Q 14: The diameters of two cones are equal. If their slant height are in the ratio 5:4,find the ratio of their curved surfaces.
Solution:
It is given that
Diameters of two cones are equal
Therefore their radius are also equal i.e
r_{1} = r_{2}
Let the ratio of slant height be x
Therefore l_{1 }= 5x
l_{2 }= 4x
Therefore Ratio of curved surface area =
=
∴ Ratio of curved surface area is 5:4.
Q 15: Curved surface area of a cone is 308 cm^{2} and its slant height is 14cm. Find the radius of the base and total surface area of the cone.
Solution:
(1) It is given that
Slant height of cone = 14cm
Let radius of circular end of cone = r
Curved surface area of cone = πrl
⇒ 308 cm^{2} = ∗r∗14
⇒ r = = 7cm
Thus radius of circular end of cone = 7 cm.
(ii) It is given that C.S.A = 308 cm^{2}
We know that total surface area of a cone
= curved surface area of a cone + Area of base
= πr_{l} + πr_{2}
= [308 + (∗7^{2})]
= 308 + 154
= 462cm^{2}
Thus total surface area of the cone is 462cm^{2}
Q 16: The slant height and base diameter of a conical tomb are 25m and 14m respectively. Find the cost of white washing its curved surface area at the rate of Rs 210 per 100m^{2}.
Solution:
It is given that
Slant height of conical tomb(l) = 25m
Base radius (r) of tomb = = 7m
Curved surface area of conical length tomb = πrl
= ∗7∗25
= 550 m^{2}
Cost of white washing 100 m^{2 }area = Rs 210
Cost of white washing 550m^{2} area = = Rs 1155
Therefore the cost of white washing the whole tomb is Rs 1155.
Q 17: A conical tent is 10 m high and the radius of it base is 24 m. Find the slant height of the tent. If the cost of 1m^{2} canvas is Rs 70, find the cost of canvas required for the tent.
Solution:
It is given that
Height of the conical tent(h) = 10m
Radius of conical tent(r) = 24m
Let slant height of conical tent be l
l^{2} = h^{2} + r^{2}
= 10^{2 }+ 24^{2} = 100 + 576
= 676 m^{2}
⇒ l = 26m
Thus,the slant height of the conical tent is 26m.
(ii) It is given that:
Radius(r) = 24 m
Slant height (l) = 26 m
C.S.A of tent = πrl
= ∗24∗26
Cost of 1m^{2} canvas = Rs 70
Cost of m^{2} canvas = Rs *70 = Rs 1,37,280
Thus the cost of canvas required to make the tent is Rs 1,37,280.
Q 18: 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 11m while the vertex of the cone is 16 m above the ground. Find the area of the canvas required for the tent.
Solution:
It is given that
Diameter of cylinder = 24 m
therefore radius =
= 12 cm
Also radius of cone = 12 m
Height of cylinder = 11 m
Height of cone = 16−11 = 5m
Slant height of cone
= 13m
Therefore area of canvas required for the tent = πrl + 2πrh
= [(12∗13) + (2∗12∗11)] = 490.286 + 829.714
= 1320m^{2}
Q19. A circus tent is cylindrical to a height of 3 m 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.
Solution:
Given diameter = 105 m,
Radius = m = 52.5 m
Therefore curved surface area of circus tent = πrl + 2πrh
= (∗52.5∗53) + (2∗∗52.5∗3)
= 8745 + 990 = 9735 m^{2}
Therefore length of the canvas required for tent
= 1947 m
Q20 The circumference of the base of a 10m height conical tent is 44m, calculate the length of canvas used in making the tent if width of canvas is 2m.
Solution:
We know that
C.S.A of cone = πrl
Given circumference = 2πr
⇒ 2∗∗r = 44
⇒ r = 7m
Therefore l =
Therefore C.S.A of tent = πrl
Therefore the length of canvas used in making the tent
= 134.2 m
Q 21: What length of tarpaulin 4 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.
Solution:
Given that,
Height of conical tent(h) = 8m
Radius of base of tent(r) = 6m
Slant height(l) =
C.S.A of conical tent = πrl
= (3.14*6*10) m^{2} = 188.4 m^{2}
Let the length of tarpaulin sheet required be l
As 20 cm will wasted ,so effective
Length will be (l – 0.2 m)
Breadth of tarpaulin = 3m
Area of sheet = C.S.A of sheet
[l*0.2*3]m = 188.4 m^{2} = l – 0.2 m = 62.8 m
Accounting extra for wastage:
⇒ l = 63 m
Thus the length of the tarpaulin sheet will be = 63 m
Q 22: A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled card board. Each cone has a base diameter of 40 cm and height 1m. If the outer side of each of the cones is to be painted and the cost of painting is Rs 12 per m^{2},what will be the cost of painting all these cones?
Solution:
The area to be painted is the curved surface area of each cone.
The formula of the curved surface area of a cone with base radius and slant height 7 is given as
Curved Surface Area = πrl
For each cone, we’re given that the base diameter is 0.40 m.
Hence the base radius r = 0.20 m.
The vertical height l 1 = 1 m.
To find the slant height ‘l’ to be used in the formula for Curved Surface Area we use the following relation
Slant height ,
l = 1.02 m
Now substituting the values of r = 0.2 m and slant height 1 = 1.02 m and using pi = 3.14 in the formula of C.S.A.
We get Curved Surface Area = (3.14)(0.2)(1.02) = 0.64056 m^{2}
This is the curved surface area of a single cone.
Since we need to paint 50 such cones the total area to be painted is,
Total area to be painted = (0.64056) (50) = 32.028 m^{2}
The cost of painting is given as Rs. 12 per m^{2}
Hence the total cost of painting = (12) (32.028) = 384.336
Hence, the total cost that would be incurred in the painting is Rs. 384.336
Q23. A cylinder and a cone have equal radii of their base and equal heights. If their curved surface area are in the ratio 8:5, show that the radius of each is to the height of each as 3:4.
Solution:
It is given that the base radius and the height of the cone and the cylinder are the same.
So let the base radius of each is ‘ r ‘ and the vertical height of each is ‘h’.
Let the slant height of the cone be ‘ l ’.
The curved surface area of the cone = πrl
The curved surface area of the cylinder = Extra close brace or missing open brace
It is said that the ratio of the curved surface areas of the cylinder to that of the cone is 8:5
So,
But we know that l =
Squaring on both sides we get:
Hence it is shown that the ratio of the radius to the height of the cone as well as the cylinder is: 3 : 4
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