Sponsor Area
(i) The area of the sheet required for making the box.
(ii) The cost of sheet for it, if a sheet measuring 1 m2 costsर 20.
(i) l = 1.5 m
b= 1.25 m
h = 65 cm = 0.65 m.
∴ The area of the sheet required for making the box = lb + 2(bh + hl)
= (1.5)(1.25) + 2 {(1.25)(0.65) + (0.65)(1.5)} = 1.875 + 2{0.8125 + 0.975}
= 1.875 + 2(1.7875) = 1.875 + 3.575 = 5.45 m2.
(ii) The cost of sheet for it = र 5.45 x 20 = र 109.
l = 5 m
b = 4 m
h = 3 m
Area of the walls of the room = 2(l + b) h
= 2(5 + 4) 3 = 54 m2
Area of the ceiling = lb
= (5) (4) = 20 m2
∴ Total area of the walls of the room and the ceiling = 54 m2 + 20 m2 = 74 m2
∴ Cost of white washing the walls of the room and the ceiling = 74 x 7.50 = र 555.
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. [Hint : Area of the four walls = Lateral surface area.]
Let the length, breadth and height of the rectangular hall be l m, b m and h m respectively.
Perimeter = 250 m
⇒ 2(l + b) = 250
⇒ l + b = 125 ...(1)
Area of the four walls
Hence, the height of the hall is 6 m.
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?
For a brick
l = 22.5 cm b = 10 cm h = 7.5 cm
∴ Total surface area of a brick = 2 (lb + bh + hl)
= 2(22.5 x 10 + 10 x 7.5 + 7.5 x 22.5)
= 2(225 + 75 + 168.75)
= 2(468.75) = 937.5 cm2 = .09375 m2
∴ Number of bricks that can be painted out
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?
Each edge of the cubical box (a) = 10 cm
∴ Lateral surface area of the cubical box = 4a2 = 4(10)2 = 400 cm2.
For cuboidal box
l = 12.5 cm b = 10 cm h = 8 cm
Lateral surface area of the cuboidal box = 2(l + b) h
= 2(12.5+ 10)(8) = 360 cm2.
∴ Cubical box has the greater lateral surface area than the cuboidal box by (400 – 360) cm2, i.e., 40 cm2.
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?
Total surface area of the cubical box = 6a2
= 6(10)2 = 600 cm2
Total surface area of the cuboidal box
= 2 (lb + bh + hl)
= 2[(12.5)(10) + (10)(8) + (8)(12.5)]
= 2[125 + 80 + 100] = 610 cm2.
Cubical box has the smaller total surface area than the cuboidal box by (610 – 600) cm2, i.e., 10 cm2.
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) For herbarium
l = 30 cm b = 25 cm h = 25 cm
Area of the glass = 2(lb + bh + hl)
= 2[(30)(25) + (25)(25) + (25)(30)] = 2[750 + 625 + 750] = 4250 cm2.
(ii) The tape needed for all the 12 edges = 4 (l + b + h)
= 4(30 + 25 + 25) = 320 cm.
For bigger box
l = 25 cm b = 20 cm h = 5 cm
Total surface area of the bigger box
= 2 (lb + bh + hl)
= 2[(25)(20) + (20)(5) + (5)(25)]
= 2[500 + 100 + 125] = 1450 cm2
Cardboard required for all the overlap
∴ Net surface area of the bigger box
= 1450 cm2 + 72.5 cm2 = 1522.5 cm2
∴ Net surface area of 250 bigger boxes
= 1522.5 x 250 = 380625 cm2
Cost of cardboard
For smaller box
l = 15 cm b = 12 cm h = 5 cm
∴ Total surface area of the smaller box = 2 (lb + bh + hl)
= 2[(15)(12) + (12)(5) + (5)( 15)] = 2[ 180 + 60 + 75] = 630 cm2
Cardboard required for all the overlaps
∴ Net surface area of the smaller box
= 630 cm2 + 31.5 cm2 = 661.5 cm2 Net surface area of 250 smaller boxes = 661.5 x250 = 165375 cm2
∴ Cost of cardboard
Cost of cardboard required for supplying 250 boxes of each kind
= र 1522.50 + र 661.50 = र 2184.
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?
For shelter
l = 4 m b = 3 m 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] = 47 m2 Hence, 47 m2 of tarpaulin will be required.
Let the height of the hall be h m.
Area of the floor = l x b
= 20 x 16 = 320 m2 Area of the flat roof = l x b
= 20 x 16 = 320 m2 Sum of the areas of the four walls = 2(l + b) h
= 2(20 + 16) h = 72 h m2 According to the question,
72 h = 320 + 320
Hence, the height of the hall is 
For tank
I = 12 m b = 9 m h = 4 m Total surface area of the tank
= 2 (I x b + b x h + h x l)
= 2(12 x 9 + 9 x 4 + 4 x 12),
= 2(108 + 36 + 48)
= 2(192) = 384 m2 Width of the iron sheet = 2 m
∴ Length of the iron sheet = 
Cost of the iron sheet = 192 x 50 = र 9600.
For table cover
l = 4 – (0.25 + 0.25)
| ∵ 25 cm = 0.25 m
= 3.5 m b = 2 – (0.25 + 0.25) = 1.5 m
∴ Area of the cover = l x b
= 3.5 x 1.5 = 5.25 m2 Cost of polishing the top of the table at र 40 per square metre = 5.25 x 40 = र 210.
Let the length and the breadth of the room be 4x m and x m respectively.
