Find the curved surface area of a cone, if its slant height is 60 cm and the radius of its base is 21 cm.
Given,
Slant height of cone (r) = 60 cm
Radius of base of cone (l) = 21 cm
Curved surface area of cone = πrl = 22/7 × 21 × 60 = 3960 cm2
The radius of a cone is 5 cm and vertical height is 12 cm. Find the area of the curved surface.
Radius of cone (r) = 5 cm
Vertical height (h) = 12 cm
Slant height of cone (l) = √(r2+h2) = √(52+122) = √25+144 = √169 = 13 cm
Curved surface area of cone = πrl = 22/7×5×13 = 204.28 cm2
The radius of a cone is 7 cm and area of curved surface is 176 cm2. Find the slant height.
Given,
Radius of base (r) = 7 cm
Area of curved surface = πrl = 176 cm
∴22/7×7×l = 176 = l = 8 cm
The height of a cone is 21 cm. Find the area of the base if the slant height is 28 cm.
Given,
Height of cone (h) = 21 cm
Slant height ( l )=28 cm
∴l2=r2 + h2 =r == =7√7 cm
∴ Area of the base = πr2=×cm2 = 1078 cm2
Find the total surface area of a right circular cone with radius 6 cm and height 8 cm.
We have,
Radius of right circular cone (r) = 6
Height of cone ( h) = 8 cm
∴ slant height (l) == l ==10 cm
Total surface area of cone = πr2+ πrl= πr(r+l)
= ×6×16 cm2=301.71 cm2
Find the curved surface area of a cone with base radius 5.25 cm and slant height 10 cm.
Given,
Base radius of cone (r )= 5.25 cm
Slant height of cone ( l )= 10 cm
∴ Curved surface area =πrl
= ×5.25×10 cm2=165 cm2
Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
We have,
Slant height of cone(l ) = 21 m
Radius of cone ( r )= 12 m
Total surface area = πr2+ πrl= πr(r+l)
= ×12(12+21) m2
=×12×33 cm2=1244.57 m2
The area of the curved surface of a cone is 60π cm2. If the slant height of the cone be 8 cm, find the radius of the base.
We have,
Curved surface area of cone = 60π cm2
Slant height l = 8 cm
So , πrl = 60π
rl = 60
= r ==7.5 cm
The curved surface area of a cone is 4070 cm2 and its diameter is 70 cm. What is its slant height? (Use π=22/7).
We have,
Area of curved surface = πrl = 4070 cm2
Radius of base r = 70/2 = 35 cm
= ×35×l =4070 = l = 37 cm
The radius and slant height of a cone are in the ratio of 4 : 7. If its curved surface area is 792 cm2, find its radius. (Use π=22/7)
We have,
Let radius of cone = r =4x,
Let slant height = l =7x
Area of curved surface = πrl = 792 cm2
Now, πrl = 792
=
∴r =4×3= 12 cm
l = 7×3 = 21 cm
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.
We have,
Based radius of conical cap =r = 7cm
Vertical height = h = 24 cm
∴slant height = l === 25 cm
Required area of sheet =
= 10×πrl=10××7×25 cm2=5500 cm2
Find the ratio of the curved surface areas of two cones if their diameters of the bases are equal and slant heights are in the ratio 4 : 3.
We have,
r1 = r2
let , l1 =4x, l2 = 3x
There are two cones. The curved surface area of one is twice that of the other. The slant height of the later is twice that of the former. Find the ratio of their radii.
We have,
A1 = 2A2 = πr1l1= 2(πr2l2)
l2 = 2l1
r1l1=2r2×2l1 , = r1 = 4r2
The diameters of two cones are equal. If their slant heights are in the ratio 5 : 4, find the ratio of their curved surfaces.
We have,
r1 = r2
l1 : l2 = 5 : 4
Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find the radius of the base and total surface area of the cone.
We have,
Curved surface area of cone =πrl= 308 cm2
Slant height =l = 14cm
=× r × 14 = 308 = r = 7 cm
Total surface area = πrl+πr2=308 +×72
= 462 cm2
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.
We have,
Radius of conical tomb = diameter /2 = 14/2 = 7 m
Slant height = l = 25 m
Curved surface area = πrl
Cost of white washing = (curved surface area × rate of white washing per m2)
=
= = Rs.1155
A conical tent is 10 m high and the radius of its base is 24 m. Find the slant height of the tent. If the cost of 1 m2 canvas is Rs. 70, find the cost of the canvas required to make the tent.
