Question

(iii) area of triangle = abc/4R​
where R is the circum radius

Answer 1

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

Answer 2

Answer:

Hii

Step-by-step explanation:

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