A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are $3.5 m$ and $2 m$ respectively, and the slant height of the top is $3.5 m$ , find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs. $1000$ per $m^{2}$ . (Note that the base of the tent will not be covered with canvas)

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Solution

$A r e a o f C a n v a s = C u r v e d A r e a o f C o n e + C u r v e d S u r f a c e A r e a o f t h e C y l i n d e r$

$L e t s c a l c u l a t e t h e C u r v e d S u r f a c e a r e a o f C o n e :$

$D i a m e t e r o f c o n e = D i a m e t e r o f C y l i n d e r = 4 m$

$⟹ r a d i u s = \frac{4}{2} = 2 m$

$S l a n t h e i g h t = l = 2.8 m$

$C u r v e d S u r f a c e a r e a o f C o n e = π r l$

$= \frac{22}{7} \times 2 \times 2.8 = 17.6 m^{2}$

$C u r v e d S u r f a c e a r e a o f C y l i n d e r = 2 π r h$

$= 2 \times \frac{22}{7} \times 2 \times 2.1 = 26.4 m^{2}$

$∴ T h e A r e a o f C a n v a s = 17.6 + 26.4 = 44 m^{2}$

Now,

$C o s t o f t h e c a n v a s o f t h e t e n t f o r 1 m^{2} = R s 500$

$C o s t o f t h e c a n v a s o f t h e t e n t f o r 44 m^{2} = 500 \times 44 = R s 22000$

$∴ T o t a l C o s t = 22000 r u p e e s$

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A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs 500 per $m^{2}$ . (Note that the base of the tent will not be covered with canvas.)

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