We have,
Radius of base of tent = r= 24 m
Vertical height = h = 10 m
∴slant height = l == = 26m
∴ Cost of canvas required = Rs.= Rs. 137280
A tent is in the form of a right circular cylinder surmounted by a cone. The diameter of cylinder is 24 m. The height of the cylindrical portion is 11 m while the vertex of the cone is 16 m above the ground. Find the area of the canvas required for the tent.
For cylindrical part, we have
Diameter = 24 m so, radius r = 24/2 = 12 m ,
Height = h = 11 m
For conical part, we have
Height of the cone = (16-11) m = 5 m
∴ Slant height = m = 13 m
Hence, Area of the canvas required
= Curved surface area of cone + Curved surface area of cylinder
= πrl+2πrh=πr(l+2h) = ×12×(13+22) m2 = 1320 m2
A circus tent is cylindrical to a height of 3 metres and conical above it. If its diameter is 105 m and the slant height of the conical portion is 53 m, calculate the length of the canvas 5 m wide to make the required tent.
For cylindrical part, we have
Height( h) = 3 m
Radius (r) = m
∴ Total curved surface area = 2πrh+2πrl
= πr(2h+l) = ×× (6+53) m2= 11×15×59 m2
Hence, length of 5 m wide canvas = m = 1947 m
The circumference of the base of a 10 m height conical tent is 44 metres. Calculate the length of canvas used in making the tent if width of canvas is 2 m. (Use π=22/7).
We have,
Circumference of circular base of cone = 2πr = 44 m
Height (h )= 10 m
= 2××r = 44 = r = 7m
Slant height (l ) = √(102+72) = √149 = 12.20 m
Curved surface area of tent = πrl = 22/7 ×7×12.20 = 268.4 m2
Hence, length of 2m wide canvas needed = 268.4/2 = 134.2 m
What length of tarpaulin 3 m wide will be required to make a conical tent of height 8 m and base radius 6 m? Assume that the extra length of material will be required for stitching margins and wastage in cutting is approximately 20 cm. (Use π=3.14).
we have,
Height of tent = 8 m
Base radius = 6 m
Slant height (l) == = 10 m
Curved surface area of the cone = πrl = 3.14×6×10 m2
∴Length of 3 m wide tarpaulin required
= = 62.8 m
Extra length required for stitching and cutting wastage = 0.2 m
Total length required = 62.8+0.2 = 63m
A bus stop is barricaded from the remaining part of the road, by using 50 hollow cone made of recycled card-board. 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 = 1.02).
We have,
Radius of base= 20 cm = 0.2 m
Height of cone = 1 m
∴ slant height l = = m = m
= 1.02 m
Curved surface area of a cone = πrl =3.14×0.2×1.02 m2
Cost of painting = Rs. [(3.14×0.2×1.02×12)×50] = Rs. 384.33
A cylinder and a cone have equal radii of their bases and equal heights. If their curved surface areas are in the ratio 8 : 5, Show that the radius of each is to the height of each as 3 : 4.
For cylinder, we have
Base radius = r
Height = h
∴ Curved surface area S1 = 2πrh
For cone, we have
Slant height l =
S2 = πrl =πr
We have,
= =
==
(squaring both side )
= =
= 100h2 = 64r2+64h2
= 36h2 = 64r2
(Square root both side )
= 6h = 8r
==
Find the volume of a right circular cone with :
(i) radius 6 cm, height 7 cm.
(ii) radius 3.5 cm, height 12 cm
(iii) height 21 cm and slant height 28 cm.
i) we have,
Radius of cone = 6 cm
Height of cone = 7 cm
So, volume of cone = πr2h = 1/3 × 22/7 ×36×7 = 264 cm3
ii) we have,
radius = 3.5 cm
height = 12 cm
volume of cone = 1/3×22/7×3.5×3.5×12 = 154 cm3
iii) we have,
height = 21 cm
slant height = 28 cm
radius r = √(282-212) = √49×7 = 7√7 cm
volume of cone = 1/3×22/7×7√7×7√7×21 = 7546 cm3
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) we have,
radius of cone =7cm
slant height =25 cm
vertical height of cone h = √(252 – 72) = √576 = 24 cm
hence capacity (volume) of cone = 1/3×22/7×49×24 = 1232 cm3
ii) we have,
height = 12 cm
slant height =13 cm
radius of cone r = √(132-122) = √25 = 5 cm
volume of conical vessel = 1/3×22/7×5×5×12 = 2200/7 = 314.28 cm3
Two cones have their heights in the ratio 1 : 3 and the radii of their bases in the ratio 3 : 1. Find the ratio of their volumes.
