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

 

 
 Nov 21, 2020
 #1
avatar+112375 
0

 

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 

 

 

cool cool cool

 
 Nov 21, 2020
 #2
avatar+452 
+2

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

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

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

 

 
 Nov 22, 2020
 #3
avatar+1421 
+1

∠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

 
 Nov 22, 2020

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