboid = l x b x h
∴ Volume of the cuboidal vessel = 10 x 8 x h m3 = 80h m3
⇒ 80h = 380
Thus, the required height of the liquid = 4.75 m
Exercise 13.5
Page Number 228
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
Length (l)= 8 mBreadth (b) = 6 m
Depth (h) = 3 m
∴ Volume of the cuboidal pit = l x b x h = 8 x 6 x 3 m3
= 144 m3
Since, rate of digging the pit is ₹ 30 per m3.
∴ Cost of digging = ₹ 30 x 144 = ₹ 4320
Exercise 13.5
Page Number 228
5
The capacity of 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
Length of the tank (l) = 2.5 mDepth of the tank (h) = 10 m
Let breadth of the tank = b m
∴ Volume (capacity) of the tank = l x b x h
= 2.5 x b x 10 m3
But the capacity of the tank = 50000 l
= 50 m3 [∵ 1000 l = 1 m3]
∴ 25b = 50 m3
Exercise 13.5
Page Number 228
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?
It has a tank measuring 20 m × 15 m × 6 m. For how many days will the water of this tank last?
Answer
Length of the tank (l) = 20 mBreadth of the tank (b) = 15 m
Height of the tank (h) = 6 m
∴ Volume of the tank = l x b x h = 20 × 15 × 6 m3 = 1800 m3
Since, 1 m3 = 1000 l
∴ Capacity of the tank = 1800 × 1000 l = 1800000 l
Village population = 4000
Since, 150 l of water is required per head per day.
∴ Amount of water is required per day = 150 x 4000 l.
Let the required number of days = x
∴ 4000×150× x = 1800000
Thus, the required number of days is 3.
Exercise 13.5
Page Number 228
7
A godown measures 60 m × 25 m × 10 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
Volume of the godown = 60x 25 × 10 m3Volume of a crate = 1.5 × 1.25 × 0.5 m3
Exercise 13.5
Page Number 228
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
Side of the given cube = 12 cm∴ Volume of the given cube = (side)3 = (12)3 cm3
Side of the smaller cube:
Let the side of the new (smaller) cube = n
⇒ Volume of smaller cube = n3
⇒ Volume of 8 smaller cubes = 8n3
n3 = (6)3
n = 6
Thus, the required side of the new (smaller) cube is 6 cm.
Ratio between surface areas:
Surface area of the given cube = 6 × (side)2
= 6 × 122 cm2
= 6 × 12 × 12 cm2
Surface area of one smaller cube = 6 × (side)2
= 6 × 62 cm2
= 6 × 6 × 6 cm2
∴ Surface area of 8 smaller cubes = 8 × 6 × 6 × 6 cm2
Now,
Thus, the required ratio = 1:4
Exercise 13.5
Page Number 228
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
The water flowing in a river can be considered in the form of a cuboid
Such that Length (l) = 2 km = 2000 m
Breadth (b) = 40 m
Depth (h) = 3 m
∴ Water volume (volume of the cuboid so formed) = l x b x h = 2000 x 40 x 3 m3
Now, volume of water fallowing in 1 hr (= 60 minutes) = 2000 x 40 x 3 m3
∴ Volume of water that will fall in 1 minute
= [ 2000 ×40×3]÷60 m3
Exercise 13.5
Page Number 228
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 cm3 = 1 l)
Answer
Let the base radius of the cylindrical vessel be ‘r’ cm.
∴ Circumference = 2πr
⇒ 2πr = 132 [∵ Circumference = 132 cm]
= 22 × 3 × 21 × 25 cm3
= 34650 cm3
∵ Capacity of the vessel = Volume of the vessel
∴ Capacity of cylindrical vessel = 34650 cm3
Since, 1000 cm3 = 1 litre
⇒ 34650 cm3
= (34650/1000) litres
= 34.65 L
Exercise 13.6
Page Number 230
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 cm3 of wood has a mass of 0.6 g.
Answer
Here, Inner diameter of the cylindrical pipe = 24 cm
⇒ Inner radius of the pipe (r) =
cm = 12 cm
Outer diameter of the pipe = 28 cm
⇒ Outer radius of the pipe (R) =
cm = 14 cm
Length of the pipe (h) = 35 cm
∵ Inner volume of the pipe = πr2h
Outer volume of the pipe = πr2h
∴ Amount of wood (volume) in the pipe = Outer volume – Inner volume
= πR2h - πr2h
= πh(R2-r2)
= πh(R+r)(R-r) [∵ a2 - b2 = (a+b)(a-b)]
= 22 × 5 × 26 × 2 cm3
Mass of the wood in the pipe
= [Mass of wood in 1 m3 of wood] × [Volume of wood in the pipe]
= [0.6g] × [22 × 5 × 26 × 2] cm3
= (6/10)× 22 × 10 × 26 g
= 6 × 22 × 26 g
Thus, the required mass of the pipe is 3.432 kg.