Let the heights of the cones be h and 3h units and the radii of their bases be 3r and r respectively. Then, their volumes are
V1= π(3r)2×h and V2= π×r2×3h
=
The radius and the height of a rightr circular cone are in the ratio 5 : 12. If its volume is 314 cubic metre, find the slant height and the radius (Use π =3.14).
We have,
Let radius of cone r=5x
Let height h = 12x
∴ Volume = 314 m3
=×3.14×(5x)2×12x =314
= x3= 1 = x = 1
∴r =5 and h = 12
Now, slant height l == l = = 13 m
The radius and height of a right circular cone are in the ratio 5 : 12 and its volume is 2512 cubic cm. Find the slant height and radius of the cone. (Use π =3.14)
Let radius r =5x
Let height h = 12x
We have,
Volume= 2512 m3
= ×3.14×25x2×12x =2512
= x3 = 8 = x = 2
∴r =10 cm and h = 24 cm
Hence, slant height l = = = 26 cm
The ratio of volumes of two cones is 4 : 5 and the ratio of the radii of their bases is 2 : 3. Find the ratio of their vertical heights.
1) We have,
V1 : V2 = 4 : 5
r1 = r2 = 2 : 3
A cylinder and a cone have equal radii of their bases and equal heights. Show that their volumes are in the ratio 3 : 1.
We have,
r1 = r2
h1 = h2
If the radius of the base of a cone is halved, keeping the height same, what is the ratio of the volume of the reduced cone to that of the original cone?
Let r be the radius of the base and h be the height of the original cone. Then,
Let V2 be the volume of the reduces cone. Then,
Hence ratio of reduce cone to original cone = 1:4
A heap of wheat is in the form of a cone of diameter 9 m and height 3.5 m. Find its volume. How much canvas cloth is required to just cover the heap? (Use π =3.14)
We have,
r adius = 9/2 = 4.5 m
height = 3.5 m
∴slant height l == = 5.70 m
∴ Volume of the heap =
= ×3.14×(4.5)2×3.5 = 74.1825 m3
Area of canvas required = πrl
= 3.14×4.5×5.7 m2 = 80.54 m2
Find the weight of a solid cone whose base is of diameter 14 cm and vertical height 51 cm, supposing the material of which it is made weighs 10 grams per cubic cm.
We have ,
Radius r = 14/2 = 7 cm,
Height h = 51 cm
∴ Weight of solid cone = × kg = 26.180 kg
A right angled triangle of which the sides containing the right angle are 6.3 cm and 10 cm in length, is made to turn round on the longer side. Find the volume of the solid, thus generated. Also, find its curved surface.
We have,
r = 6.3 cm, h = 10 cm
∴l = cm= cm
= cm = 11.82 cm
Now, Volume = × ×(6.3)2×10 cm3
= 415.8 cm3
Curved surface area = ×6.3×11.82 cm2
= 234.03 cm2
Find the volume of the largest right circular cone that can be fitted in a cube whose edge is 14 cm.
We have,
Radius of the base of the cone =14/2 = 7 cm
Height of the cone = 14 cm
∴ Volume of the cone = ××72×14 cm3
= 718.66 cm3
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.
We have,
Volume of right circular cone = 9856 cm3,
Radius of cone = 14 cm
h = 48 cm
slant height l = = = 50 cm
Curved Surface area of cone = πrl =
A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?
We have,
Radius of conical pit =
Depth of pit 'h' = 12 m
Volume =
= * * 1.75 * 1.75 * 12
= 38.5 m3
Since, 1 m3 = 1 kiloliter
Capacity of the pit = (38.5 × 1)
= 38.5 kilolitres
Monica has a piece of Canvas whose area is 551 m2. She uses it to have a conical tent made, with a base radius of 7 m. Assuming that all the stitching margins and wastage incurred while cutting, amounts to approximately 1 m2. Find the volume of the tent that can be made with it.
