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Question
A right triangle whose sides are 15 cm and 20 cm (other than hypotenuse), is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of π as found appropriate)
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Solution
We have,
In ΔABC, ∠B = 90° , AB = l1 =15 cm and BC = l2 = 20 cm
Let OD = OB = r , AO = h1 and CO = h2
Using Pythagoras therom,
`"AC" = sqrt("AB"^2 + "BC"^2)`
`=sqrt(15^2 + 20^2)`
`= sqrt(225 + 400)`
`= sqrt(625)`
⇒ h = 25 cm
As, ar(ΔABC)` = 1/2xxACxxBO=1/2xxABxxBC `
⇒ AC × BO = AB × BC
⇒ 25r = 15 × 20
`rArr r = (15xx20)/25`
⇒ r = 12 cm
Now,
Volume of the double cone so formed = Volume of cone 1 +Volume of cone 2
`= 1/3 pir^2h_1 + 1/3pir^2h_2`
`= 1/3 pir^2 ("h"_1 + "h"_2)`
`= 1/3pi"r"^2"h"`
`= 1/3xx3.14xx12xx12xx25`
= 3768 cm3
Also,
surace area of the solid so formed = CAS of cone 1 + CSA of cone 2
= πrl1 + πrl2
= πr ( l1 + l2 )
`= 22/7xx12xx(15+20)`
`=22/7xx12xx35`
= 1320 cm2
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