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1. Triangle ABC has circumcenter O. If AB=10 and Area of OAB=30 find the circumradius of triangle ABC

 

2. Triangle ABC is an isosceles right triangle where Angle A=90 degrees. O is the circumcenter of Triangle ABC. What is Angle AOB in degrees?

 

3. In the diagram below, O is the circumcenter of Triangle ABC. If Angle BAC=28 degrees and Angle OAC=32 degrees, then what is the degree measure of Angle AOC?

 

 

4. Point O is the circumcenter of Triangle ABC. The distance between O and Line AC is 7, and the distance between O and Line AB is 15. If AC=48, what is AB?

 

5. Point H is the orthocenter of Triangle ABC. If Angle BAC=115 degrees, what is angle BHC in degrees?

 

6.  In Triangle ABC, the circumcenter and orthocenter are collinear with vertex A. Which of the following statements must be true?

(1)  Traingle ABC must be an isosceles triangle.
(2)  Traingle ABC must be an equilateral triangle.
(3)  Traingle ABC must be a right triangle.
(4)  Traingle ABC must be an isosceles right triangle.

Enter your answer as a comma-separated list. If there is no correct option, write "none".

 

7. Let H be the orthocenter of the equilateral triangle ABC. We know the distance between the orthocenters of Traingle AHC and Triangle BHC is 12. What is the distance between the circumcenters of Triangle AHC and Triangle BHC?

 

8. As shown in the diagram, points B and D are on different sides of line AC. We know that Angle B= 2*Angle D=60 Degrees and that AC=4sqrt(3) . What is the distance between the circumcenters of Triangle ABC and Triangle ADC?

 Nov 26, 2018
 #1
avatar+128475 
+3

1. Triangle ABC has circumcenter O. If AB=10 and Area of OAB=30 find the circumradius of triangle ABC

 

AB is a chord of the circumcircle

 

And we can find the altitude of triangle oAB thusly

 

30  =  (1/2)AB * altitude       multiply through by 2

 

60 = 10 * altitude     divide both sides by 10

 

6 = the altitude

 

But.... because the altitude is perpendicular to AB, the altitude will bisect chord AB

 

Call the bisection point, M

 

So....we have right triangle   MOA

 

And OA  is  the hypotenuse of this triangle = the circumradius

And AM is one leg = 6

And MO is the other leg = 5

 

So.....using the Pythagorean Theorem

 

OA = √ [ AM^2 + MO^2] = √ [ 6^2 + 5^2 ] = √ [ 61 ]  = the circumradius

 

 

cool cool cool

 Nov 26, 2018
 #2
avatar+128475 
+1

2. Triangle ABC is an isosceles right triangle where Angle A=90 degrees. O is the circumcenter of Triangle ABC. What is Angle AOB in degrees?

 

In a right triangle.....the circumcenter is at the midpoint of the hypotenuse

 

So.....OB = OC

 

Since ABC is isosceles, Angle ABC = Angle ACB =  45°

And AC = AB

And AO = AO

And by SAS, triangle AOC is cogruent to triangle AOB

But angle CAO = angle BAO  ....so AO bisects angle BAC

So angle BAO = 45°

And angle ABO = angle ABC = 45°

 

So....in triangle AOB....angle AOB = 90°

 

 

cool cool cool

 Nov 26, 2018
 #3
avatar+128475 
+2

3. In the diagram below, O is the circumcenter of Triangle ABC. If Angle BAC=28 degrees and Angle OAC=32 degrees, then what is the degree measure of Angle AOC?

 

OA and OC will be the radii of the circumcircle.....so....they are equal

 

So....in triangle AOC.....the angles opposite OA and OC will also be equal

 

So....angles AOC and OCA are equal = 32°

 

So.....angle AOC   =  180  - 2(32)   =  180 - 64   = 116°  

 

 

cool cool cool

 Nov 26, 2018
 #4
avatar+128475 
+2

4. Point O is the circumcenter of Triangle ABC. The distance between O and Line AC is 7, and the distance between O and Line AB is 15. If AC=48, what is AB?

 

OD is a perendicular bisector of AC

So.....DA = (1/2)AC = 24

 

And triangle ODA is a 7 - 24 - 25  Pythagorean Right Triangle

So OA = 25

And OE = 15

 

So....since OE is perpendicular to AB....triangle AOE is a right triangle

 

And   EA = √ [ OA^2 - OE^2 ] = √ [ 25^2 - 15^2] = √ [ 400]   = 20

 

And since OE is a perpendicular isector to AB, then AB = 2EA = 2 * 20  = 40

 

 

cool cool cool

 Nov 26, 2018
 #5
avatar+128475 
+3

5. Point H is the orthocenter of Triangle ABC. If Angle BAC=115 degrees, what is angle BHC in degrees?

 

angle BAC = 115....so angle BAF = 180 - 115 = 65°

 

And angle AFB = 90

 

So angle ABF = 180 - 65 - 90 = 25°

 

And in triangle  BHE, angle BEH = 90°

 

So....angle BHE = angle BHC = 180 - 90 - 25 = 65°

 

 

cool cool cool

 Nov 26, 2018

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