Exercise 13.6
Page Number 230
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?
(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
For rectangular pack: Length (l) = 5 cmBreadth (b) = 4 cm Height (h) = 15 cm
∴ Volume = l x b x h
= 5 x 4 x 15 cm3
= 300 cm3
⇒ Capacity of the rectangular pack = 300 cm3 ...(1)
For cylindrical pack: Base diameter = 7 cm
= 11 x 7 x 5 cm3 = 385 cm3
⇒ Volume of the cylindrical pack = 385 cm3 ...(2)
From (1) and (2),
we have 385 cm3 – 300 cm3 = 85 cm3
⇒ The cylindrical pack has the greater capacity by 85 cm3.
Exercise 13.6
Page Number 230
4
If the lateral surface of a cylinder is 94.2 cm2 and its height is 5 cm, then find:
(i) radius of its base
(ii) its volume. (Use π = 3.14)
(i) radius of its base
(ii) its volume. (Use π = 3.14)
Answer
Height of the cylinder (h) = 5 cm Let the base radius of the cylinder be ‘r’.(i) Since lateral surface of the cylinder = 2 πrh
But lateral surface of the cylinder = 94.2 cm2
2πrh = 94.2
Thus, the radius of the cylinder = 3 cm
(ii) Volume of a cylinder = πr2h
⇒ Volume of the given cylinder = 3.14 x (3)2 x 5 cm3
Exercise 13.6
Page Number 230
5
It costs ₹ 2200 to paint the inner curved surface of cylindrical vessel 10 m deep. If the cost of painting is at the rate of ₹ 20 per m2; find:
(i) inner curved surface of the vessel
(ii) radius of the base
(iii) capacity of the vessel.
(i) inner curved surface of the vessel
(ii) radius of the base
(iii) capacity of the vessel.
Answer
(i) To find inner curved surfaceTotal cost of painting = ₹ 2200
Rate of painting = ₹ 20 per m2
⇒ Inner curved surface of the vessel = 110 m2
(ii) To find radius of the base Let the base radius of the cylindrical vessel.
∵ Curved surface of a cylinder = 2 πrh
∴ 2πrh = 110
⇒ The required radius of the base = 1.75 m
(iii) To find the capacity of the vessel
Since, volume of a cylinder = πr2h
∴ Volume (capacity) of the vessel
Since, 1 m3 = 1000000 cm3 = 1000 l = 1 kl
∴ 96.5 m3 = 96.5 kl
Thus, the required volume = 96.25 kl
Exercise 13.6
Page Number 231
6
The capacity of 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
Capacity of the cylindrical vessel = 15.4L= 15.4×1000 cm3
Exercise 13.6
Page Number 231
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 graphite is 1 mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite.
Answer
Since, 10 mm = 1cm∴ 1 mm = (1/10) cm
For graphite cylinder
Thus,
the required volume of the graphite = 0.11 cm3
For the pencil Diameter of the pencil
Volume of the wood Volume of the wood = [Volume of the pencil] – [Volume of the graphite]
= 5.39 cm3 – 0.11 cm3 = 5.28 cm3
Thus, the required volume of the wood is 5.28 cm3.
Exercise 13.6
Page Number 231
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
The bowl is cylindrical.Diameter of the base = 7 cm
Thus, the hospital needs to prepare 38.5 litres of soup daily for 250 patients.
Exercise 13.6
Page Number 231
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.
(i) radius 6 cm, height 7 cm
(ii) radius 3.5 cm, height 12 cm.
Answer
(i) Here, radius of the cone r = 6 cm Height (h) = 7 cm
Exercise 13.7
Page Number 233
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
(i) radius 7 cm, slant height 25 cm
(ii) height 12 cm, slant height 13 cm
Answer
(i) Here, r = 7 and l = 25 cm
Thus, the required capacity of the conical vessel is 1.232 l.
(ii) Here, height (h) = 12 cm and l = 13 cm
Exercise 13.7
Page Number 233
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)
Answer
Here, height of the cone (h) = 15 cmVolume of the cone (v) = 1570 cm3
Let the radius of the base be ‘r’ cm
⇒ r2 = 102
⇒ r = 10 cm
Thus, the required radius of the base is 10 cm.