Area of the four walls of the room = 2(l + b) h
= 2(4x + x) 5 = 50 x m2
Cost of papering the four walls of the room @ 70 paise per square metre
= र 50x x 0.70 = र 35x According to the question,
35x = 157.50
∴ Breadth of the room = 4.5 m
∴ Length of the room = 4 x 4.5 m = 18 m.
For classroom
l = 1 m b = 6.5 m h = 4 m Area to be colour washed
= 2(l + b) h – [3 x 1.4 + 3 {2 x 1}] = 2(7 + 6.5) 4 – (4.2 + 6)
= 108 – 10.2 = 97.8 m2
Cost of colour washing = 97.8 x 3.50 = र 34.30
For resulting cuboid Length (l) = 6 + 6 = 12 cm Breadth (b) = 6 cm Height (h) = 6 cm
∴ Surface area
= 2 (lb + bh + hl)
= 2(12 x6 + 6x6 + 6x 12)
= 360 cm2
Let the dimensions of the box be 2k, 3k and 4k.
Total surface area = 2(lb + bh + hl)
= 2(2k ⋅ 3k + 3k ⋅ 4k + 4k ⋅ 2k)
= 52k2 m2
Cost of covering at the rate of र 8 per m2 = 52k2 x 8 = र416 k2
Cost of covering at the rate of र 9.50 per m2 = 52k2 x 9.50 = र 494k2
Difference between the costs = र 494k2 – र 416k2 = र 78k2
According to the question,
78k2 = 1248 ⇒ k2 = 16 ⇒ k = 4 Hence, the dimensions of the box are 8 m, 12 m and 16 m.
Solution not provided.
Ans. 1300 cm2
Sponsor Area
Solution not provided.
Ans. 11200 cm2
Solution not provided.
Ans. र 5400
Solution not provided.
Ans. 126 cm2
Solution not provided.
Ans. 8 m, 12 m, 16 m
Solution not provided.
Ans. 6.615 cm2
Solution not provided.
Ans. 6 cm
Solution not provided.
Ans. 10 : 3
Solution not provided.
Ans. 28 cm
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.
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?
h = 1 m = 100 cm
2r = 140 cm
Total surface area of the closed cylindrical tank = 2πr(h + r)
Hence, 7.48 square metres of the sheet are required.
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.
h = 77 cm
2r = 4 cm
⇒ r = 2 cm
2R = 4.4 cm
⇒ R = 2.2 cm
2 r = 84 cm
⇒ r = 42 cm
h = 120 cm
∴ Area of the playground levelled in taking 1 complete revolution
Area of the playground = 31680 x 500
Hence, the area of the playground is 1584 m2.
2r = 50 cm
∴ r = 25 cm = 0.25 m h = 3.5 m
∴ Curved surface area of the pillar = 2πrh
Cost of painting the curved surface of the pillar at the rate of र 12.50 per m2
= र 5.5 x 12.50 = र 68.75.
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.
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 र 40 per m2.
(ii) Cost of plastering the curved surface at the rate of र 40 per m2
= र 110 x 40 = र 4400.
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.
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.
(i) 2r = 4.2 m
h = 4.5 m
Lateral or curved surface area = 2πrh
Find:
(ii) how much steel was actually used if of the steel actually used was wasted in making the tank?
(ii) Total surface area = 
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 which has gone into the tank = 11/12 of x.
Steel actually used = 95.04 m2
In Fig. 13.12, 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.
2r = 20 cm ⇒ r = 10 cm h = 30 cm
Cloth required = 2πr(h + 2.5 + 2.5)
= 2200 cm2.
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?
r = 3 cm
h = 10.5 cm
∴ Cardboard required for 1 competitor
∴ Cardboard required for 35 competitors
Hence, 7920 cm2 of cardboard was required to be bought for the competition.
Sponsor Area
External radius (r) = 
Length of the pipe (h)
= 4.2 m = 4.2 x 100 cm = 420 cm Area of the outer surface = 2πrh
= 3.14)
Height (h) = 1.4 m
= 1.4 x 100 cm = 140 cm
Surface area to be tin-coated = 2(2πrh + πr2)
= 2[2 x 3.14 x 50 x 140 + 3.14 x (50)2]
= 2 [43960 + 7850] = 2(51810)
= 103620 cm2
∴ Cost of tin-coating at the rate of र 50 per 1000 cm2
Height (h) = 4 m
∴ Curved surface area of 1 pillar = 2
rh
= र 880.
Let the base radius of the cylinder be r cm.
Then,
h = 3 m
Toatal surface area = 
= 66 + 38.5
= 104.5 m2
= 3.14)
Height (h) = 4 m
∴ Curved surface area = 2πrh
= 2 x 3.14 x 0.25 x 4 = 6.28 m2
∴ Curved surface area of 20 pillars
= 6.28 x 20 m2 = 125.6 m2
∴ Cost of cleaning them
= 125.6 x 3 = र 376.80
Curved surface area
Curved surface area = 
∴ Area of the ground levelled in 1 revolution = 3.96 m2
∴ Area of the ground levelled in 100 revolutions
= 3.96 x 100 m2 = 396 m2
∴ Cost of levelling
= 396 dm2
Let the radius of the base of the tank be r dm. Then, height = 6 r dm
∴ Curved surface area = 2πRH
According to the question,
= 198 x 3.24 = 641.52 dm3
Solution not provided.
Ans. 550 cm2
Solution not provided.
Ans. 5 : 6
Solution not provided.
Ans. र 2200
Solution not provided.
Ans. 7 cm
Solution not provided.
Ans. 66 m2
Solution not provided.
Ans. र 23760
A metal pipe is 77 cm long. The inner diameter of a cross-section is 4 cm and outer diameter is 5.0 cm. Find its
(i) Inner curved surface area
(ii) Outer curved surface area
Solution not provided.
