Question

A right triangle having sides 15cm and 20cm is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed..please answer ASAP

Answer 1

hope it helped you , if yes then please mark as brainliest

Answer 2

Hope it helps you
plz mark it as brainliest 

Let the right angled triangle be ABC, with the right angle at C, AC = 20 cm and BC = 15 cm. The hypotenuse will be [20^2 + 15^2]^0.5 = 25 cm.
Next draw a perpendicular CD on to AB.
Cos A = 20/25 = 4/5.
Cos B = 15/25 = 3/5.
Triangles ABC and CDB are similar, Therefore CD = 12 cm, AD = 9 cm and BD = 16 cm.
When you revolve the triangle about the hypotenuse there will be two cones attached, base to base. So it will be double cone.
The radius of the common base is 12 cm (CD above).
The upper cone (UC) will have a height of 9 cm (AD above) and the lower cone (LC) will have a height of 16 cm (BD above).
Volume of a cone = (1/3)(22/7)R^2*H
Volume of UC = (1/3)(22/7)12^2*9 = 1357.7143 cc
Volume of LC = (1/3)(22/7)12^2*16 = 2413.7143 cc
The total volume of U C and LC = (1/3)(22/7)12^2*(9+16) = 3771.4286 cm3
-------------------------------------------------------------------------------------------