Answer 1

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