Question

Find the volume of the double cone which is formed when the right angled triangle with sides 3cm, 4cm and 5cm is revolved about its hypotenuse

Answer 1

The volume of the double cone is 30.14 cm ²

Step-by-step explanation:

the double cone is shown in the figure

we need to find the volume of the double cone

it has two cones  ABD, CBD

for cone ABD , radius = BO

height = A0

for cone CBD , radius = BO

height = OC

first we need to find BO,AO,OC

area of triangle ABC

$\frac{1}{2}\times base \times height \\\\\frac{1}{2}\times AB \times BC \\\\\frac{1}{2}\times 3 \times 4 \\\\6 cm ^2$..(1)

again , area of triangle ABC  

$\frac{1}{2}\times base \times height \\\\\frac{1}{2}\times AC \times BO \\\\\frac{1}{2}\times 5 \times BO \\\\\frac{5}{2} BO cm ^2$...(2)

From 1 and 2

$BO = \frac{12}{5}$

Now, using pythagoras theorem

AB² =AO²+BO²

placing all the values we get

AO = $\frac{9}{5}$ cm

similarly in triangle BOC

using pythagoras theorem

BC² =BO²+CO²

we get CO = $\frac{16}{5} cm$

now ,

$BO = \frac{12}{5}cm \\\\AO = \frac{9}{5}cm \\\\CO = \frac{16}{5}cm$

Volume of double cone = volume of ABD + volume of cone CBD

volume of cone ABD

$\frac{1}{3}\pi r^2h\\\\\frac{1}{3}\pi(BO)^2\times AO\\\\\frac{1}{3}\pi\times(\frac{12}{5})^2\times\frac{9}{5}\\\\= \frac{432}{125}\pi$

Volume of cone CBD

$\frac{1}{3}\pi r^2h\\\\\frac{1}{3}\pi(BO)^2\times CO\\\\\frac{1}{3}\pi\times(\frac{12}{5})^2\times\frac{16}{5}\\\\= \frac{768}{125}\pi$

volume of double cone

$\frac{432}{125}\pi+\frac{768}{125}\pi\\\\\frac{48}{5}\times 3.14\\\\30.14 cm ^2$

hence , The volume of the double cone is 30.14 cm ²

#Learn more:

a right triangle with sides 3cm,4cm,5cm is rotated about the side of 3cm to formed a cone. what is the volume of cone?

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