Exercise 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 to be open at the top. Ignoring the thickness of the plastic sheet, determine:
Question 1- A plastic box 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.
(ii) The cost of sheet for it, if a sheet measuring 1 m2 costs Rs 20.
Answer - 
It is given that, length (l) of box = 1.5 m
Breadth (b) of box = 1.25 m
Depth (h) of box = 0.65 m
(i) Box is to be open at top.
Area of sheet required
= 2lh + 2bh + lb
= [2 × 1.5 × 0.65 + 2 × 1.25 × 0.65 + 1.5 × 1.25] m2
= (1.95 + 1.625 + 1.875) m2 = 5.45 m2
(ii) Cost of sheet per m2 area = Rs 20
Cost of sheet of 5.45 m2 area = Rs (5.45 × 20)
= Rs 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 Rs 7.50 per m2.
Answer - It is given that
Length (l) of room = 5 m
Breadth (b) of room = 4 m
Height (h) of room = 3 m
It can be observed that four walls and the ceiling of the room are to be white-washed. The floor of the room is not to be white-washed.
Area to be white-washed = Area of walls + Area of ceiling of room
= 2lh + 2bh + lb
= [2 × 5 × 3 + 2 × 4 × 3 + 5 × 4] m2
= (30 + 24 + 20) m2
= 74 m2
Cost of white-washing per m2 area = Rs 7.50
Cost of white-washing 74 m2 area = Rs (74 × 7.50)
= Rs 555
Question 3- The floor of a rectangular hall has a perimeter 250 m. If the cost of panting 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.]
Answer - Let length, breadth, and height of the rectangular hall be l m, b m, and h m respectively.
Area of four walls = 2lh + 2bh
= 2(l + b) h
Perimeter of the floor of hall = 2(l + b)
= 250 m
∴ Area of four walls = 2(l + b) h = 250h m2
Cost of painting per m2 area = Rs 10
Cost of painting 250h m2 area = Rs (250h × 10) = Rs 2500h
However, it is given that the cost of paining the walls is Rs 15000.
∴ 15000 = 2500h
h = 6
Therefore, the height of the hall is 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 - Total surface area of one brick = 2(lb + bh + lh)
= [2(22.5 ×10 + 10 × 7.5 + 22.5 × 7.5)] cm2
= 2(225 + 75 + 168.75) cm2
= (2 × 468.75) cm2
= 937.5 cm2
Let n bricks can be painted out by the paint of the container.
Area of n bricks = (n ×937.5) cm2 = 937.5n cm2
Area that can be painted by the paint of the container = 9.375 m2 = 93750 cm2
∴ 93750 = 937.5n
n = 100
Therefore, 100 bricks can be painted out by the paint of the container.
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 - (i) Edge of cube = 10 cm
Length (l) of box = 12.5 cm
Breadth (b) of box = 10 cm
Height (h) of box = 8 cm
Lateral surface area of cubical box = 4(edge)2
= 4(10 cm)2
= 400 cm2
Lateral surface area of cuboidal box = 2[lh + bh]
= [2(12.5 × 8 + 10 × 8)] cm2
= (2 × 180) cm2
= 360 cm2
Clearly, the lateral surface area of the cubical box is greater than the lateral surface area of the cuboidal box.
Lateral surface area of cubical box − Lateral surface area of cuboidal box = 400 cm2 − 360 cm2 = 40 cm2
Therefore, the lateral surface area of the cubical box is greater than the lateral surface area of the cuboidal box by 40 cm2.
(ii) Total surface area of cubical box = 6(edge)2 = 6(10 cm)2 = 600 cm2
Total surface area of cuboidal box
= 2[lh + bh + lb]
= [2(12.5 × 8 + 10 × 8 + 12.5 × 100] cm2
= 610 cm2
Clearly, the total surface area of the cubical box is smaller than that of the cuboidal box.
Total surface area of cuboidal box − Total surface area of cubical box = 610 cm2 − 600 cm2 = 10 cm2
Therefore, the total surface area of the cubical box is smaller than that of the cuboidal box by 10 cm2.
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) Length (l) of green house = 30 cm
Breadth (b) of green house = 25 cm
Height (h) of green house = 25 cm
Total surface area of green house
= 2[lb + lh + bh]
= [2(30 × 25 + 30 × 25 + 25 × 25)] cm2
= [2(750 + 750 + 625)] cm2
= (2 × 2125) cm2
= 4250 cm2
Therefore, the area of glass is 4250 cm2.
(ii)
It can be observed that tape is required along side AB, BC, CD, DA, EF, FG, GH, HE, AH, BE, DG, and CF.