1) We have,
Area of canvas = 551 m2
Area of canvas lost in wastage = 1 m2
∴ Area of canvas used in making tent
= (551-1) m2 = 550 m2
= Surface area of the cone = 550 m2
We have,
r = radius of the base of the cone = 7 m
∴ Surface area = 550 m2 = πrl = 550
=×7×l=550 = l = 25 m
Let h be the height of the cone. Then,
l2 = r2+h2 = h = = = 24 m
∴ Volume of the cone =
= m3 = 1232 m3
The height of a cone is 15 cm. If its volume is 500π cm3, then find the radius of its base.
1) We have,
Height of cone = 15 cm
Volume of cone = 500π cm3
=1/3 πr2h = 500π
= r2 =100 r =√100
Hence, radius of base r = 10 cm ,
The number of surfaces of a cone has, is
A. 1
B. 2
C. 3
D. 4
A cone has two surfaces,
(i) Curved surface
(ii) Circular base
If the volume of a right circular cone of height 9 cm is 48π cm3, find the diameter of its base.
We have ,
Volume of cone = 48π cm3
Height of cone h = 9 cm
= 1/3πr2h = 48π
= r2= 16 r = 4 cm
Hence diameter of its base = 2 × radius = 2×4 = 8 cm
The area of the curved surface of a cone of radius 2r and slant height , is
A. πrl
B. 2πrl
C. πrl
D. π(r+l)r
Radius of cone = 2r
Slant hight =
Area of curved surface
=
The total surface area of a cone of radius and length 2l, is
A. 2πr(l+r)
B. πr
C. πr(l+r)
D. 2πrl
Radius of cone =
Length of cone =2l
Total surface area of cone =
=
=
If the height and slant height of a cone are 21 cm and 28 cm respectively. Find its volume.
We have ,
Height of cone h = 21 cm
Slant height l = 28 cm
Radius of cone = √(l2-h2) = √( 282 – 212) = √343 = 7√7 cm
Volume of cone = 1/3×22/7×7√7×7√7×21
= 2401π cm3
A solid cylinder is melted and cast into a cone of same radius. The heights of the cone and cylinder are in the ratio
A. 9 : 1
B. 1 : 9
C. 3 : 1
D. 1 : 3
We have radius of cylinder = radius of base of cone
Let height of cylinder and cone is respectively h1 and h2
=
=
The height of a conical vessel is 3.5 cm. If its capacity is 3.3 litres of milk. Find the diameter of its base.
We have
Height of conical vessel = 3.5 cm
Volume = 3.3 litre = 3300 cm3
= 1/3πr2h = 3300
=r2 = 3300×3×7 / 22×3.5 = 900
= r =√900 = 30 cm
Hence diameter of its base = 2× radius = 2×30 = 60 cm
The slant height of a cone is increased by 10%. If the radius remains the same, the curved surface area is increases by
A. 10%
B. 12.1%
C. 20%
D. 21%
Let radius of cone = r
Let slant height = x
Curved surface area =
New slant height =
New volume =
Increase in volume = =
Percentage increase in volume =
If the radius and slant height of a cone are in the ratio 7 : 13 and its curved surface area is 286 cm2, find its radius.
We have
Let radius of cone = 7x
Let slant height = 13x
Curved surface area of cone = 286 cm2
= πrl = 286 , 22/7×7x×13x = 286
= x2 = 1
= x = 1
Hence radius of cone = 7×1 = 7 cm
Find the area of canvas required for a conical tent of height 24 m and base radius 7 m.
We have,
Height of conical tent = 24 m
Radius of its base = 7 m
Slant height of cone l = √(r2+h2) = √(242+72) = √625 = 25 m
Area of canvas required = πrl = 22/7×7×25 = 550 m2
The height of a solid cone is 12 cm and the area of the circular base is 64π cm2. A plane parallel to the base of the cone cuts through the cone 9 cm above the vertex of the cone, the area of the base of the new cone so formed is
A. 9π cm2
B. 16π cm2
C. 25π cm2
D. 36π cm2
We have,
Height of cone = 12cm
Area of circular base = 64π cm2
Height from the vertex of small cone = 9 cm
π r2 = 64π
r2= 64
Radius of base = 8cm
From similarity triangle =
= R =
R = radius of small cone
Area of small cone = πr2 = π(6)2 = 36π cm2
If the radius of the base of a right circular cone is 3r and its height is equal to the radius of the base, then its volume is
A. πr3
B. πr3
C. 3πr3
D. 9πr3
We have,
Radius of base of cone = 3r
Height of cone = 3r
Volume of cone =
= 9πr3
Find the area of metal sheet required in making a closed hollow cone of base radius 7 cm and height 24 cm.