Ans. (i) 968 cm2 (ii) 1210 cm2
Solution not provided.
Ans. 209 m2
Solution not provided.
Ans. 3.5 cm
Solution not provided.
Ans. र 137155.20
Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.
Radius of the base (r) = 
cm.
Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
Radius of base (r) = 
= 12 m
(i) Slant height (l) = 14 cm Curved surface area = 308 cm2
⇒ πrl = 308
Hence, the radius of the base is 7 cm.
(ii) Total surface area of the cone = πr(l + r)
Hence, the total surface area of the cone is 462 cm2.
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 र 70.
= 26 m
Hence, the slant height of the tent is 26 m.
(ii) Curved surface area of the tent = 
rl
Cost of the canvas required to make the tent, if the cost of 1 m2 canvas is र 70.
= र 137280.
Hence, the cost of the canvas is र 137280.
= 3.14)
For conical tent
h = 8 m r = 6 m
∴ Curved surface area = πrl
= 3.14 x 6 x 10 = 188.4 m2. Width of tarpaulin = 3 m
Length of tarpaulin = 
Extra length of the material required = 20 cm = 0.2 m
∴ Actual length of tarpaulin required = 62.8 m + 0.2 m = 63 m.
Base radius (r) = 
rl
Base radius (r) = 7 cm
Height (h) = 24 cm
Slant height (l) = 
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 5500 cm2.
= 3.14 and take 
Base diameter = 40 cm
Base radius (r) = 
cm = 20 cm
= 
Height (h) = 1 m
Curved surface area = πrl
= 3.14 x 0.2 x 1.02 = 0.64056 m2
∴ Curved surface area of 50 cones = 0.64056 x 50 m2 = 32.028 m2
∴ Cost of painting all these cones = 32.028 x 12
= 384.336 = र 384.34 (approximately).
Let the height of the right circular cone be h cm. r = 56 cm Curved surface area = 12320 cm2 ⇒ πrl = 12320
⇒ (56)2 + h2 = (70)2
| Squaring both sides
⇒ h2 = (70)2 – (56)2
⇒ h2 = (70 – 56)(70 + 56)
⇒ h2 = (14)(126)
⇒ h2= (14)(14 x 9)
⇒ h = 14 x 3 = 42
| Extracting square root
Hence, the height of the right circular cone is 42 cm.
Total surface area
= 3.14)
Let the base radius and height of the cone be r cm and h cm respectively. Then,
πrl = 113.04 ⇒ 3.14 x r x 12 = 113.04 ⇒ r = 3 cm
l2 = r2 + h2 ⇒ (12)2 = (3)2 + h2
⇒ 144 = 9 + h2
⇒ h2 = 135
∴ Curved surface area = πrl = π(5)(13) = 65π cm2
h = 24 cm l = 25 cm
l2 = r2 + h2
⇒ (25)2 = r2 + (24)2 ⇒ r = 7 cm
∴ Area of the metal sheet required = 2(
rl)
= 600 cm2
For cone
Base radius (r) = 8 m
Height (h) = 6m
= 10 m
∴ Curved surface area = πrl = π(8)(10) = 80π m2
For cylinder
Base radius (R) = 8 m Height (H) = 14 m
∴ Curved surface area = 2πRH = 2 ⋅ π ⋅ 8 ⋅ 14 = 224π m2 Total curved surface area = Curved surface area of the cone + Curved surface area of the cylinder
= 80π + 224π = 304π m2 = 304 x 3.14 m2 = 954.56 m2
∴ Cost of canvas = 954.56 x 50 = र 47728
Solution not provided.
Ans. 220 cm2
= 3.14)
Solution not provided.
Ans. 753.6cm2, 1205.76cm2
.
Solution not provided.
Ans. 192.5 cm2
Sponsor Area
Solution not provided.
Ans. 
Solution not provided.
Ans. 
Solution not provided.
Ans. 550 m2
wide will be required to make a conical tent whose base diameter is 10 m and whose vertical height is 12 cm?
Solution not provided.
Ans. 130 m
= 3.14)
Solution not provided.
Ans. 62.8 m
= 3.14)
Solution not provided.
Ans. 213.52 m
Solution not provided.
Ans. 7 cm, 616 cm2
Find the surface area of a sphere of radius:
(i) 10.5 cm (ii) 5.6 cm (iii) 14 cm.
= 1386 cm2.
(ii) r = 5.6 cm
∴ Surface area = 4πr2
= 394.24 cm2.
(iii) r = 14 cm
Surface area = 4πr2
Radius (r) = 
Radius (r) = 
Diameter = 3.5 m
Radius (r) = 
Surface area = 
Find the total surface area of a hemisphere of radius 10 cm. (Use 
= 3.14)
r = 10 cm.
∴ Total surface area of the hemisphere = 3πr2
= 3 x 3.14 x (10)2 = 942 cm2.
Case I. r = 7 cm
∴ Surface area = 4
r2
Case II. r = 14 cm
∴ Surface area = 4πr2
= 2464 cm2
∴ Ratio of surface areas of the balloon = 616 : 2464
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 ` 16 per 100 cm2.
Inner radius (r) = 
Find the radius of a sphere whose surface area is 154 cm2.
Radius of the earth = 
Surface area of the earth = 
Inner radius of the bowl = 5 cm
Thickness of steel = 0.25 cm
∴ Outer radius of the bowl
= 5 + 0.25 = 5.25 cm
∴ Outer curved surface of the bowl = 4πr2
= 346.5 cm2.
