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?
+129895 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
+129895 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°
+129895 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°
+129895 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
+129895 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°
