web2.0calc
 
+0  
 

Geometry -- Pls answer quick!! --

+1
2719
2
avatar

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?

 Nov 10, 2017

2+0 Answers

 #1
avatar
+3

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 $\dfrac{AC}{2}=\boxed{\dfrac{13}{2}}$.

 Nov 10, 2017
 #2
avatar+128474 
+2

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

 Nov 11, 2017

2 Online Users

Top Users

avatar+128474 CPhill moderator
avatar+118609 Melody moderator
avatar+33615 Alan moderator
avatar+14913 asinus moderator
avatar+9519 MaxWong
avatar+3144 admin administrator
avatar+2440 GingerAle
avatar+1791 tomtom
avatar+1484 bader
avatar+1085 parmen
avatar+1032 kittykat

Sticky Topics