A right circular cylinder just encloses a sphere of radius r. 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 = 4πr2
(ii) For cylinder Radius of the base = r
Height = 2 r Curved surface area of the cylinder = 2π(r)(2r) = 4πr2
(iii) Ratio of the areas obtained in (i) and (ii)
If the radius of a sphere is halved then what is the decrease in its surface area?
Let the radius of the sphere be r.
Then surface area of the sphere = 4πr2
New radius of the sphere = 
New surface area of the sphere 
Decrease in surface area = 
= 
( surface area of the original sphere).
Surface area of sphere of radius 5 cm = 4π (5)2 cm2
Area of the curved surface of cone of radius 4 cm = π(4) l cm2 where I cm is the slant height of the cone.
According to the question,
4
(5)2 = 5[
(4)l}
l = 5 cm
r2 + h2 = 25 
(4)2 + h2 = 25
16 + h2 + 25 
h2 = 9
h = 3
Hence, the height of the cone is 3 cm.
External radius of (R) = 
Internal radius (r) = 
∴ Total surface area
= 2πR2 + 2πr2 + π(R2 – r2)
= 27π(8)2 + 2π(6)2 + π(82 – 62) = 128π + 72π + 28π = 228π
= र 1433.14
Inner Radius (r) = 6 cm
Outer Radius (R) = 6 cm + 1 cm = 7 cm
Total surface area of the bowl
= 2πr2 + 2πR2 + π(R2 – r2)
= 2π(6)2 + 2π(7)2 + π(72 – 62)
= 72π + 98π+ 13π
= 183π cm2
Solution not provided.
Ans. (i) 2772 cm2 (ii) 4158 cm2
Solution not provided.
Ans. 154 m2
The internal and external diameters of a hollow hemispherical vessel are 24 cm and 25 cm respectively. The cost to paint 1 sq. cm of surface is र 1.75. Find the total cost to the nearest rupee to paint the vessel all over. Ignore the area of the edge. (Take 
= 3.14)
Solution not provided.
Ans. र 269
Solution not provided.
Solution not provided.
Ans. 
Solution not provided.
Ans. 50.18 cm
A matchbox measures 4 cm × 2.5 cm × 1.5 cm. What will be the volume of a packet containing 12 such boxes?
Volume of a matchbox = 4 x 2.5 x 1.5 cm3 = 15 cm3
∴ Volume of a packet containing 12 such boxes = 15 x 12 cm3 = 180 cm3.
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)
Capacity of the tank = 6 x 5 x 4.5 m3 = 135 m3
∴ Volume of water it can hold = 135 m3 = 135 x 1000 l = 135000 l.
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?
Let the height of the cuboidal vessel be h m. l = 10 m b = 8 m
Capacity of the cuboidal vessel = 380 m3
lbh = 380
(10)(8)h = 380
Hence, the cuboidal vessel must be made 4.75 m high.
l = 8 m
b = 6 m h = 3 m
Volume of the cuboidal pit = Ibh = 8 x 6 x 3 m3 = 144 m3 Cost of digging the cuboidal pit @ र 30 per m3 = र 144 x 30
= र 4320.
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.
Let the breadth of the cuboidal tank be b m. l = 2.5 m h = 10 m
Capacity of the cuboidal tank = 50000 litres
Hence, the breadth of the cuboidal tank is 2 m.
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?
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
= 600 m3
For tank
l = 20 m b = 15 m h = 6 m Capacity of the tank
= 20 x 15 x 6 m3 = 1800 m3
∴ Number of days for which the water of this tank last
= 600 m3
For tank
l = 20 m b = 15 m h = 6 m Capacity of the tank
= 20 x 15 x 6 m3 = 1800 m3
∴ Number of days for which the water of this tank last
For godown
l = 40 m b = 25 m h = 10 m
∴ Capacity of the godown = Ibh
= 40 x 25 x 10 m3 = 10000 m3
For a wooden crate
l = 1.5 m b = 1.25 m h = 0.5 m
∴ Capacity of a wooden crate = Ibh
= 1.5 x 1.25 x 0.5 m3 = 0.9375 m3
We have,
Hence, the maximum number of wooden crates that can be stored in the godown is 10666.
Sponsor Area
Side of the solid cube (a) = 12 cm
∴ Volume of the solid cube = a3
= (12)3 = 12 x 12 x 12 cm3 = 1728 cm
∵ It is cut into eight cubes of equal volume.
∴ Volume of a new cube
Let the side of the new cube be x cm.
Then, volume of the new cube = x3 cm3. According to the question, x3 = 216
∴ x = (216)1/3
∴ x = (6 x 6 x 6)1/3
∴ x = 6 cm
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
∴ Ratio between their surface areas
Hence, the ratio between their surface areas is 4: 1.
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?
In one hour
l = 2 km = 2 x 1000 m = 2000 m b = 40 m h = 3 m
∴ Water fell into the sea in one hour = Ibh = 2000 x 40 x 3 m3
∴ Water fell into the sea in a minute
= 4000 m3
Hence, 4000 m3 of water will fall into the sea in a minute.
For water tank l = 48 m b = 36 m h = 28 m
∴ Volume of the tank = Ibh
= 48 x 36 x 28 m3 = 48384 m3.
Let the length, breadth and height of the cuboid be l, b and h units respectively. Then, p = lb q = bh r = hl
∴ pqr = (lb)(bh)(hl) = l2b2h2 ...(1) Again, v = Ibh
∴ v2 = (Ibh)2 = l2b2h2 ...(2)
(1) and (2) give
v2 = pqr.
