The slant height of the frustum of a cone is $4 c m$ . If the perimeters of its circular bases be $18 c m$ and $6 c m$ , find the curved surface area of the frustum and also find the cost of painting its total surface at the rate of Rs. $12.50$ per $100 {c m}^{2}$

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Solution

R.E.F image
$2 π R = 18$ $2 π r = 4$
$π R = 9$ $π r = 2$
$R = \frac{9}{π}$ $r = \frac{2}{π}$
Curved surface area $= π (R + 2) l$
$= (\frac{9}{π} + \frac{2}{π}) 4$
$= 44 c m^{2}$
Total cost $= \frac{12.50}{100} \times 44$
$= 5.5 R s$

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The slant height of the frustum of a cone is 4cm. If the perimeters of its circular bases be 18cm and 6cm, find the curved surface area of the frustum and also find the cost of painting its total surface at the rate of Rs. 12.50 per 100cm2

R.E.F image 2πR=18 2πr=4πR=9 πr=2R=9π r=2πCurved surface area =π(R+2)l=(9π+2π)4=44cm2Total cost =12.50100×44 =5.5Rs

The slant height of the frustum of a cone is $4 c m$ . If the perimeters of its circular bases be $18 c m$ and $6 c m$ , find the curved surface area of the frustum and also find the cost of painting its total surface at the rate of Rs. $12.50$ per $100 {c m}^{2}$

R.E.F image
$2 π R = 18$ $2 π r = 4$
$π R = 9$ $π r = 2$
$R = \frac{9}{π}$ $r = \frac{2}{π}$
Curved surface area $= π (R + 2) l$
$= (\frac{9}{π} + \frac{2}{π}) 4$
$= 44 c m^{2}$
Total cost $= \frac{12.50}{100} \times 44$
$= 5.5 R s$