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# 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

In ΔABC right-angled at B using Pythagoras theorem,

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

We know that, Volume of the cone = 1/3πr²h

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 × AC

= 1/3 × 3.14 × 2.4 cm × 2.4 cm × 5 cm

= 30.144 cm³

We know that, CSA of frustum of a cone = πrl

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 × 7 cm

= 52.752 cm²

= 52.75 cm²

**☛ Check: **Class 10 Maths NCERT Solutions Chapter 13

**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

**Summary:**

A right triangle, whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. The volume and surface area of the double cone so formed are 30.144 cm³ and 52.75 cm² respectively.

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