For cuboid
Length (l) = 12 + 12 + 12 = 36 cm Breadth (b) = 12 cm Height (h) = 12 cm
∴ Volume of the resulting cuboid = Ibh
= 36 x 12 x 12 = 5184 cm3
Capacity of the box
= [25 – (2 + 2)} x [18 – (2 + 2)} x {15 – (2 + 2)} = 21 x 14 x 11 cm3 = 3234 cm3
Volume of the wood used
= 25 x 18 x 15 cm3 – 3234 cm3
= 6750 cm3 – 3234 cm3 = 3516 cm3
For open box
Length (l) = 48 – (8 + 8) = 32 cm Breadth (b) = 36 – (8 + 8) = 20 cm Height (h) = 8 cm
∴ Volume of the box = Ibh
= 32 x 20 x 8 = 5120 cm3
A cubical box has each edge of length 10 cm and another cuboidal box is 12.5 cm long, 10 cm wide and 8 cm high. Find the volume of the box which has greater lateral surface area.
Lateral surface area of cubical box = 4a2 = 4(10)2 = 400 cm2Lateral surface area of the cuboidal box = 2(l + b)h = 2(12.5 + 10)8 = 360 cm2
Clearly, the cubical box has greater lateral surface area.
∴ Volume of the cubical box = a3
= 103
= 1000 cm3
Find the volume of the cube.
Let the length of each edge of the cube be a m.
Total surface area of the cube = 
Volumen of the cube = a3
.
For room
l = 12m b = 9 m h = 8 m
Length of the longest rod that can be placed in the room
= Length of the diagonal
L. H. S. = 
R. H. S. = 
(1) and(2) given
Let the dimensions of the cuboid be ft m, 2k m, 4k m. Then,
Volume of the cuboid
= (k) (2k) (4k) m3 = 8k3 m3
Volume of the cube = 8k3 m3
Side of the cube = 
Whole surface area of the cube = 6(2k)2 m2 = 24k2 m2 Page 240(2)
Whole surface area of the cuboid = 2 (lb + bh + hl)
= 2{k ⋅2k + 2k ⋅ 4k + 4k ⋅ k}
= 28k2 m2 Cost of painting the cube = (24k2) (5)
= र 120k2 Cost of painting the cuboid = (28k2) (5)
= र 140k2
Difference between the cost of painting the cuboid and cube
= र 140k2 – र 120k2 = र 20 k2 According to the question,
20k2 = 80 ⇒ k2 = 4 ⇒ k = 2
∴ Volume of the cube = 8k3 = 8(2)3 = 64 m3
= volume of the cuboid
l = 20 m b = 16 m According to the question, 2 lb = 2(l + b)h
∴ lb = (I + b)h
∴ 20 x 16 = (20 + 16)h
For original cuboid
Length (l) = 20 m Breadth (b) = 24 m Height (h) = 12 m Volume (V1) =lbh
= 20 x 24 x 12 m3 = 5760 m3
For new cuboid
Length (L) = 20m + 20 X 
= 23 m
Breadth (B) = 24 m + 24 X 
= 30 m
Height (H) = 12m + 12 X 
= 18 m
Volume (V2) = LBH
= 23 x 30 x 18
= 12420 m3
Required ratio = 
Volume of wall occupied by bricks
= 32.4 m3 - 3.24 m3
Number of bricks
Cost of bricks
= र 30375
Solution not provided.
Ans. 4167 bricks
. What is the length of the edge of the cube?
Solution not provided.
Ans. 2 cm
Solution not provided.
Ans. 24 dm, 18 dm, 6 dm2 cm
Solution not provided.
Ans. 2058 m3
Solution not provided.
Ans. 1728 m3
Solution not provided.
Ans. 350 cm2
Solution not provided.
Ans. 8 cm
Solution not provided.
Ans. 270, 1 : 43
Solution not provided.
Ans. 729 cm3, 486 cm2
Solution not provided.
Ans. 8000 cm3, 2400 cm2, 1600 cm2, 
Solution not provided.
Ans. 2058 dm3
Solution not provided.
Ans. 216 m3
Let the base radius of the cylindrical vessel be r cm.
Then, circumference of the base of the cylindrical vessel = 2πr cm.
According to the question,
r = 21 cm; h = 25 cm
∴ Capacity of the cylindrical vessel = 
Hence, the cylindrical vessel can hold 34.65 l of water.
R2h
r2h
Volume of the wood used = Outer volume – Inner volume
= 21560 cm3 – 15840 cm3 = 5720 cm3
∴ Mass of the pipe = 5720 x 0.6 g
= 3432 g = 3.432 kg.
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
(i) For tin can
l = 5 cm b = 4 cm h = 15 cm
∴ Capacity = l x b x h
= 5 x 4 x 15 cm3 = 300 cm3.
(ii) For plastic cylinder Diameter = 7 cm
Radius (r) = 
Height (h) = 10 cm
∴ Capacity = 
r2h
Clearly the second container i.e., a plastic cylinder has greater capacity than the first container i.e., a tin can by 385 – 300 = 85 cm3.
= 3.14)
2 x 3.14 x r x 5 = 94.2
Hence, the radius of the base is 3 cm.
(ii) r = 3 cm h = 5 cm
∴ Volume of the cylinder = πr2h
= 3.14 x (3)2 x 5 = 141.3 cm3.
It costs र 2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of painting is at the rate of र 20 per m2, find
(i) inner curved surface area of the vessel,
(ii) radius of the base,
(iii) capacity of the vessel.
Hence, the radius of the base is 1.75 m.
(iii) r = 1.75 m
h = 10 m π Capacity of the vessel = 
r2h
For solid cylinder of graphite
Diameter = 1 mm
Radius (r) = 
Length of the pencil (h) = 14 cm = 140 mm
∴ Volume of the graphite = 
r2h
For cylinder of wood
Diameter = 7 mm
Radius (R) = 
mm
Length of the pencil (h) = 14 cm = 140 mm
Volume of the wood 
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?
