Question

A right triangle whose base and hiegtht are 15cm and 20cm respectively made to revolve about its hypotenuse find volume and surface area of double cone so formed​

Answer 1

Answer

Volume = 3771.428 cm³

Surface area = 1320 cm²

$\rule{50}3$

Solution

(Refer the attachment for figure)

First of all, we will find the hypotenuse for the given right triangle

Apply Pythagoras theorem,

AB² + BC² = AC²

→ 15² + 20² = AC²

→ 225 + 400 = AC²

→ AC² = 625

Taking square root,

→ AC = 25 cm

For finding out volume, we need to find the radius of the cones.

For a double cone, the radius is same.

Consider ∆ABC,

The radius, OB, will work as altitude for hypotenuse AC

Then, ar(∆ABC) = ½ × OB × AC

but, ar(∆ABC) = ½ × AB × BC

Equating the two,

½ × OB × AC = ½ × AB × BC

→ OB = AB × BC/AC

→ OB = 15 × 20/25

→ OB = 12 cm.

Hence, the radius for the cones is 12 cm.

Now, as we can see in the figure,

The volume of double cone formed = volume of cone at the top + volume of the cone at bottom.

The height for first cone is OA and for second is OC

→ volume of double cone

= ⅓πr² × OA + ⅓πr² × OC

→ ⅓πr²(OA + OC)

→ ⅓πr²(AC)

= ⅓ × 22/7 × (12)² × 25

= ⅓ × 22/7 × 144 × 25

= 79200/21

= 3771.428 cm³ (approx)

$\rule{100}2$

Now, Surface area of double cone = CSA of cone at top + CSA of cone at bottom

Slant height for first cone is 15 and for second is 20

→ 15πr + 20πr

= 35πr

= 35 × 22/7 × 12

= 1320 cm²

Answer 2

Refer to the attachments..