Univ. of Wisconsin
J.D. Univ. of Wisconsin Law school
Brian was a geometry teacher through the Teach for America program and started the geometry program at his school
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Univ. of Wisconsin
J.D. Univ. of Wisconsin Law school
Brian was a geometry teacher through the Teach for America program and started the geometry program at his school
Given a triangle with two segments parallel to the third side that together divide each side of the triangle into three congruent segments, it is possible to find the lengths of the segments using what is known about trapezoid and triangle midsegments.
The segment closest to the third side of the triangle is the trapezoid midsegment of the trapezoid formed by the third side and the upper parallel segment. From this, you know that the length of this trapezoid midsegment is half the length of the sum of the bases (the parallel lines of the trapezoid).
From knowledge of triangle midsegments, the triangle midsegment is half the length of the side it is parallel to. The upper parallel segment is the triangle midsegment of the triangle formed by the sides of the larger triangle and the lower parallel line.
In this problem we have a triangle with two segments that are drawn in between it. But these segments aren’t the midsegments of the large triangle and why is that? Well because if I look at this point, it has divided these three pieces congruently, so it's divided this side into three congruent pieces, not the definition of a midsegment. But what I can do is say well I might have a triangle within a triangle here. So I’m going to redraw it right over here.
I’m going to look at just the top part of this drawing and I’m going to draw in that this side is parallel and that we have midpoints right here and right here. So I can use what I know about a triangle mid-segment to find my value of x.
So let’s see what we know about triangle midsegments. Well we said that it’s related to a trapezoid midsegment, but since this point up here a has no length, we’re going to substitute in 0 which means x is just going to be half of the third side. So the midsegment is half the length of the third side.
So let’s go back to our problem and apply that. Well if x is 12 we know that 12 is equal to 1/2 of x. So if I want to solve for x, I’m going to have to multiply by the reciprocal. The reciprocal of ½ is 2, so I’m going to multiply both sides by 2 here and 2 times 2 is 24, so we find that x must be 24, so I’m going to write that right up here and my units I see are centimetres.
Now the question is how do I find y since I don’t have a midsegment in this large triangle. Well what I can do is I can is I can say forget about this top part and if you ignore this top part, what type of shape do you see? Well you should see a trapezoid which I’m going to redraw down here. So we have a trapezoid and we have a midsegment which I know because we have two congruent pieces here and two congruent pieces on that other side. We have parallel bases and the mid segment is parallel to the two bases, so we know that we have 12 as our top base, 12cm we know what x is, x we said was 24 and now we have to find our what y is.
So if we apply what we know which is that the mid-segment in a trapezoid is the average of the two bases, whereas say that x which is your midsegment which is 24. So all I’m doing is substituting into this equation 24 is equal to 12 plus y, which we don’t know, and I’m going to have to divide that by 2.
So the key right now is realizing that if you ignore the top part that we have a trapezoid where we know 2 of the unknowns, so we can solve for the third. So to solve this for y, I’m going to need to undo dividing by 2, so I’m going to multiply both sides by 2, 2 times 24 is 48. These 2s are going to cancel out because 2 divided by 2 is 1, so we have 12 plus y.
The last step is to subtract 12 from both sides and 48 minus 12 is 36, so y equals 36. So if I wrote these out here x equals 24cm, y equals 36. It’s always a good step before you say that you have a final answer to make sure that it makes sense and we see that each of these midsegments are getting larger as we go down, so that our answer makes sense and the key thing here was realizing that we had a triangle with a mid-segment and a trapezoid with a mid-segment.
Unit
Polygons