Exercise 13.7
Page Number 233
4
If the volume of a right circular cone of height 9 cm is 48p cm3, find the diameter of its base.
Answer
Volume of the cone = 48 πcm3Height of the cone (h) = 9 cm
Let ‘r’ be its base radius.
∵ Diameter = 2 × Radius
∴ Diameter of the base of the cone = 2 × 4 = 8 cm
Exercise 13.7
Page Number 233
5
A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?
Answer
Here, diameter of the conical pit = 3.5 m
= 38.5 m3
∴ 1000 cm3 = 1 l and 1000000 cm3 = 1m3
∴ 1000 x 1000 cm3 = 1000 l = 1 kl
Also 1000 x 1000 cm3 = 1 m3
⇒ 1 m3 = 1 kl
⇒ 38.5 m3 = 38.5 kl
Thus, the capacity of the conical pit is 38.5 kl.
Exercise 13.7
Page Number 233
6
The volume of 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
(iii) curved surface area of the cone.
(i) height of the cone
(ii) slant height of the cone
(iii) curved surface area of the cone.
Answer
Volume of the cone (v) = 9856 cm3Diameter of the base = 28 cm
⇒ Radius of the base =(28/2)cm = 14 cm
(i) To find the height Let the height of the cone be ‘h’ cm.
= 16 x 3 cm = 48 cm Thus, the required height is 48 cm.
(ii) To find the slant height Let the slant height be ‘ℓ’ cm.
∵ (Slant height)2 = (Radius)2 + (Height)2
∴ ℓ2 = 142 + 482
= 196 + 2304
= 2500 = (50)2
⇒ ℓ = 50
Thus, the required height = 50 cm.
(iii) To find the curved surface area
∵ The curved surface area of a cone is given by πrl
∴ Curved surface area = (22/7) x 14 x 50 cm2
= 22 x 2 x 50 cm2
= 2200 cm2
Thus, the curved surface area of the cone is 2200 cm2.
Exercise 13.7
Page Number 233
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
Sides of the right triangle are 5 cm, 12 cm and 13 cm.∵ The right angled triangle is revolved about the 12 cm side.
∴ Its height is 12 cm and base is 5 cm.
Thus, we have Radius of the base of the cone so formed (r) = 5 cm
Height (h) = 12 cm
Slant height = 13 cm
= 100 πcm3
Thus, the required volume of the cone is 100 πcm3.
Exercise 13.7
Page Number 233
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
Since the right triangle is revolved about the side 5 cm.∴ Height of the cone so obtained (h) = 5 cm
Radius of the cone (r) = 12 cm
Thus, the required ratio is 5 : 12.
Exercise 13.7
Page Number 233
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 πrotect it from rain. Find the area of the canvas required.
Find its volume. The heap is to be covered by canvas to πrotect it from rain. Find the area of the canvas required.
Answer
Here the heap of wheat is in the form of a cone such that Base diameter = 10.5 m⇒ Base radius (r) = (10.5/2)m = (105/20) m
Height (h) = 3 m Volume of the heap
Thus, the required volume = 86.625 m3
Area of the canvas
∵ The area of the canvas to cover the heap must be equal to the curved surface area of the conical heap.
∴ Area of the canvas = πrl
= 11 × 1.5 × 6.05 m2
= 99.825 m2
Thus, the required area of the canvas is 99.825 m2.
Exercise 13.7
Page Number 233
1
Find the volume of a sphere whose radius is
(i) 7 cm
(ii) 0.63 m
(i) 7 cm
(ii) 0.63 m
Answer
(i) Here, radius (r) = 7 cm
= 1.047816 m3 = 1.05 m3 (apπrox)
Thus, the required volume is 1.05 m3 (apπrox).
Exercise 13.8
Page Number 236
2
Find the amount of water displaced by a solid spherical ball of diameter (i) 28 cm (ii) 0.21 m
Answer
(i) Diameter of the ball = 28 cm∴ Radius of the ball (r) =(28/2) cm = 14 cm
Exercise 13.8
Page Number 236
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?
Answer
Diameter of the metallic ball = 4.2 cm⇒ Radius (r) = (4.2/2) cm = 2.1 cm
∴ Volume of the metallic ball =
= 345.39 g (approx.)
Thus, the mass of ball is 345.39 g (aprrox.)
Exercise 13.8
Page Number 236
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?
What fraction of the volume of the earth is the volume of the moon?
Answer
Let diameter of the earth be 2r.
Exercise 13.8
Page Number 236
5
How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?
