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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 surface area of the double cone so formed. (Choose value of π as found appropriate)

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Solution

When a right angled triangle revolves around its hypotenuse, a double cone is formed.

Slant height ABB' = 20 cm
and slant height of cone BCB' = 15 cm

In ABC,B=90o

AC=202+15=400+225=625=25 cm

Let in ABC,AO=x cm

BO2=AB2AO2=400x2(i)

In BOC, OC=ACAO=25xBO2=BC2OC2=225(25x)2(ii)

From (i) and (ii), we get,

400x2=225(25x)2400x2=225(62550x+x2)400x2=225625+50xx250x=800x=80050=16

Subtitute the value in (i),

BO2=400(16)2=400256=144BO2=144BO=144=12 cm

Rdaius of each cone, r = 12 cm

Surface area of double cone = surface area of ABB' + surface area of CBB'

=πr×AB+πr×BC=πr(AB+BC)=3.14×12×(20+15)=3.14×12×35=1318.8 cm2


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