Nice puzzle ... could be a worksheet

Step 1: reduce bounding rectangle by 2 × 2', to become 6' by 12'

Triangle on the left is triangle A, triangle on the right is triangle B.  Angle x is at the top of triangle A and angle y is at the bottom right of triangle A.  I have to assume both triangles are  similar, so angle x is also at the bottom of triangle B and angle y is at the top left of triangle B.

Draw a line from the point of angle y in triangle B to the hypotenuse of triangle A, such that it hits the hypotenuse at a 90 degree angle.  Then, draw a line which connects the very top of triangle A to that of B.  Call this new triangle, triangle C.

The bottom most angle of triangle C is 90 degrees.  Since the hypotenuses in triangles A and B are parallel (assumed), the top left angle in triangle C is equal to angle y in triangle B.  Now add the top right angle of triangle C, the 90 degree angle just to the right of it, and angle y in triangle B.  Since all these angles put together form a line, they add up to 180 degrees.  But we also know that x + y + 90 = 180.  Thus, the top right angle in C is equal to angle x.  Therefore, triangle C is similar to triangle B and A (assuming these two are congruent).

However, we know that the side of triangle C is 2'.  Since the triangles are similar: 6 / 12 = 2 / x, 6x = 24, x = 4.  So the other side of triangle C is 4'.  This means the hypotenuse of triangle C is about 4.47'.

The length of the entire bottom is 12 feet, 12 - 4.47 = 7.53'.

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Last edited by Ricky (2006-01-13 17:17:49)

Then I give up.  I'm not quite sure if it is possible since you don't know any of the angles.

Of course, if your father measured one single angle, he would have saved me 30 minutes of work.

Ok, here's me thinking outside the box.

Take a protractor, measure the angle of the triangle, use this and tan to find A.

Problem solved.

I am confused ... are you saying the two triangles are not congruent?

If they are congruent, then we need to find the gap "B" (=12-A)

And the gap "B" will be a function of the angle θ

And θ will be a function of "B"

Round and round we go ...

OK, how about

Well I"ve been thinking about it, but unless I'm mistaken, it appears that only produces another unknown dimension, so that doesn't help.

I shall reveal the solution to the problem momentarily.

You are right, my new method is not as good as it looks

But I can't help but think that a there may be some obvious way to lay it out.

I know - give the job to a carpenter!

Let me see ... they would mark lines 2' in from the outer border, and form a rectangle. Then they would probably draw a circle around one corner 2' in radius, then put a nail in the other corner, attach a string, and run the string so it was tangent to the circle at the first corner!

So, it would look a bit like your solution, then!

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