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Carolyn S. Ashley E. Hour 3 Honors Geometry Quarter 2

First, we created an acute triange, as shown below. Then we found midpoints E and G, by calculating the following:

C (0,4) __+ A (4,8)__ (4, 12) which we divided by two, and found E as (2,6)

A (4,8) __+ B (6,4)__ (10,12) which we divided by two, and found G as (5,6)

A (4,8) __- C (0,4)__ -4,-4 therefore it has a slop of 1

The slope of line segment CA is 1, so the perpendicular bisector has a slope perpendicular to the original one, so the perpendicular bisector's slope is -1.

A (4,8) __- B (6,4)__ -2,4 therefore it has a slope of -2

The slope of line AB is -2, so the slop of the perpendicular bisector is 1/2, because it must be perpendicular to the slope of -2.

y=-1x+b 6=-1(2)+b 6= -2 +b 6+2 =-2+2+b 8=b The equation for the perpendicular bisector for line CA is y=-1x+8

y=1/2x+b 6=1/2(5)+b 6=2.5+b 6+-2.5=2.5+-2.5+b 3.5=b The equation for the perpendicular bisector for line AB is y=1/2x+3.5



Now, we solve the system of equations. y=-1x+8 y=1/2x+3.5

-1x+8=1/2x+3.5 +1x +1x 8=1 1/2x+3.5 -3.5 -3.5 4.5=1 1/2x 3=x

y=-1(3)+8 y=-3+8 y=5

Now we have x=3 and y=5, so the circumcenter's coordinates are (3,5). We'll call this point H.



To find the radius of the circle, we'll use the Pythagorean Theorm. We will creat an imaginary triangle using the points C, H, and the point at (3,4). Now we take the theorm and plug in the numbers. a ²+b ²=c ²

We'll make H and the point directly beneath it (3,4) form one side, which will be a. Point (3,4) to point C will form side b, and the hypotenuse will be from C to H. Now let's plug in the numbers. a=1, b=3, and we want to find c.

1 ²+3 ²=c ² When solved, we took c square rooted it, which gave us our radius. Then is was just a matter of constructing the circle, which was easily done.

That is how we found the circumcenter and constructed the cirlce.