Results 1 – 24 of This worksheet is designed to replace a lecture on constructing the midsegments, perpendicular bisectors, and the circumcenter of a. Worksheet – constructing the circumcenter of a given triangle with compass and straightedge. Students will locate circumcenter of a triangle; Students will discover that the circumcenter is equidistant TI-Nspire calculator; Pencil; Straight-edge; Worksheet.
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Worksheet – Construct the circumcenter of a triangle
Let’s say that we find some point that is equidistant from A and B. Obviously, any segment is going to be equal to itself. So we’ve drawn a triangle here, and crcumcenter done this before.
This workshedt right over here is equal to that distance right over there is equal to that distance over there. And so what we’ve constructed right here is one, we’ve shown that we can construct something like this, but we call this thing a circumcircle, and this distance right here, we call it the circumradius.
a Circumcenters Worksheet ∆ ∆
Area circumradius formula proof. And we’ll see what special case I was referring to. And I don’t want it to make it ccircumcenter intersect in C because that’s not necessarily going to be the case. So we can just use SAS, side-angle-side congruency. OC must be equal to OB. And we know if this is a right angle, this is also a right angle. We know that since O sits on AB’s perpendicular bisector, we know that the distance from O to B is going to be the same as the distance from O to A.
This is my B, and let’s throw out some point. It’s at a right angle. Now, let’s go the other way around. So let me pick an arbitrary point on this perpendicular bisector. So let’s just drop an altitude right over here. This is what we’re going to start off with.
So our circle would look something like this, my best attempt to draw it. So let’s say that’s a triangle of some ciecumcenter. This one might be a little bit better. Although we’re really not dropping it. So these two things must be congruent. Then you have an angle in between that corresponds to this angle over here, angle AMC corresponds to angle BMC, and they’re both 90 degrees, so they’re congruent.
We have a leg, and we have a hypotenuse. Now, let me take this point right over here, which is the midpoint of A and B and draw the perpendicular bisector. You could call this triangle ABC. Or another way to think of it, we’ve shown that the perpendicular bisectors, or the three sides, intersect at a unique point that is equidistant from the vertices.
So to prove that C lies on the circumecnter bisector, we really have to show that CM is a segment on the perpendicular bisector, and the way we’ve constructed it, it is already perpendicular. So that’s point A.
workssheet This length must be the same as this length right over there, and so we’ve proven circumcenrer we want to prove. If we construct a circle that has a center at O and whose radius is this orange distance, whose radius is any of these distances over here, we’ll have a circle that goes through all of the vertices of our triangle centered at O.
So it will be both perpendicular and it will split the segment in two.
This is going to be B. And what’s neat about this simple little proof that we’ve set up in this video is we’ve shown that there’s a unique point in this triangle that is equidistant from all of the vertices of the triangle and it sits on the perpendicular bisectors of the three sides. And essentially, if we can prove that CA is equal to CB, then we’ve proven what we want to prove, that C is an equal distance from A as it is from B.
So this distance is going to be equal to this distance, and it’s going to be perpendicular. And now there’s some interesting properties of point O.