(iii) area of triangle = abc/4R
where R is the circum radius
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Let O be the circumcentre of the Δ ABC. Join OB and OC. Triangle OBC is isosceles since OB=OC radii.
Draw OD perpendicular to BC. So OD bisects BC
Angle BOC = 2 angle BAC = 2 A.
OD is also the bisector of angle BOC. So angle BOD = A.
We have Sin A = Sin BOD = BD/OB = (a/2)/R = a/2R. So, Sin A = a/2R……..(1)
If we draw BE perpendicular to AC, Sin A = BE/AB or BE = AB sin A = c Sin A
Area of ΔABC = (1/2) AC. BE = (1/2) bc Sin A = (1/2) bc a/2R from (1)
So area of ΔABC = abc/4R
Draw OD perpendicular to BC. So OD bisects BC
Angle BOC = 2 angle BAC = 2 A.
OD is also the bisector of angle BOC. So angle BOD = A.
We have Sin A = Sin BOD = BD/OB = (a/2)/R = a/2R. So, Sin A = a/2R……..(1)
If we draw BE perpendicular to AC, Sin A = BE/AB or BE = AB sin A = c Sin A
Area of ΔABC = (1/2) AC. BE = (1/2) bc Sin A = (1/2) bc a/2R from (1)
So area of ΔABC = abc/4R
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