# A right triangle, whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of π as found appropriate)

**Solution:**

A figure is drawn below to visualize the double cone formed.

In order to find the volume and surface area, we need to find BD or radius of the double cone

From the figure it can be seen that BD ⊥ AC

To find BD

(i) We first find AC using Pythagoras theorem

AC² = AB² + BC²

AC = √AB² + BC²

(ii) Using AA criterion of similarity

Prove

Δ ABC ∼ Δ BDC

AB/BD = AC/BC (Corresponding Sides of similar triangles are in proportion)

Radius or BD = AB/AC × BC

As we know AB, AC and BC; BD can be found out

Since double cone is made by joining 2 cones by their bases

Therefore, Volume of double cone = Volume of Cone ABB’ + Volume of Cone BCB’

Let us find the volume of the cone by using formulae;

Volume of the cone = 1/3πr²h

where r and h are the radius and height of the cone respectively.

Visually from the figure it’s clear that CSA of double cone includes CSA of both the cones.

Therefore, CSA of double Cone = CSA of cone ABB’ + CSA of cone BCB’ We will find the CSA of the cone by using formulae;

CSA of frustum of a cone = πrl

where r and l are the radius and slant height of the cone respectively.

In ΔABC right-angled at B

AC² = AB² + BC²

AC = (3 cm)² + (4 cm)²

= 9 cm² + 16 cm²

= 25 cm²

= 5 cm

Consider ΔABC and ΔBDC

∠ABC = ∠CDB = 90° (BD ⊥ AC)

∠BCA = ∠BCD (common)

By AA criterion of similarity ΔABC ∼ ΔBDC

Therefore,

AB/BD = AC/BC (Corresponding sides of similar triangles are in proportion)

BD = (AB × BC)/AC

= (3 cm × 4 cm)/5 cm

= 12/5 cm

= 2.4 cm

Volume of double cone = Volume of Cone ABB’ + Volume of Cone BCB’

= 1/3 × π(BD)² × AD + 1/3π (BD)² × DC

= 1/3 × π(BD)² [AD + DC]

= 1/3 × π(BD)² × AD

= 1/3 × 3.14 × 2.4 cm × 2.4 cm × 5-cm

= 90.432 cm³

= 30.144 cm³

= 30.14 cm³

CSA of double Cone = CSA of cone ABB’ + CSA of cone BCB’

= π × BD × AB + π × BD × BC

= π × BD [ AB + BC]

= 3.14 × 2.4 cm × [3 cm + 4 cm]

= 3.14 × 2.4 cm x 7 cm

= 52.752 cm²

= 52.75 cm²

**Video Solution:**

## A right triangle, whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed

### NCERT Solutions for Class 10 Maths - Chapter 13 Exercise 13.5 Question 2 :

A right triangle, whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed

The volume and surface area of the double cone formed by revolving a right triangle having sides 3 cm and 4 cm around its hypotenuse are 30.14 cm^3 and 52.75 cm^2 respectively