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=AB2−AO2=400−x2−−−−−(i)
In △BOC, OC=AC−AO=25−x⇒BO2=BC2−OC2=225−(25−x)2−−−−−(ii)
From (i) and (ii), we get,
400−x2=225−(25−x)2400−x2=225−(625−50x+x2)400−x2=225−625+50x−x250x=800x=80050=16
Subtitute the value in (i),
BO2=400−(16)2=400−256=144BO2=144⇒BO=√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