Three circles are drawn with the vertices of a triangle as centres such that each circle touches the other two. If the sides of the triangle are 2 cm, 3 cm, and 4 cm, find the diameter of the smallest circle.

(a) 1 cm

(b) 3 cm

(c) 5 cm

(d) 4 cm.

**Please give the working also..**

Let ABC be the triangle and three circles are drawn such that each circle touches the other two.

Let *x* be the radius of circle drawn at A. *y* be the radius of circle drawn at C and *z* be the radius of the circle drawn at B.

Let the sides

AC = 2 cm

AB = 3 cm

BC = 4 cm

∴ AC = *x* + *y* = 2 ........ (1)

AB = *x* + *z* = 3 ........ (2)

BC = *y* + *z* = 4 ........ (3)

On adding (1), (2) & (3), we get

2 (*x* + *y* + *z*) = 9

⇒ *x* + *y* + *z* = 4.5 ...... (4)

Since *x* + *y* + *z* = 4.5

and *x* + *y* = 2

⇒ *z* = 4.5 – 2

*z *= 2.5 cm

*y* + *z* = 4

⇒ *x* = 4.5 – 4

*x* = 0.5 cm

*z* + *x* = 3

⇒ *y* = 1.5 cm

Hence the radius of smallest circle is 0.5 cm

∴ Diameter of the smallest circle is 0.5 × 2 = 1 cm

**
**