Due to sudden floods some welfare Association jointly requested the government to get 100 tents fixed immediately and offered to contribute 50% of the cost if the lower part of the each tent is of the form of a cylinder of diameter 4.2 M and height 4 m with the conical upper part of the same diameter but of height 2.8 M and the Canvas to be used cost Rs 100 per square metre find the amount the association will have to pay and what values are shown by the situations?
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Answered by
2
Given height of cylindrical part $$=4$$ m
And height of conical pert $$=2.8$$ m
Diameter of cylindrical part $$=$$ Diameter of conical part $$=4.2$$ m
Radius of cylindrical part $$=$$ Radius of conical part $$=2.1$$ m
The surface area of cylindrical part $$=2\times \dfrac{22}{7}\times 2.1\times 4=2\times 22\times 0.3\times 4=52.80\ m^{2}$$
Surface area of conical part
$$=\dfrac{22}{7}\times 2.1\times \sqrt{(2.8)^{2}+(2.1)^{2}}=22\times 0.3\times \sqrt{7.84+4.41}=6.6\times \sqrt{12.25}=6.6\times 3.5=23.10\ m^{2}$$
The surface area of tent $$=$$ Surface area of cylindrical part $$+$$ Surface area of conical part $$=52.80+23.10=75.90\ m^2$$
Surface area of $$100$$ tent $$=75.90\times 100=7590\ m^{2}$$
Cost of $$100$$ tent $$=100\times 7590=Rs\ 759000$$
But $$50\%$$ cost have to paid by association.
Association have to pay : $$\dfrac{759000}{100}\times 50= Rs\ 379500$$
PLEASE MARK MY ANSWER AS A BRAINLIEST.
And height of conical pert $$=2.8$$ m
Diameter of cylindrical part $$=$$ Diameter of conical part $$=4.2$$ m
Radius of cylindrical part $$=$$ Radius of conical part $$=2.1$$ m
The surface area of cylindrical part $$=2\times \dfrac{22}{7}\times 2.1\times 4=2\times 22\times 0.3\times 4=52.80\ m^{2}$$
Surface area of conical part
$$=\dfrac{22}{7}\times 2.1\times \sqrt{(2.8)^{2}+(2.1)^{2}}=22\times 0.3\times \sqrt{7.84+4.41}=6.6\times \sqrt{12.25}=6.6\times 3.5=23.10\ m^{2}$$
The surface area of tent $$=$$ Surface area of cylindrical part $$+$$ Surface area of conical part $$=52.80+23.10=75.90\ m^2$$
Surface area of $$100$$ tent $$=75.90\times 100=7590\ m^{2}$$
Cost of $$100$$ tent $$=100\times 7590=Rs\ 759000$$
But $$50\%$$ cost have to paid by association.
Association have to pay : $$\dfrac{759000}{100}\times 50= Rs\ 379500$$
PLEASE MARK MY ANSWER AS A BRAINLIEST.
Answered by
27
Radius of the cylinder, r = 2.1 m
Height of the cylinder, h = 4 m
Radius of the cone = radius of the cylinder = 2.1 m
Height of the cone, H = 2.8 m
Slant height of the cone, I = VP+H?
(21)' +(28) m
4.41 +784 m
- 12.25 m 35 m.
Area of the canvas required for each tent
= curved surface area of the cylinder
+ curved surface area of the cone
22
= 27th + ferl 2x x 2.1 x 4 + x 21 x 35
2 1 m
1 m
m
(52.8 + 23.1) m = 75.9 m
Total area of the canvas required for 100 tents
(759 x 100) m =7590 m
Total cost of 100 tents (7590 x 100) 759000.
Amount to be paid by the associations
50
50% of 759000 = x 759000379500.
100
Hence, the associations will have to pay 379500.
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