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Question
A right triangle whose sides are 3 cm and 4 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.)
Solution
The double cone so formed by revolving this right-angled triangle ABC about its hypotenuse is shown in the figure.
Hypotenuse AC = `sqrt(3^2+4^2)`
=`sqrt25` = 5 cm
Area of ΔABC = 1/2 x AB x AC
`1/2xx AC xx OB = 1/2xx4xx3`
`1/2xx5xxOB = 6`
OB = 12/5 = 2.4 cm
Volume of double cone = Volume of cone 1 + Volume of cone 2
`=1/3 pir^2h_1 + 1/3 pir^2h_2`
`=1/3pir^2(h_1+h_2)=1/3pir^2(OA+OC)`
`=1/3xx3.14xx(2.4)^2(5)`
=30.14 cm3
Surface area of double cone = Surface area of cone 1 + Surface area of cone 2
= πrl1 + πrl2
= πr[4+3] = 3.14 x 2.4 x 7
= 52.75 cm2
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