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Triangle ABC has circumcenter O. If AB = 10 and [OAB] = 50 find the circumradius of triangle ABC.

 May 16, 2021
 #2
avatar+146 
-3

Well, the circumcenter is where the perpendicular bisectors of the sides of the triangle meet. (right?)
If the Circumradius (x, as we shall call it) shall be Cosine (50) = 1/2 AB / X

Which is 5/Cos 50.

Which, by a calculator, is 7.77862

or 7.78

Hopefully, that's the answer. 

 May 16, 2021
 #3
avatar+556 
+1

Let $OA=OB=OC=x$. Then $x^2\sin\theta=100$ by sine area formula (I used $\angle AOB=\theta$). By Law of Cosines, $2x^2-2x^2\cos\theta=100\implies x^2(1-\cos\theta)=50$. So dividing, $\frac{1-\cos\theta}{\sin\theta}=\frac{1}{2}\implies \tan(\theta/2)=1/2$. So it's not hard to get $\sin\theta=\frac{4}{5}$ (it should remind you of the 3-4-5 triangle). Then $x^2=100*5/4=125\implies x=\sqrt{125}=\boxed{5\sqrt{5}}$

 

The other answer is utter nonsense. 

 May 16, 2021
 #4
avatar+1318 
+4

Triangle ABC has circumcenter O. If AB = 10 and [OAB] = 50 find the circumradius of triangle ABC.

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

AB = 10           [OAB] = 50

Let M be a midpoint of AB

Altitude   OM = (2 * 50) / 10 = 10

OA = OB => R   (Circumradius)

R = sqrt(AM2 + OM2) = √125

 May 16, 2021

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