Answer
Diameter of the hemisphere = 10.5 cm
= 0.3031875 l
= 0.303 l (approx.) [∵ 1000 cm3 = 1 L]
Thus, the capacity of the bowl = 0.303 l (approx.)
Exercise 13.8
Page Number 236
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
Inner radius (r) = 1m
∴ Outer radius (R) = 1 m + 0.01 m = 1.01 m
Now, outer volume of the hemispherical bowl
= 0.063487776 m3
= 0.06348 m3 (apπrox.)
Thus, the required volume of the iron = 0.06348 m3
Exercise 13.8
Page Number 236
7
Find the volume of a sphere whose surface area is 154 cm2
Answer
Let ‘r’ be the radius of the sphere.∴ Its surface area = 4πr2
⇒ 4πr2 = 154
Exercise 13.8
Page Number 236
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 cost of white washing is ₹ 20 per square metre, find the (i) inside surface area of the dome, (ii) volume of the air inside the dome.
Answer
Total cost of white washing = ₹ 4989.60Rate of white washing = ₹ 20 per m
(i) To find the inside surface area of the dome:
∵ Radius of the hemisphere (r) = 6.3 m
Surface area of a hemisphere = 2πr2 m2
∴ Surface area of the dome
= 249.48 m2
Thus, the required surface area of the dome = 249.48 m2
(ii) To find the volume of air in the dome:
= 523.9 m3(approx.)
Thus, the required volume of air inside the dome is 523.9 m3 (approx.)
Exercise 13.8
Page Number 236
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 sphere, (ii) ratio of S and S'.
Answer
(i) To find r ':∵ Radius of a small sphere = r
(r’)3 = (3r)3
r’=3r
(ii) To find ratio of S and S ':
∵ Surface area of a small sphere = 4πr2
∴ S= 4πr2 and S ' = 4p(3r)2
Exercise 13.8
Page Number 236
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?
Answer
Diameter of the spherical capsule = 3.5 mm∴ Volume of the spherical capsule
=22.46 mm3 (approx.)
Thus, the required quantity of medicine = 22.46 mm3 (approx).
Exercise 13.8
Page Number 236
1
A wooden bookshelf has external dimensions as follows: Height = 110 cm, Depth = 25 cm, Breadth = 85 cm (see the 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 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
Here, Height = 110 cmDepth = 25 cm
Breadth = 85 cm
Thickness of the plank = 5 cm
Cost of polishing
∵ Area to be polished =
[(110 x 85)] + [(85 x 25) x 2] + [(110 x 25) x 2] + [(5 x 110) x 2] + [(75 x 5) x 4] cm2
= [9350] + [4250] + [5500] + [1100] + [1500] cm2
= 21700 cm2
∴ Cost of polishing at the rate 20 paise per cm2
Cost of painting
∵ Area to be painted = [(75 x 20) x 6] + [(90 x 20) x 2] + [90 x 75] cm2
= [9000 + 3600 + 6750] cm2 = 19350 cm2
∴ Cost of painting at the rate of 10 paise per cm2
Total expenses = (Cost of polishing) + (Cost of painting)
= ₹ 4340+₹ 1935
= ₹ 6275
Thus, the total required expense = ₹ 6275.
Exercise 13.9
Page Number 236
2
The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in Fig. 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 cm2and black paint costs 5 paise per cm2.
Answer
For spherical part
= 7.071 cm2
∴ Surface area of a wooden sphere to be painted silver = [1386 – 7.071] cm2
= 1378.93 cm2
⇒ Surface area of 8 wood spheres to be painted = 8 x 1378.93 cm2
= 11031.44 cm2
Now, the cost of silver painting at the rate of 25 paise per cm2
For cylindrical part Radius of the base of the cylindrical part (R) = 1.5 cm
Height of the cylindrical part (h) = 7 cm
∴ Curved surface area of the cylindrical part (pillar) = 2πRh
⇒ Total curve surface area of 8 pillars = 8 x 66 cm2 = 528 cm2
∴ Cost of painting black for 8 pillars at the rate of 5 paise per cm2
Total cost of painting Total cost of painting = [Cost of silver painting] + [Cost of black painting]
= ₹ 2757.86 +₹ 26.40
= ₹ 2784.26
Exercise 13.9
Page Number 237
3
The diameter of a sphere is decreased by 25%. By what per cent does its curved surface area decrease?
Answer
Let the original diameter = d [Case I]
Per cent decrease in surface area Decrease in curved surface area = [Original surface area] – [Decreased surface area]
Thus, the required per cent decrease in curved surface area is 43.75%
Exercise 13.9
Page Number 237