A circle is iscribed in an equilateral triangle of side 12 cm,touching its sides. Find the radius of the inscribed circle?

The triangle of side 12cm side (each) is inscribed in a
circle.

Solution:

To determine
the radius.

The inscribed triangle has each side has 12cm.
So the angles of an equilateral triangle are 60 deg
each.

We can detrmine the radius by (i) applying the sine
rule  or (ii) By considering  distance of the mid point from the centre O of the
cicle.e

By sinerule  we
have:

  side/sine of the angle oppsite to the side =
2R.

12/sin(60) = 2R.

2R =
12/((sqrt3)/2)

2R = 12*2/sqrt3 =
24*sqrt3/3

R =
4sqrt3cm.

(ii)

Let D be the
mid point of a side AB of the eqilateral triangle.

Now
consider the triangle AOB in which angle AOB = 2angle ACB , as angle subtended by the
chord AB at the centre ) is twice the angle subtended by AB at C on the qpposite part of
circumference. So angle AOB = 120 degree.

Triangle OAD and
OBD are congruent as OD common and ODA = ODB = 90 degree each and AB= BD, as the the
line joining the mid point of the chord and the centre is perpendicular to the chord. So
Angle AOD = Angle BOD = 120/2 = 60. Therefore triangle DBO is a right angled triangle
with angle DOB = 69de and angle ABD = 30 degree.

Therefore
2OD =  OB  = R the radius  Or

OD =
R/2.

QB = R.

and DB = 12/2 =
6cm

Applying pythagorus
theorem,

OD^2+DB^2 =
OD^2.

(R/2)^2 +6^2 = R^2

6^2 =
R^2-(R/2)^2

6^2 = (3/4)R^2

R^2
= 4*6^2/3

R^2 = 48

R =
4sqrt3