Radius (r) = 
Volume of soup in the cylindrical bowl = 
Volume of soup to be prepared daily to serve 250 patients
= 154 x 250 cm3 = 38500 cm3 (or 38.5l)
Hence, the hospital has to prepare 38500 cm3 (or 38.5l) of soup daily to serve 250 patients.
Let the radius of the base of the cylinder be r cm.
h = 7 cm
Volume = 448
cm3
Lateral surface area = 2
rh
Total surface area = 
Radius of the base (r) = 
rh
Outer radius (R) = 
Inner diameter = 6 cm
Inner radius (r) = 
Volume of copper used in making the pipe = volume of the outer cylinder
– volume of the inner cylinder = 
R2h – 
r2h
A powder tin has a square base with side 8 cm and height 13 cm. Another is cylindrical with the radius of its base 7 cm and its height 15 cm.
Find the difference in their capacities. 
For a powder tin with a square base
Side of the square base = 8 cm Height = 13 cm
∴ Volume (v1) = 8 x 8 x 13 = 832 cm3 For a cylindrical powder tin
Radius of the base (r) = 7 cm Height (h) =15 cm
∴ Volume (v2) = 
r2h
∴ Difference in their capacities = v2 – v1 = 2310 – 832 = 1478 cm3.
A square piece of paper of side 22 cm is rolled to form a cylinder. Find the volume of the cylinder.
Let the base radius and height of the cylinder be r cm and h cm respectively.
Then,
Volume of the cylinder = 
Let the rise in the level of water be h cm.
Then,
Volume of water displaced
= Volume of block
Hence, the rise in the level of water is 4 cm.
0.31 km/hour
= 0.31 x 1000 m/hour = 310 m/hour
∴ Volume of water that falls into the sea in a minute
When the rectangular piece of paper is rolled along its length, then the length of the piece forms the circumference of the base and the breadth of the piece becomes the height of the cylinder.
Let the base radius and height of the cylinder be r cm and h cm respectively.
Then,
Volume of the cylinder = 
= 
= 4622
Depth of the wall (h) = 10 m Volume of the earth dug out = Volume of the well = π r2h = π(5)2(10) = 250π m3 Radius of the well with embankment (R)
= 5 + 5 = 10 m π Area of the embankment
= Area of the well with embankment
– Area of the well without embankment = πR2 – πr2 = π(R + r) (R – r)
= π(10 + 5) (10 – 5)
= 75π m2
∴ Height of the embankment
Internal radius (r) = 
Thickness = 5 mm = 
External radius (R) = 5 + 0.5 = 5.5 cm
Length (h) 24 cm
Volume = 
= 3.14)
Let the height of water in the cylindrical vessel be h cm, Then,
Volume of water in the cylindrical vessel = π(20)2 h cm3 According to the question,
Solution not provided.
Ans. र 300
The volume of a right circular cylinder is 1100 cm3 and the radius of its base is 5 m. Find its curved surface area.
Solution not provided.
Ans. 220 cm2
Solution not provided.
Ans. 240 p cm2
Solution not provided.
Ans. 396 cm3
Solution not provided.
Ans. 440 cm2
Solution not provided.
Ans. 20 : 27
Solution not provided.
Ans. 99
The volume of a right circular cylinder is 3850 cm3. Find its height if its, diameter is 14 cm.
Solution not provided.
Ans. 25 cm
Solution not provided.
Ans. 13 cm, 2002 cm3.
(i) r = 6 cm
h = 7 cm
Volume of the right circular cone = 
(ii) r = 3.5 cm
h = 12 cm
Volume of the right circular cone = 
(i) r = 7 cm
l = 25 cm r2 + h2 = l2 ⇒ (7)2 + h2 = (25)2 ⇒ h2 = (25)2 – (7)2
h2 = 625 - 49
h2 = 576
h = 
h = 24 cm
Capacity = 
(ii) h = 12
l = 13 cm
r2 + h2 + l2 
= 3.14)
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.
Let the radius of the base of the right circular cone be r cm.
h = 9 cm Volume = 48π cm3
Hence, the diameter of the base of the right circular cone is 8 cm.
For conical pit
Diameter = 3.5 cm
Radius (r) = 
Depth (h) = 12 m
Capacity of the conical pit = 
= 38.5 m3 = 38.5 x 100 l
= 38.5 kl.
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,
(iii) curved surface area of the cone.
Radius of the base (r) = 
Hence, the slant height of the cone is 50 cm.
(iii) r = 14 cm
I = 50 cm
∴ Curved surface area = πrl
Hence, the curved surface area of the cone is 2200 cm2.
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 = 12 cm
Volume = 
= 
Hence, the volume of the solid so obtained is 100 ∴ cm3.
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 = 5 cm
Ratio of the volumes of the two solids obtained = 100
∴: 240
∴ = 5 : 12
For heap of wheat
Diameter = 10.5 m
Radius (r) = 
Height (h) = 3 m
Volume = 
Curved surface area = 
Hence, the area of the canvas required is 99.825 m2. .
(3.2)3 cm3
(3.2)2 cm2.
D.
4 (3.2)2 cm2.
is:
r (l + r)
r (4 r + l)
r
C.
r (4 r + l)
B.
2πr (r + h) unit2
Volume of the cone = 
If the radius of the base of a right circular cone is halved keeping the height same, what is the ratio of the volume of the reduced cone to that of the original one?
Let the radius of the base and the height of the original cone be r and h respectively.
