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.

Explanation

Let `DeltaAOB` be the given right triasngle in which `/_AOB=90^(@)`. When `DeltaAOB` is rotated about the hypotenus AB, the two cones generated are AO'O and BO'O.
Clearly, `OA = 3 cm, OB = 4 cm`.
In right-angled `DeltaAOB`, we have
`AB=sqrt((OA)^(2)+(OB)^(2))=sqrt(3^(2)+4^(2))cm`
`=sqrt(25)cm=5cm`.
Let OP = x cm. Then,
`1/2xxOAxxOB=1/2xxABxxOP`
`rArr" "1/2xx3xx4=1/2xx5xxOPrArrOP=((3xx4)/5)cm=2.4cm.`
In rihgt-angled `DeltaAPO`, we have
`AP=sqrt((OA)^(2)-(OP)^(2))=sqrt(3^(2)-(2.4)^(2))cm`
`=sqrt(9-5.76)cm=sqrt(3.24)cm=1.8cm.`
`BP=AB-AP=(5-18)cm=3.2 cm.`
`:." radius of each of the cones AO'O and BO'O, r = OP = 2.4 cm"`
`"Height of the coneAO'O, h = AP = 1.8 cm".`
`"Height of the cne BOO', H=BP=3.2 cm".`
Volume of the double cone
= volume of the the cone AO'O + volume of the cone BOO'
`1/3pir^(2)h+1/3pir^(2)H=1/3pir^(2)(h+H)`
`=(1/3xx3.14xx2.4.xx2.4xx5)cm^(3)=30.144cm^(3)`.
Slant height of the cone AO'O, `l_(1)=OA=3 cm`.
Slant height of the cone BO'O `l_(2)=OB=4 cm`.
Surface area of the double cone
=curved cusrface of the cone AO'O + curved surface area of the cone BOO'
`=pirl_(1)+pirl_(2)+pirl_(2)=pir(l_(1)+l_(1))=[22/7xx2.4xx(3+4)]cm^(2)`
`=(22/7xx2.4xx7)cm^(2)=52.8cm^(2).`