Total length of tape = 4(l + b + h)
= [4(30 + 25 + 25)] cm
= 320 cm
Therefore, 320 cm tape is required for all the 12 edges.
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 Rs 4 for 1000 cm2, find the cost of cardboard required for supplying 250 boxes of each kind.
Answer - Length (l1) of bigger box = 25 cm
Breadth (b1) of bigger box = 20 cm
Height (h1) of bigger box = 5 cm
Total surface area of bigger box = 2(lb + lh + bh)
= [2(25 × 20 + 25 × 5 + 20 × 5)] cm2
= [2(500 + 125 + 100)] cm2
= 1450 cm2
Extra area required for overlapping
= 72.5 cm2
While considering all overlaps, total surface area of 1 bigger box
= (1450 + 72.5) cm2 =1522.5 cm2
Area of cardboard sheet required for 250 such bigger boxes
= (1522.5 × 250) cm2 = 380625 cm2
Similarly, total surface area of smaller box = [2(15 ×12 + 15 × 5 + 12 × 5] cm2
= [2(180 + 75 + 60)] cm2
= (2 × 315) cm2
= 630 cm2
Therefore, extra area required for overlapping
cm2
cm2
Total surface area of 1 smaller box while considering all overlaps
= (630 + 31.5) cm2 = 661.5 cm2
Area of cardboard sheet required for 250 smaller boxes = (250 × 661.5) cm2
= 165375 cm2
Total cardboard sheet required = (380625 + 165375) cm2
= 546000 cm2
Cost of 1000 cm2 cardboard sheet = Rs 4
Cost of 546000 cm2 cardboard sheet 
Therefore, the cost of cardboard sheet required for 250 such boxes of each kind will be Rs 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 - Length (l) of shelter = 4 m
Breadth (b) of shelter = 3 m
Height (h) of shelter = 2.5 m
Tarpaulin will be required for the top and four wall sides of the shelter.
Area of Tarpaulin required = 2(lh + bh) + l b
= [2(4 × 2.5 + 3 × 2.5) + 4 × 3] m2
= [2(10 + 7.5) + 12] m2
= 47 m2
Therefore, 47 m2 tarpaulin will be required.
Exercise 13.2
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. Assume π =
Answer - Height (h) of cylinder = 14 cm
Let the diameter of the cylinder be d.
Curved surface area of cylinder = 88 cm2
⇒ 2πrh = 88 cm2 (r is the radius of the base of the cylinder)
⇒ πdh = 88 cm2 (d = 2r)
⇒
⇒ d = 2 cm
Therefore, 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 meters of the sheet are required for the same?
Answer - Height (h) of cylindrical tank = 1 m
Base radius (r) of cylindrical tank
Therefore, it will require 7.48 m2 area of sheet.
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.
(i) Inner curved surface area,
(ii) Outer curved surface area,
(iii) Total surface area. 
Answer - Inner radius 
of cylindrical pipe 
of cylindrical pipe
Outer radius 
of cylindrical pipe 
of cylindrical pipe
Height (h) of cylindrical pipe = Length of cylindrical pipe = 77 cm
(i) CSA of inner surface of pipe
(ii) CSA of outer surface of pipe 
(iii) Total surface area of pipe = CSA of inner surface + CSA of outer surface + Area of both circular ends of pipe
Therefore, the total surface area of the cylindrical pipe is 2038.08 cm2.
Question 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 - It can be observed that a roller is cylindrical.
Height (h) of cylindrical roller = Length of roller = 120 cm
Radius (r) of the circular end of roller = 
CSA of roller = 2πrh
Area of field = 500 × CSA of roller
= (500 × 31680) cm2
= 15840000 cm2
= 1584 m2
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 Rs.12.50 per m2. 
Answer- Height (h) cylindrical pillar = 3.5 m
Radius (r) of the circular end of pillar = 
= 0.25 m
CSA of pillar = 2πrh
Cost of painting 1 m2 area = Rs 12.50
Cost of painting 5.5 m2 area = Rs (5.5 × 12.50)
= Rs 68.75
Therefore, the cost of painting the CSA of the pillar is Rs 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 - Let the height of the circular cylinder be h.
Radius (r) of the base of cylinder = 0.7 m
CSA of cylinder = 4.4 m2
2πrh = 4.4 m2
h = 1 m
Therefore, 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 Rs 40 per m2.
Answer - Inner radius (r) of circular well
Depth (h) of circular well = 10 m
Inner curved surface area = 2πrh
= (44 × 0.25 × 10) m2
= 110 m2
Therefore, the inner curved surface area of the circular well is 110 m2.
