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