Triangle ABC has circumcenter O. If AB = 10 and [OAB] = 50 find the circumradius of triangle ABC.
+150 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.
+605 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.
+1693 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
