Question

A right triangle whose sides are 15 cm and 20 cm is made to revolve about its hypotenuse. Find the volume and the surface area of the double cone so formed. (Use π =3.14)

Answer 1

When a right angled triangle revolves around its hypotenuse, a double cone is formed.

Slant height of cone ABB' = 20 cm

and slant height of cone BCB' = 15 cm

In △ABC,

∠B = 90^∘

AC = \(\sqrt{20^2 + 15}\)

\(= \sqrt{400 + 225}\)

\(= \sqrt{625}\)

= 25 cm

In △ABC,

AO = x cm

BO² = AB² − AO² = 400 − x²   .......(i)

In △BOC,

OC = AC − AO = 25 − x

⇒ BO² = BC² − OC²

= 225 − (25−x)²    .......(ii)

From (i) and (ii), we get

400 − x² = 225 − (25−x)²

400 − x² = 225 − (625 − 50x + x²)

400 − x² = 22 5− 625 + 50x − x²

50x = 800

x = \(\frac{800}{50} \) = 16

Substitute the value in (i), we get

BO² = 400 − (16)²

= 400 − 256

= 144

BO² = 144

⇒ BO = √144 = 12 cm

Radius of each cone, r = 12 cm

The volume of double cone = Volume of ABB' + Volume of BCB'

= \(\frac 13\)πBO² × AO + \(\frac 13\)πBO² × CO

= \(\frac 13 \times \frac{22}7\) × BO²[AO + CO]

= \(\frac 13\) × 12² × 25 × \(\frac{22}7\)

= 12 × 4 × 25 × \(\frac{22}7\)

= 3771.42 cm³

Surface area of the double cone = CSA of ABB' + CSA of BCB'

= πr × AB + πr × BC

= πr(AB + BC)

= 3.14 × 12 × (20 + 15)

= 3.14 × 12 × 35

= 1318.8 cm²

Answer 2

Volume of double cone = Volume of upper cone + Volume of lower cone