 ###### Brian McCall

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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# Trapezoid Arch and Panel Design Problem - Problem 1

Brian McCall ###### Brian McCall

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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First examine the top isosceles triangle. In order to find the base angles, recall that the larger triangle is part of a 10-panel design. This means that the vertex angle of the larger triangle (also the vertex of the smaller top isosceles triangle) is 360°/10 = 36°. As a result of the triangle angle sum theorem, and properties of isosceles triangles, the base angles of this small triangle are each (180° - 36°)/2 = 72°. The angle on the other side of the top parallel line is supplementary to the base angle of the triangle, so it is 180° - 72° = 108°. Then, using what is known about trapezoid midsegments, it is possible to solve for the base angle of the small trapezoid within the larger triangle.

A common application of what you know about trapezoids and triangles is the panel design problem. In this problem you’re told that this design is part of a 10-panel total design, so the key is how are we going to find x and y. And I’m going to look at this top isosceles triangle, so I’m going to redraw that triangle over here just so I can mark it up.

Now I know that’s an isosceles because these two legs are labelled as congruent and I want to find these three angles here. Well the only piece of information that I know is that this is a 10-panel design, so this isosceles triangle is part of one of 10. So to find this vertex angle, I’m going to do, I’m going to write it down a little bit lower. I’m going to do 360 divided by 10, because all of these triangles are congruent, which means this vertex angle must be 36 degrees.

So to find my two base angles, I guess I could call those x, I can say that x plus x plus 36 must be 180. So I’m going to write that x plus x plus 36 degrees is 180 degrees, so this is just the triangle angle sum. So 2x I’m going to subtract 36, I’m going to do two steps in 1 here, so 2x is equal to 180 minus 36 is 144, so if I divide by 2, I see that x must be 72 degrees.

So in my original drawing, I’m going to label these two angles as 72 degrees. Now in my head when I saw this problem I said well working backwards if I want y and x, I first need to know these two base angles because 72 degrees and x form a linear pair which means they have to sum to 180 degrees, so I could write that x plus 72 equal 180. So I’m going to subtract 72 degrees from both sides of that equation and I’m going to say that x must be 108 degrees.

So I’m going to write that in, x is 108 degrees and I’m also going to try and find y by saying if this is 108 degrees, that corresponding angles, since I have lines marked parallel here, corresponding angles must be congruent. So this is 108 degrees and that these two must be supplementary since we have a trapezoid, so y is going to be 72 degrees.

So the key to this problem was dividing it up into two pieces; the first finding the angles in your isosceles triangle, the second using what you know about trapezoids to find missing angles.