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.

Author: Tezilkree Menris
Country: Guyana
Language: English (Spanish)
Genre: Video
Published (Last): 4 December 2007
Pages: 491
PDF File Size: 11.35 Mb
ePub File Size: 17.42 Mb
ISBN: 240-7-25757-913-3
Downloads: 82642
Price: Free* [*Free Regsitration Required]
Uploader: Shakajar

And so you can construct this line so it is at a right angle with AB, and let me call this the point at which it intersects M.

Obviously, any segment is going to be equal to itself. Now, let me just construct the perpendicular bisector of segment AB.

I’ll try to draw it fairly large. So just to review, we found, hey if any point sits on a perpendicular bisector of a segment, it’s equidistant from the endpoints of a segment, and we went the other way.

But we also know that because of the intersection of this green perpendicular bisector and this yellow perpendicular bisector, we also know because it sits on the perpendicular bisector of AC that it’s equidistant from A as it is to C. So this line MC really is on the perpendicular bisector.

Worksheet – Construct the circumcenter of a triangle –

So that tells us that AM must be equal to BM because they’re their corresponding sides. Just for fun, let’s call that point O.

If any point is equidistant from the endpoints of a segment, it sits on the perpendicular bisector of that segment. And it will be perpendicular. Now this circle, because it goes through all of the vertices of our triangle, we say that it is circumscribed about the triangle. If this is a right angle here, this one clearly has to be the way we constructed it.


Let’s prove that it has to sit on the perpendicular bisector. And so you can imagine we like to draw a triangle, so let’s draw a triangle where we draw a line from C to A and then another one from C to B. So this is C, and we’re going to start with the assumption that C is equidistant from A and B. So that’s fair enough. This is point B right over here. Video transcript Let’s start off with segment AB. So we also know that OC must be equal to OB. Well, if a point is equidistant from two other points that sit on either end of a segment, then that point must sit on the perpendicular bisector of that segment.

Three points defining a circle. OA is equal to OB.

To log in and use all the features of Khan Academy, circumcenetr enable JavaScript in your browser. We really just have to show that it bisects AB.

Let’s say that we find some point that is equidistant from A and B. So thus we could call that line l. This is going to be B. We have a leg, and we have a hypotenuse. And let circmucenter do the same thing for segment AC right over here. We’re kind of lifting an altitude in this case.

Circumcenter Worksheet

This is my B, and let’s throw out some point. So let’s say that C right over here, and maybe I’ll draw a C right down here. So we know that OA is equal to OC. And once again, we know we can construct it because there’s a point here, and it is centered at O.


This length must be the same as this length right over there, and so we’ve proven what we want to prove.

That’s that second proof that circumcebter did right over here. That’s point A, point B, and point C. And now there’s some interesting properties of point O. So it’s going to bisect it. This is what we’re going to start off with. So we’ve drawn a triangle here, and we’ve done this before.

Worksheet – Construct the circumcenter of a triangle

So let’s call that arbitrary point C. Let me give ourselves some labels to this triangle. So this is my A. Now, let’s go the other way around.

And this unique point on a triangle has a special name. So to prove that C lies on the perpendicular bisector, circumcetner really have to show that CM is a segment wogksheet the perpendicular bisector, and the way we’ve constructed it, it is already perpendicular. And then you have the side MC that’s on both triangles, and those are congruent.

Well, that’s kind of neat.