Cost of plastering 1 m2 area = Rs 40
Cost of plastering 100 m2 area = Rs (110 × 40)
= Rs 4400
Therefore, the cost of plastering the CSA of this well is Rs 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 - Height (h) of cylindrical pipe = Length of cylindrical pipe = 28 m
Radius (r) of circular end of pipe = 
= 2.5 cm = 0.025 m
= 2.5 cm = 0.025 m
CSA of cylindrical pipe = 2πrh
= 4.4 m2
The area of the radiating surface of 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 
of the steel actually used was wasted in making the tank. 
of the steel actually used was wasted in making the tank.
Answer - Height (h) of cylindrical tank = 4.5 m
Radius (r) of the circular end of cylindrical tank = 
(i) Lateral or curved surface area of tank = 2πrh
= (44 × 0.3 × 4.5) m2
= 59.4 m2
Therefore, CSA of tank is 59.4 m2.
(ii) Total surface area of tank = 2πr (r + h)
= (44 × 0.3 × 6.6) m2
= 87.12 m2
Let A m2 steel sheet be actually used in making the tank.
Therefore, 95.04 m2 steel was used in actual while making such a tank.
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 - 
Height (h) of the frame of lampshade = (2.5 + 30 + 2.5) cm = 35 cm
Radius (r) of the circular end of the frame of lampshade = 
Cloth required for covering the lampshade = 2πrh
= 2200 cm2
Hence, for covering the lampshade, 2200 cm2 cloth will be required.
Question 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 - Radius (r) of the circular end of cylindrical penholder = 3 cm
Height (h) of penholder = 10.5 cm
Surface area of 1 penholder = CSA of penholder + Area of base of penholder
= 2πrh + πr2
Area of cardboard sheet used by 1 competitor 
Area of cardboard sheet used by 35 competitors
=
= 7920 cm2
= 7920 cm2
Therefore, 7920 cm2 cardboard sheet will be bought.
Exercise 13.3
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- Radius (r) of the base of cone =
= 5.25 cm
= 5.25 cm
Slant height (l) of cone = 10 cm
CSA of cone = πrl
Therefore, the curved surface area of the cone is 165 cm2.
Question 2- Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m. 
Answer - Radius (r) of the base of cone =
= 12 m
= 12 m
Slant height (l) of cone = 21 m
Total surface area of cone = πr(r + l)
Question 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.
Answer - (i) Slant height (l) of cone = 14 cm
Let the radius of the circular end of the cone be r.
We know, CSA of cone = πrl
Therefore, the radius of the circular end of the cone is 7 cm.
(ii) Total surface area of cone = CSA of cone + Area of base
= πrl + πr2
Therefore, the total surface area of the cone is 462 cm2.
Question 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 Rs 70.
Answer - 
(i) Let ABC be a conical tent.
Height (h) of conical tent = 10 m
Radius (r) of conical tent = 24 m
Let the slant height of the tent be l.
In ΔABO,
AB2 = AO2 + BO2
l2 = h2 + r2
= (10 m)2 + (24 m)2
= 676 m2
∴ l = 26 m
Therefore, the slant height of the tent is 26 m.
(ii) CSA of tent = πrl
Cost of 1 m2 canvas = Rs 70
Cost of 
canvas =
canvas =
= Rs 137280
Therefore, the cost of the canvas required to make such a tent is
Rs 137280.
Question 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 approximately 20 cm. [Use π = 3.14]
Answer - Height (h) of conical tent = 8 m
Radius (r) of base of tent = 6 m
Slant height (l) of tent =
CSA of conical tent = πrl
= (3.14 × 6 × 10) m2
= 188.4 m2
Let the length of tarpaulin sheet required be l.
As 20 cm will be wasted, therefore, the effective length will be (l − 0.2 m).
Breadth of tarpaulin = 3 m
Area of sheet = CSA of tent
[(l − 0.2 m) × 3] m = 188.4 m2
l − 0.2 m = 62.8 m
l = 63 m
Therefore, the length of the required tarpaulin sheet will be 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 white-washing its curved surface at the rate of Rs 210 per 100 m2.
Answer - Slant height (l) of conical tomb = 25 m
Base radius (r) of tomb = 
7 m
7 m
CSA of conical tomb = πrl
= 550 m2
Cost of white-washing 100 m2 area = Rs 210
Cost of white-washing 550 m2 area =
= Rs 1155
Therefore, it will cost Rs 1155 while white-washing such a conical tomb.
