Question

Right triangle whose sides are 15cm and 20cm is is revolve around the hypotenuse find the volume of the double cone so formed. *​

Answer 1

Answer:

Volume of the double cone formed = $3,771 cm^3$

Step-by-step explanation:

Please, see attached.  AB = 20 cm, BC = 15 cm.

Hypotenuse squared = (15x15 + 20x20) square cm = (225 + 400) square cm = 625 square cm = (25 cm) squared

Therefore, hypotenuse, AC  = 25 cm.

When ΔABC revolve around its hypotenuse AC, two cones represented by ABD and BCD are formed.  O is the mid point of BD.  Let's assume that, BO = OD = x, AO = y and OC = z.

y + z = AC = 25  . . . . . . . . . . . . . . . (i)

$x^2 + y^2 = 20^2 = 400$ . . . . . . . . . . . . (ii)

$x^2 + z^2 = 15^2 = 225$ . . . . . . . . . . . .(iii)

Subtracting (iii) from (ii):

$y^2 - z^2 = 175$ . . . . . . . . . . . . . . . . (iv)

Dividing (iv) by (i):

y - z = 7 . . . . . . . . . . . . . . . (v)

From (i) and (v):

y = 16 and z = 9

From (iii), $x^{2} = 225 - 81 = 144 = 12^2$

Now Volume of a cone = Area of the base x One third of its height

Volume of the double cone formed = π$x^{2} (y + z).\frac{1}{3}$ = $\frac{22}{7} . x^2 . 25. \frac{1}{3}$ = $\frac{22}{7} . 144 . \frac{25}{3} cm^3$

= $3,771 cm^3$

Answer 2

Answer:

Check your answer please