# A metallic right circular cone 20 cm high and whose vertical angle is 60° is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter 1/16 cm, find the length of the wire

**Solution:**

A figure is drawn below to visualize the cross-section of a cone.

As mentioned above, a metallic right circular cone is cut into two parts at the middle of its height by a plane parallel to its base.

To get the values of radii of both ends of the frustum formed compare ΔADB and ΔADC

Since the frustum obtained is drawn into the wire which will be cylindrical in the shape. Then the volume of the wire will be same as the volume of the frustum of the cone.

Therefore, Volume of the wire = Volume of frustum of the cone

We will find the volume of the frustum by using formula;

Volume of frustum of a cone = 1/3 πh (r_{1}^{2} + r_{2}^{2} + r_{1}r_{2}), where r_{1}, r_{2,} and h are the radii and height of the frustum of the cone respectively.

We will find the volume of the wire by using formula;

Volume of cylinder = πr^{2}h, where r and h are radius and height of the cylinder

In ΔABC , EF parallel to BC and

AD = 20 cm

AG = 10 cm

∠BAC = 60°

To get the values of radii of both ends of the frustum formed compare ADB and ADC

AD = AD (common)

AB = AC (Slant height )

∠ADB = ∠ADC = 90 (Right circular cone)

ΔADB ≅ ΔADC (RHS criterion of congruency)

∠BAD = ∠DAC (CPCT)

Then,

∠BAD = ∠DAC = 1/2

∠BAC = 1/2 × 60° = 30°

In ΔADB

BD/AD = tan 30°

BD = AD tan 30°

BD = 20 cm × 1/√3

BD = 20√3/3 cm

Similarly, in ΔAEG

EG/AG = tan 30°

EG = AG tan 30°

EG = 10 cm × 1/√3

EG = 10√3/3 cm

Height of the frustum of the cone, h = 10 cm

Radius of lower end, r_{1} = (20√3) / 3 cm

Radius of upper end, r_{2} = (10√3) / 3 cm

Diameter of the cylindrical wire, d = 1 / 16 cm

Radius of the cylindrical wire, r = 1 / 2 × 1 / 16 cm = 1 / 32 cm

Let the length of the wire be H

Since the frustum is drawn into wire

Volume of the cylindrical wire = Volume of frustum of the cone

πr^{2}H = 1/3 πh(r_{1}^{2} + r_{2}^{2} + r_{1}r_{2})

H = [h(r_{1}^{2} + r_{2}^{2} + r_{1}r_{2})]/ 3r^{2}

= [ 10 × {(20√3) /3)^{2} + (10√3/ 3)^{2} + (20√3) /3 × (10√3) /3] / 3 × (1/32)^{2}

= [10 × {400/3 + 100/3 + 200/3}] / 3 × (1/1024)

= 10240 × 700 / 9

= 7168000 / 9

= 796444.44 cm

= 7964.4 m

Thus the length of the wire is 7964.4 m.

**Video Solution:**

## A metallic right circular cone 20 cm high and whose vertical angle is 60° is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter 1/16 cm, find the length of the wire

### NCERT Solutions for Class 10 Maths - Chapter 13 Exercise 13.4 Question 5 :

A metallic right circular cone 20 cm high and whose vertical angle is 60° is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter 1/16 cm, find the length of the wire

The length of the wire drawn out from a frustum obtained from cutting a metallic right circular cone into two parts at the middle of its height by a plane parallel to its base is 7964.4 m