Question 7- A joker’s cap is in the form of 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 - Radius (r) of conical cap = 7 cm
Height (h) of conical cap = 24 cm
Slant height (l) of conical cap =
CSA of 1 conical cap = πrl
CSA of 10 such conical caps = (10 × 550) cm2 = 5500 cm2
Therefore, 5500 cm2 sheet will be required.
Question 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
= 1.02).
= 1.02).
Answer - Radius (r) of cone =
= 0.2 m
= 0.2 m
Height (h) of cone = 1 m
Slant height (l) of cone =
CSA of each cone = πrl
= (3.14 × 0.2 × 1.02) m2 = 0.64056 m2
CSA of 50 such cones = (50 × 0.64056) m2
= 32.028 m2
Cost of painting 1 m2 area = Rs 12
Cost of painting 32.028 m2 area = Rs (32.028 × 12)
= Rs 384.336
= Rs 384.34 (approximately)
Therefore, it will cost Rs 384.34 in painting 50 such hollow cones.
Exercise 13.4
Question 1- Find the surface area of a sphere of radius:
(i) 10.5 cm (ii) 5.6 cm (iii) 14 cm
Answer - (i) Radius (r) of sphere = 10.5 cm
Surface area of sphere = 4πr2
Therefore, the surface area of a sphere having radius 10.5cm is 1386 cm2.
(ii) Radius(r) of sphere = 5.6 cm
Surface area of sphere = 4πr2
Therefore, the surface area of a sphere having radius 5.6 cm is 394.24 cm2.
(iii) Radius (r) of sphere = 14 cm
Surface area of sphere = 4πr2
Therefore, the surface area of a sphere having radius 14 cm is 2464 cm2.
Question 2- Find the surface area of a sphere of diameter:
(i) 14 cm (ii) 21 cm (iii) 3.5 m
Answer - (i) Radius (r) of sphere = 
Surface area of sphere = 4πr2
Therefore, the surface area of a sphere having diameter 14 cm is 616 cm2.
(ii) Radius (r) of sphere =
Surface area of sphere = 4πr2
Therefore, the surface area of a sphere having diameter 21 cm is 1386 cm2.
(iii) Radius (r) of sphere = 
m
m
Surface area of sphere = 4πr2
Therefore, the surface area of the sphere having diameter 3.5 m is 38.5 m2.
Question 3- Find the total surface area of a hemisphere of radius 10 cm. [Use π = 3.14]
Answer - 
Radius (r) of hemisphere = 10 cm
Total surface area of hemisphere = CSA of hemisphere + Area of circular end of hemisphere
Therefore, the total surface area of such a hemisphere is 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 - Radius (r1) of spherical balloon = 7 cm
Radius (r2) of spherical balloon, when air is pumped into it = 14 cm
Therefore, the ratio between the surface areas in these two cases is 1:4.
Question 5- A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of Rs 16 per 100 cm2.
Answer - Inner radius (r) of hemispherical bowl 
Surface area of hemispherical bowl = 2πr2
Cost of tin-plating 100 cm2 area = Rs 16
Cost of tin-plating 173.25 cm2 area 
= Rs 27.72
= Rs 27.72
Therefore, the cost of tin-plating the inner side of the hemispherical bowl is Rs 27.72.
Question 6- Find the radius of a sphere whose surface area is 154 cm2. 
Answer - Let the radius of the sphere be r.
Surface area of sphere = 154
∴ 4πr2 = 154 cm2
Therefore, the radius of the sphere whose surface area is 154 cm2 is 3.5 cm.
Question 7- The diameter of the moon is approximately one-fourth of the diameter of the earth. Find the ratio of their surface area.
Answer- Let the diameter of earth be d. Therefore, the diameter of moon will be
.
.
Radius of earth = 
Radius of moon =
Surface area of moon = 
Surface area of earth = 
Required ratio 
Therefore, the ratio between their surface areas will be 1:16.
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 - Inner radius of hemispherical bowl = 5 cm
Thickness of the bowl = 0.25 cm
∴ Outer radius (r) of hemispherical bowl = (5 + 0.25) cm
= 5.25 cm
Outer CSA of hemispherical bowl = 2πr2
Therefore, the outer curved surface area of the bowl is 173.25 cm2.
Question 9- A right circular cylinder just encloses a sphere of radius r (see 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 - 
(i) Surface area of sphere = 4πr2
(ii) Height of cylinder = r + r = 2r
Radius of cylinder = r
CSA of cylinder = 2πrh
= 2πr (2r)
= 4πr2
(iii) 
Therefore, the ratio between these two surface areas is 1:1.
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