∴ Volume of the original cone (v1)
For the reduced cone
Radius = 
Height = h
Volume of the reduced cone (v2)
Hence, the ratio of the volume of the reduced cone to that of the original one is 1 : 4.
Let the base radii of the two right circular cones be 3x and 5x respectively.
Let their common height be h. Then,
Volume of the first cone (v1) = 
and, volume of the second cone (v2) = 
Ratio of their volumes
If h, c, v are respectively the height, curved surface and the volume of a cone, prove that 3πvh3 – c2h2 + 9v2 = 0
Let the base radius and the height of the cone be r and h respectively.
Let the slant height of the cone be l.
Then,
c = 
rl
= π2r2h2 – π2r4h2 – π2r4 + π2r4h4 = 0
A vessel is in the shape of a cone. Radius of the broader end is 2.1 cm and height is 20 cm. Find the volume of the vessel.
r = 2.1 cm
h = 20 cm
∴ Volume of the vessel
= 3.14)
Let the radius and height of the cone be 5k and 12k respectively. Then,
r = 5k
h = 12k
Hence, the slant height and radius of the cone are 13 cm and 5 cm respectively.
A conical tent is to accommodate 11 persons. Each person must have 4 square metre of the space on the ground and 20 cubic metres of air to breath. Find the height of the cone.
For conical vessel
r = 5 cm h = 63 cm
∴ Volume of the oil
Let the required height be 4 cm. Then, according to the question, π(10)2H = 1650
Let the base radius of the cup be r cm. Then,
2nr = π x 14
Let the height of the cup be h cm. Then\
l2 + r2 + h2
Capacity of the cup
cm
Volume = 
Solution not provided.
Ans. 7546 cm3
Solution not provided.
Ans. 1232 m3
Solution not provided.
Ans. 9 : 5
= 3.14)
Solution not provided.
Ans. 6 cm, 10 cm
= 3.14)
Solution not provided.
Ans. 26 cm, 10 cm
Solution not provided.
Ans. 1232 cm3
Solution not provided.
Ans. 1.232l
Solution not provided.
Ans. 5 cm
Solution not provided.
Ans. 10 cm.
= 3.14)
Solution not provided.
Ans. 13 cm, 65
Solution not provided.
Ans. 49.125 m3
= 3.14)
Solution not provided.
Ans. 314 cm3
Radius (r) = 
Amount of water displaced = 
Radius (r) = 
Amount of water displaced = 
Radius (r) = 
Volume = 
Density = 8.9 g per cm3
Mass of the ball = Volume x Density
= 38.808 x 8.9 = 345.39 g (approx.).
Let the radius of the earth be r.
Then, diameter of the earth = 2r
Diameter of the moon 
Radius of the moon = 
Volume of the earth (v1) = 
Volume of the moon (v2)
(volume of the earth)
Hence, the volume of the moon is 
th fraction of the volume of the earth.
Diameter = 10.5 cm
Radius (r) = 
Amount of milk = 
Inner radius (r) = 1 m Thickness of iron sheet = 1 cm = 0.01 m
∴ Outer radius (R) = Inner radius (r) + Thickness of iron sheet = 1 m + 0.01 m = 1.01 m
∴ Volume of the iron used to make the tank
Volume of the sphere = 
A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of र 498.96. If the cost of white-washing is र2.00 per square metre, find the
(i) inside surface area of the dome,
(ii) volume of the air inside the dome.
Volume of the air inside the dome = 
Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S'. Find the
(i) radius r´ of the new sphere,
(ii) ratio of S and S’.
⇒ h = 2r
⇒ Height of the cone = 2r cm.
Height of the hemisphere = r cm.
∴ Ratio of their heights = 2r : r = 2 : 1.
Let the radius of the sphere be r cm.
Then,
4πr2 = 55.44
Volume of the sphere = 
= 38.808 cm3
A cube of side 4 cm contains a sphere touching its sides. Find the volume of the gap in between.
Volume of the sphere = 
Volume of the gap in between 
Outer radius (R) = 
Inner radius (r) = 
∴ Volume of the metal contained in the shell
Let the radius of the solid sphere be R cm.
Then, according to the question,
Hence, the radius of the solid sphere is 9 cm.
cm and 2 cm, find the diameter of the third ball.
cm.
Difference of their surface areas
Volume = 
For bottle
r = 3 cm h = 6 cm
∴ Volume = 
= π (3)2 (6) cm3 Let n bottles be required. Then,
Thickness of bowl = 0.25 cm Inner radius of bowl = 5 cm
∴ Outer radius of bowl
= 5 cm + 0.25 cm = 5.25 cm
∴ Outer curved surface area of the bowl = 2πr2
= 33 cm2
A spherical iron shell with 8 cm external diameter weighs 
Find the sick ness of the shell if the density of metal is 12 g/cm3 .
Mass of metal = 
Thickness of the shell = 4 cm - 3 cm= 1 cmSolution not provided.
Ans. 3.85 kg (nearly)
A hemispherical bowl has a radius of 3.5 cm. What would be the volume of water it would contain?
Solution not provided.
Ans. 89.8 cm3
Solution not provided.
Ans. 12 cm, 9 cm
Solution not provided.
Ans. 19404 cm3
Solution not provided.
Solution not provided.
Solution not provided.
Solution not provided.
Solution not provided.
Ans. 7 cm, 10.5 cm
.
Solution not provided.
Ans. 672 
cm2, 230
cm3
Solution not provided.
Ans. (i) 7cm (ii) 462 cm2
Solution not provided.
Ans. 1331 cm3
Solution not provided.
Ans. 60 cm3
Solution not provided.
Ans. 
Solution not provided.
