Find the area of triangle OBC .

a = 7   b  =  5    c  = 5

The circumradius length is  given by:

     abc

_________________________________________  =

√[ (a + b + c) ( a + b - c) (a + c - b ) ( b + c - a)]

7* 5 * 5

________________________________________ =

√ [ ( 7 + 5 + 5)  ( 7) ( 7)  ( 3) ] 

     175

_____________   =

√[ (17) (49) (3) ]  

     175

_______________  =

  7 √51

25 / √51

Using the Law of Cosines, we can  calculate  angle BAC

7^2  - 5^2  - 5^2

_____________   = cos BAC

     -2(5)(5)

1 / 50  =cos BAC

arccos (1/50) =  BAC  = 88.85°

And  since BAC  is an inscribed angle, then central angle BOC  is twice this  = 177.7°

So the area of   BOC =  (1/2) ( 25/ √71)^2 sin (177.7)  ≈   .177  units^2 

In triangle ABC,  AB= AC=5 and BC=7. Let O be the circumcenter of triangle ABC. Find the area of triangle OBC.

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Let M be a midpoint on AB.

Angle BAO ==>    β = arcsin(3.5 / 5) ==>  44.427004º

Area of triangle ABC ==>     A₁ [sqrt(5² - 3.5²) * 3.5 ==>   A₁ = 12.49749975 u²

Area of polygon ABOC ==>  A₂ = 2[(tanβ * 2.5) * 2.5]       A₂ = 12.25245073 u²

Δ OBC area ==>             A = A₁ - A₂ = 0.24504902 u²

∠BAO = sin^-1(3.5 / 5) = 44.427 deg.

BO = 2.5 / cos(∠BAO) = 3.50070021

OM = sqrt(BO² - BM²) = 0.070014           ( M is the midpoint on BC)

Area OBC = BM * OM = 0.245049744 units squared