Due to sudden floods, some welfare associations jointly requested the government to get $100$ tents fixed immediately and offered to contribute $50 %$ of the cost. If the lower part of 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 height $2.8$ m, and the canvas to be used costs Rs. $100$ per sq. m, find the amount, the associations will have to pay. What values are shown by these associations $[U s e π = \frac{22}{7}]$

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Solution

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 $= 2 π r h$
$= 2 \times \frac{22}{7} \times 2.1 \times 4 = 2 \times 22 \times 0.3 \times 4 = 52.80 m^{2}$

Surface area of conical part $= π \times r \times \sqrt{r^{2} + h^{2}}$

$= \frac{22}{7} \times 2.1 \times \sqrt{(2.8)^{2} + (2.1)^{2}} = 22 \times 0.3 \times \sqrt{7.84 + 4.41} = 6.6 \times \sqrt{12.25} = 6.6 \times 3.5 = 23.10 m^{2}$

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 $= 100 \times 7590 = R s 759000$

But $50 %$ cost have to paid by association.
Association have to pay : $\frac{759000}{100} \times 50 = R s 379500$

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