# Perimeter of a trapeze and triangle

### Perimeter of a trapeze and triangle

1. Around the circumference with radius r = 1cm is prescribed a trapeze ABCD where AD is perpendicular to AB and the angle ABC = 30$$^\circ$$. Find the perimeter of the trapeze.
2 Around a circumference with radius 3cm is prescribed a rectangular triangle with hypotenuse 15cm. Find the perimeter of the triangle.
kate

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### Re: Perimeter of a trapeze and triangle

I presume that by "around the circumference" you mean that four vertices all lie on the circumference of a circle. At first it wasn't at all clear to me that there could be such a trapezoid!

Here's how I analyzed the first problem:
Since AD is perpendicular to AB, we can set up a coordinate system with origin at A, x-axis along AB and y-axis along AD. Take B to be (a, 0) and D to be (0, b). A circle passing through A, B, and D must have center at (a/2, b/2). A circle with that center and radius 1 has equation $$(x- a/2)^2+ (y- b/2)^2= 1$$. Taking (x, y)= (0, 0), we must have $$(0- a/2)^2+ (0- b/2)^2= a^2/4+ b^2/4= 1$$ so $$a^2+ b^2= 4$$. You can check that (a,0) and (0,b) lead to that same equation.

Dropping a line from the fourth point perpendicular to the x-axis gives a right triangle with height b and angle opposite that side 30 degrees. Since tangent is "rise over run", and tan(30)= 1/2, letting "u" be the length of the side of that triangle along the x-axis we have tan(30)= 1/2= b/u so u= 2b. The fourth vertex has coordinates (a- 2b, b). Does that point satisfy the equation of the circle? (Does it lie on the circle?)

$$(a- 2b- a/2)^2+ (b- b/2)^2= (a/2- 2b)^2+ (b/2)^2= a^2/4- 2ab+ 4b^2+ b^2/4$$. Since $$a^2/4+ b^2/4= 1$$ that reduces to $$1- 2ab+ 4b^2= 1$$ or $$4b^2- 2ab= 2b(2b- a)= 0$$. b cannot be 0 so we must have a= 2b. Then $$a^2+ b^2= 4b^2+ b^2= 5b^2= 4$$. $$b= \frac{\sqrt{5}}{2}$$ and $$a= \sqrt{5}$$

The fourth vertex, then, is at $$(a- 2b, b)= (\sqrt{5}- \sqrt{5}, \sqrt{5}/2)= (0, \sqrt{5}/2)$$.

No, there is no such trapezoid because that "fourth vertex", C, lies on the same line as A and D.
Guest

### Re: Perimeter of a trapeze and triangle

For the second problem, "2 Around a circumference with radius 3cm is prescribed a rectangular triangle with hypotenuse 15cm. Find the perimeter of the triangle."
Three points always define a triangle so that should be possible.

Starting in the same way, set up a coordinate system so the right angle is at (0, 0), the two legs lie along the x and y axes with the other two vertices at (a, 0) and (0, b). We can take one vertex to be (a, 0) and the other to be (0, b). Since the right triangle has hypotenuse of length 15, we have $$a^2+ b^2= 15^2= 225$$. Since the circle in inscribed in the triangle its center is at (a/2, b/2) and we must have $$(x- a/2)^2+ (y- b/2)^2= 9$$. Taking (x, y)= (0, 0) that gives $$a^2/4+ b^2/= 9$$ or $$a^2+ b^2= 36$$.

The equations, $$a^2+ b^2= 225$$ and $$a^2+ b^2= 36$$ can't both be true so this is also impossible.
Guest

### Re: Perimeter of a trapeze and triangle

I presume that by "around the circumference" you mean that four vertices all lie on the circumference of a circle. At first it wasn't at all clear to me that there could be such a trapezoid!

I wouldn't. If the vertices are on the circumference of the circle, then the trapezoid itself is inside the circle, not "around" it. I would interpret this as meaning that the sides of the trapezoid are tangent to the circle.

Since I have always preferred "algebra" to "geometry" I would set up a coordinate system so that the center of the circles is at (1, 1) (the radius of the circle is 1 cm so this circle is tangent to the two axes). A is the origin, AB is along the x-axis and AC is along the y-axis. The total interior angles of any four sided polygon is 360 degrees, one angle is 90 degrees, and another is 30 degrees, the other two must total 360- 90- 30= 240 degrees. A trapezoid has two sides parallel so we can take the angle the "top" side makes with the y-axis to be 90 which makes the line y= 2. The fourth angle, then is 240- 90= 150 degrees. Taking the fourth vertex to be at $$(x_0, 0)$$ for some $$x_0$$, the fourth side has equation $$y= tran(150)(x- x_0)==\frac{\sqrt{3}}{3}(x- x_0)$$. We need that to be tangent to the circle $$(x- 1)^2+ (y- 1)^2= 1$$. Setting $$y= \frac{\sqrt{3}}{3}(x- x_0)$$, we must have $$(x- 1)^2+ (\frac{\sqrt{3}}{3}(x- x_0)- 1)^2= 1$$. In order that the line be tangent to the circle, not only must x satisfy that equation, it must be a double root: $$x_0$$ must be suchy that the left side is a perfect square.
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