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 2

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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Given five identical isosceles trapezoids in a 5-block arch, it is possible to find the angles of the trapezoid. The arch only makes up half of a polygon; the entire polygon is then a 10-sided polygon, a decagon. Recall the formula (n - 2)*180°/n is the measure of each angle of the polygon. Thus, the base angles of each trapezoid is half of the measure of the angle of the decagon (find this by substituting n = 10).

Recall that the upper angle of a trapezoid is supplementary to the base angle. So, subtract the measure of the base angle from 180° to find this angle.

A common application of isosceles trapezoids is the arch problem. In this problem I’m telling you that all trapezoids are congruent and isosceles, so I’m going to mark all of these sides as being congruent and we’re trying to find angles x and y. So to do this we need a game plan. So let’s take a look at a good game plan of how to find these angles.

Well the first step to is to realize that you only see half of a polygon here. That there is actually a whole bunch of sides down below because when you build an arch, you’re not going to build part of it underground, but you do have to count these as part of a whole polygon.

The reason why that’s important is that if you zoomed in on one of these intersections here, it would look something like this and you see that if we can find this interior angle of the polygon, then we can find our base angles x.

So the key thing here is to use our formula for finding the interior angle of the polygon which is 180 times n minus 2 divide by n so we can find this angle, then we can use the sum of these being 360 to find x, then we can use what we know about the angles in an isosceles trapezoid.

So let’s get started. The first step is to ask yourself well how many more sides are there below here that we didn’t see. Well there is 1, 2, 3, 4, 5 trapezoids which means this is part of a 10 sided figure or a decagon. So I’m going to write that n equals 10. Our formula we said was 180 times n minus 2 all divided by n. So that would give us the measure of one of those congruent since all of thee angles are congruent, one of those interior angles, so let’s substitute in 10.

So you have 180, 10 minus 2 is 8, I can do that in my head okay. So 180 times 8 divided 10, if I type that into my calculator here I’m going to see that it is 144 degrees. So I’m going to redraw this intersection just right down below here. So we’ve got, I’m going to mark this with my congruence marks so I remember that this right here is part of that trapezoid and we said that this angle here was the interior angle of a regular decagon which is 144 degrees and if I look at where x is, x is the base angle that’s on the part of the inside of the arch. So both of these have to be x since all of these trapezoids are congruent.

So the equation I can write here is that 360 degrees is equal to x plus x plus 144 degrees. If I solve this for x then I’ll know one of my base angles. So 360 is equal to 2x plus 144, so I’m going to subtract 144 from both sides and I see that 2x must be 360 minus 144 is 216 and if I divide that by 2, I see that, I think I have enough room here, that x must be 108 degrees. So in this diagram right here, I can write that both of these angles are 108 degrees.

Okay so I’m going to keep track of my answers right over here x equals 108 degrees. So if I drew in the rest of this trapezoid, I know that since it’s isosceles, this angle right here has to be 108 degrees. I know that consecutive angles between the bases have to be supplementary. So if I call this y since I see that y is one of these congruent top base angles, I can write that y plus the same side interior angle 108 must equal 180 degrees. So if I subtract 108 degrees, then I see that y must be 72 degrees. So I’m going to draw a box around y equals 72 degrees and x equals 108.

So the key to this problem if we walk back was coming up with a game plan. We said that if we find this interior angle using our interior angle sum formula that we can kind of work backwards find x and once we know x, we can find y because same side interior angles are always supplementary.

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