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1)

The sides of $\triangle ABC$ have lengths $\sqrt{10}$$\sqrt{10}$, and $2$. What is the circumradius of $\triangle ABC$?

 

2)

Let $O$ and $H$ be the circumcenter and orthocenter of acute triangle $\triangle ABC$, respectively. If $\angle AHC+\angle AOC=240^\circ$, what is $\angle ABC$ in degrees?

 

3)

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

(a) $\triangle ABC$ must be an isosceles triangle.
(b) $\triangle ABC$ must be an equilateral triangle.
(c) $\triangle ABC$ must be a right triangle.
(d) $\triangle ABC$ must be an isosceles right triangle.
Enter your answer as a comma-separated list. If there is no correct option, write "none".

Guest Nov 10, 2017
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3+0 Answers

 #1
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1)   The circumradius is given by

 

[ Product of the sides ] / [ 4 * Area ]

 

The area is given by  (1/2) Base * Height

 

This ttriangle is isosceles....let the base  = 2 units

 

The height [altitude ]  is given  by   √ [  (√ 10)^2   - 1^2 ] =  √ [  10 - 1 ]  = √  [9]  =  3 units

 

So...the area is   (1/2) 2  (3)   =   3   units^2

 

So....the circumradius is      [ √10 * √10 * 2 ]  [ 4 * 3] =   [ 20/12] =  5/3  units  ≈  1.67 units  

 

 

 

cool cool cool

CPhill  Nov 10, 2017
 #2
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Wht is the answer to the other questions

Guest Nov 12, 2017
 #3
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number 1 is 5/3 because of this special formula. To find the circumradius of any triangle with sides a,b,c the formula is abc/4A where A is the area of the triangle.

Guest Nov 13, 2017

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