We have,
Base radius of cone = 7 cm
Height of cone = 24 cm
Slant height of cone l = √( r2+h2) = √(242+72) = √625 = 25 cm
Hence area of metal sheet required = total surface area of cone
= πr(l+r) = 22/7 ×7 (25+7)
=704 cm2
If the volume of two cones are in the ratio 1 : 4 and their diameters are in the ratio 4 : 5, then the ratio of their heights, is
A. 1 : 5
B. 5 : 4
C. 5 : 16
D. 25 : 64
We have,
Ratio of volume of two cones =
ratio of their diameter =
ratio of their radius =
so,
=
Find the length of cloth used in making a conical pandal of height 100 m and base radius 240 m, if the cloth is 100π m wide.
We have ,
Height of conical pandal = 100m
Base radius of pandal = 240 m
Slant height l = √( r2+h2) = √2402+1002 =√67600 = 260 m
Curved surface area of cloth = πrl = π×260×240
Hence length of 100π m wide cloth = curved surface area of cloth/ width of cloth
= π×260×240 / 100π
= 26×24 = 624m
The curved surface area of one cone is twice that of the other while the slant height of the latter is twice that of the former. The ratio of their radii is
A. 2 : 1
B. 4 : 1
C. 8 : 1
D. 1 : 1
We have,
A1 = 2 A2 ,
2l1 = l2 ,
=
=
=
If the height and radius of a cone of volume V are doubled, then the volume of the cone is,
A. 3 V
B. 4 V
C. 4 V
D. 8 V
We have,
Let radius of cone = r
Let height of cone = h
Volume = 1/3πr2h
New radius of cone = 2r
New height of cone = 2h
New volume =
=
The ratio of the volume of a right circular cylinder and a right circular cone of the same base and height, is
A. 1 : 3
B. 3 : 1
C. 4 : 3
D. 3 : 4
Let radius of both = r cm
Let height = h cm
Ratio of volume =
A right circular cylinder and a right circular cone have the same radius and the same volume. The ratio of the height of the cylinder to that of the cone is
A. 3 : 5
B. 2 : 5
C. 3 : 1
D. 1 : 3
Let radius of both = r cm
Let volume of both = V cm
Ratio of height =
=
=
If the base radius and the height of a right circular cone are increased by 20%, then the percentage increase in volume is approximately
A. 60
B. 68
C. 73
D. 78
Let base radius of cone = X
Let height of cone = Y
Volume of cone =
New radius of cone =
New height of cone =
New volume =
Increase in volume =
Percentage increase in volume =
Approx 73%
The diameters of two cones are equal. If their slant heights are in the ratio 5 : 4, the ratio of their curved surface areas, is
A. 4 : 5
B. 25 : 16
C. 16 : 25
D. 5 : 4
We have,
Ratio of slant height of cones =
Ratio of curved surface area =
If h, S and V denote respectively the height, curved surface area and volume of a right circular cone, then 3πVh3-S2h2+9V2 is equal to
A. 8
B. 0
C. 4π
D. 32π2
We have,
h= hight
s = curved surface area
v = volume
Then,
If a cone is cut into two parts by a horizontal plane passing through the mid-point of its axis, the ratio of the volumes of upper and lower part is
A. 1 : 2
B. 2 : 1
C. 1 : 7
D. 1 : 8
We have, (by mid point theorem)
Radius of complete cone = 2r
Radius of small cone after cutting = r
Hight of complete cone = 2h
Hight of small cone = h
So ratio of volume of lower and upper parts =
If the heights of two cones are in the ratio of 1 : 4 and the radii of their bases are in the ratio 4 : 1, then the ratio of their volumes is
A. 1 : 2
B. 2 : 3
C. 3 : 4
D. 4 : 1
Given,
=