When the right angle triangle is revolved about the hypotenuse AC, the figure so formed is a double cone.

Now, from the figure,

In triangle ABC,

     BC² = AB² + AC²

=> BC² = (15)² + (20)²

=> BC² = 225 + 400

=> BC² = 625

=> BC = √625

=> BC = 25

Now, let OB = x

then OC = BC - OB

=> OC = 25 - x

Now, in triangle AOB,

     AB² = OA² + OB²

=> (15)² = OA² + x²

=> 225 =  OA² + x²

=> OA² = 225 - x² .................1

Again, in triangle AOC,

     AC² = OA² + OC²

=> (20)² = OA² + (25 - x)²

=> 400 =  OA² + (25 - x)²

=> OA² = 400 - (25 - x)² .................2

Now, from equation 1 and 2, we get

     225 - x²  = 400 - (25 - x)²

=> 225 - x²  = 400 - (625 + x² - 50x)

=> 225 - x²  = 400 - 625 - x² + 50x

=> 225 = 400 - 625 + 50x

=> 225 = -225 + 50x

=> 225 + 225 = 50x

=> 50x = 450

=> x = 450/50

=> x = 9

=> OB = 9

and OC = 25 - 9 = 16

From equation 1, we get

     OA² = 225 - 9²

=> OA² = 225 - 81

=> OA² = 144

=> OA = √144

=> OA = 12

Now, volume of the double cone = volume of the smaller cone + volume of the larger cone

=> volume of the double cone = (1/3) *π*(OA)² *(OB) + (1/3) *π*(OA)² *(OC)

=> volume of the double cone = (1/3) *π*(OA)² *(OB + OC)

=> volume of the double cone = (1/3) *(22/7)*(12)² * BC

=> volume of the double cone = (1/3) *(22/7)*12*12*25

=> volume of the double cone = (22/7)*4*12*25

=> volume of the double cone = 26400/7 cm³

Agian, surface area of the double cone = couved surface area of the smaller cone + couved surface area of the larger cone

=> surface area of the double cone = π *OA*AB + π *OA*AC

=> surface area of the double cone = π *OA*(AB + AC)

=> surface area of the double cone = (22/7) *12*(15 + 20)

=> surface area of the double cone = (22/7) *12*35

=> surface area of the double cone = 22 *12*5

=> surface area of the double cone = 1320 cm²