Ans. 12 m
A hemispherical bowl made of brass has inner radius 10.5 cm. Find the cost of polishing it on the inside at the rate of र 0.12 per sq. cm.
Solution not provided.
Ans. र 8316
Solution not provided.
Ans. 6 m, 14 m
Solution not provided.
Ans. 4.2 m3
Solution not provided.
Ans. 0.77 m3
Solution not provided.
Ans. 0.77 m3
A hollow right circular cylindrical copper pipe is 21 cm long. Its outer and inner diameters are 10 cm and 6 cm respectively. Find the volume of copper used in making the pipe.
.
Solution not provided.
Ans. 10560 cm3
Solution not provided.
Ans. 14 m, 7 m
Solution not provided.
Ans. 
Solution not provided.
Ans. 1232 m3
Solution not provided.
Ans. 180 cm3
Solution not provided.
Ans. 2512 m3
Solution not provided.
Ans. 100000
Surface area to be polished = [(110 x 85) + 2(110 x 25) + 2(85 x 25) + 2(110 x 5) + 4(75 x 5)]
= (9350 + 5500 + 4250 + 1100 + 1500) cm2 = 21700 cm2
∴ Expenses required for polishing @ 20 paise per cm2
= 21700 x 20 paise
Surface area to be painted
= [2(20 x 90) + 6(75 x 20) + (75 x 90)]
= (3600 + 9000 + 6750) cm2 = 19350 cm2
∴ Expenses required for painting @ 10 paise per cm2
= 19350 x 10 paise
∴ Total expenses required for polishing and painting the surface of the bookshelf
= र 4340 + र 1935 = र 6275.
The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in figure. Eight such spheres are used for this purpose, and are to be painted silver. Each support is a cylinder of radius 1.5 cm and height 7 cm and is to be painted black. Find the cost of paint required if silver paint costs 25 paise per cm2 and black paint costs 5 paise per cm2.
For a wooden sphere
Diameter = 21 cm
Radius (r) = 
∴ Surface area of a wooden sphere = 4
∴ Surface area of a wooden sphere to be painted = 1386 – 
(1.5)2
∴ Surface area of eight wooden spheres = 1378.93 x 8 = 11031.44 cm2 Cost of painting silver @ 25 paise per cm2 = 11031.44 x 25 paise
For a cylindrical support
Radius (r) = 1.5 cm Height (h) = 7 cm
Surface area of a cylindrical support = 2
rh
∴ Surface area of eight cylindrical supports = 66 x 8 = 528 cm2
∴ Cost of painting black @ 5 paise per cm2 = 528 x 5 paise
∴ Cost of paint required = 2757.86 + 26.40 = र 2784.26 (approx.)
cm.
Find the mean of each of the following distributions :
(i)
|
xi |
10 |
15 |
20 |
25 |
30 |
35 |
40 |
Total |
|
fi |
4 |
6 |
8 |
18 |
6 |
5 |
3 |
50 |
(ii)
|
xi |
12 |
13 |
14 |
15 |
16 |
17 |
18 |
Total |
|
fi |
1 |
3 |
4 |
8 |
10 |
3 |
1 |
30 |
(iii)
|
xi |
50 |
75 |
100 |
125 |
150 |
175 |
200 |
Total |
|
fi |
12 |
18 |
50 |
70 |
25 |
15 |
10 |
200 |
(i)
|
xi |
fi |
fixi |
|
10 |
4 |
40 |
|
15 |
6 |
90 |
|
20 |
8 |
160 |
|
25 |
it |
450 |
|
30 |
6 |
180 |
|
35 |
5 |
175 |
|
40 |
3 |
120 |
|
Total |
50 |
1215 |
|
xi |
fi |
fixi |
|
12 |
1 |
12 |
|
13 |
3 |
39 |
|
14 |
4 |
56 |
|
15 |
8 |
120 |
|
16 |
10 |
160 |
|
17 |
3 |
51 |
|
18 |
1 |
18 |
|
Total |
30 |
456 |
(iii)
|
xi |
fi |
fixi |
|
50 |
12 |
600 |
|
75 |
18 |
1350 |
|
100 |
50 |
5000 |
|
125 |
70 |
8750 |
|
150 |
25 |
3750 |
|
175 |
15 |
2625 |
|
200 |
10 |
2000 |
|
Total |
200 |
24075 |
The following data have been overanged in ascending order. If the median of these data is 63, then find the value of x : 29, 32, 48, 50, x, x + 2, 72, 78, 84, 95
2x + 2 = 126
2x = 124
x = 62Find the median of the following data:
95, 65, 75, 70, 75, 100, 50, 40.
Arranging the given data in the ascending order, we have
40, 50, 65, 70, 75, 75, 95,
100 Number of observations (n) = 8 which is even.
∵ Mean of 40 observations = 160
∴ Sum of 40 observations = 160 x 40 = 6400 Corrected sum of 40 observations
= 6400 - 125 + 165
= 6440
Correct mean = 
Sponsor Area
Sponsor Area
Curved surface area = 
rh 
rh
rh
r(r + h)
rh 
rh
r(r + h) 
rl 
rl
rl
r(l + r)
rl 
r(l + r)
r
r
r
r
r
r
r
r
r
r
r
r
rh 
r
r(h + r)
r
x 6 x 6 x 7 cm
x 12 x 12 x 7 cm
x 12 x 12 x 700 cm
x 12 x 12 x 700 cm
(r
(r
(r
(r
cm
cm
cm
cm
Volume = 
Volume = 
th of a cylindrical can contains milk. The height of the can is 1.4 m and radius is 0.4 m. This milk is poured into small cylindrical glasses of height 10 cm and radius 5 cm. How many small glasses are needed to empty the can?
