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1)

Triangle $ABC$ has circumcenter $O.$ If $AB = 10$ and $[OAB] = 30,$ find the circumradius of triangle $ABC.$

 

2)

Let $AB = 5$$BC = 12$, and $AC = 13$. What is the circumradius of $\triangle ABC?$

 

3)

$\triangle ABC$ is an isosceles right triangle where $\angle A=90^\circ.$ $O$ is the circumcenter of $\triangle ABC.$ What is $\angle AOB$ in degrees?

 
Guest Nov 10, 2017
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2+0 Answers

 #1
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+1

2)

Since $AB^2 + BC^2 = AC^2$, we know that $\triangle ABC$ is a right triangle with hypotenuse $\overline{AC}$ by the converse of the Pythagorean Theorem. The circumcenter of a right triangle is the midpoint of its hypotenuse, so the circumradius is $\dfrac{AC}{2}=\boxed{\dfrac{13}{2}}$.

 
Guest Nov 10, 2017
 #2
avatar+78577 
+1

1) The circumcenter is where the perpendicular bisectors of  of the sides of the triangle meet

 

So......the circumradius, R ,  can be found as

 

cos (30)  = (1/2)AB / R    ⇒  R  = (1/2)AB / cos (30)  =

 

5 / [√3 / 2 ] =  10 / √3 ≈    5.77 units

 

 

cool cool cool

 
CPhill  Nov 11, 2017

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