Real life architectual problem

mikau · 2006-01-12 08:24:20

This was a real architectual problem my dad encountered, and it took him forever to solve. Lets see if anyone can get it:

Find A.

Last edited by mikau (2006-01-12 08:24:40)

MathsIsFun · 2006-01-12 11:14:34

Nice puzzle ... could be a worksheet

mikau · 2006-01-12 12:15:25

I don't even remember how to solve it. :-P

MathsIsFun · 2006-01-12 17:18:03

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

mikau · 2006-01-12 17:24:59

Step 2. Eat a banana, and think what to do in step 3.

Ricky · 2006-01-12 18:10:04

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'.

Last edited by Ricky (2006-01-12 18:17:49)

mikau · 2006-01-12 18:44:52

Forgive me if I'm wrong but I think you messed up. It apppears you scalled the inside triangle to a 12 by 6 triangle. You can't, they are not congruent:

Ricky · 2006-01-12 18:49:32

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.

mikau · 2006-01-12 18:54:59

Lol. Theres actually a trick to it. And a hint.

The hint is litterally this: "think outside the box".

Ricky · 2006-01-12 19:06:32

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.

mikau · 2006-01-12 19:09:28

This is achictecture, you can't measure it untill its either built or drawn, which requires you know the dimensions. (evil laughter)

MathsIsFun · 2006-01-12 21:04:17

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 ...

mikau · 2006-01-13 03:20:54

The two triangles are congruent to eachother. But they are obviously not congruent to the red triangle I drew. Ricky incorrectly scaled from this triangle.

MathsIsFun · 2006-01-13 10:00:27

OK, how about

mikau · 2006-01-13 11:04:45

(long low whistle) That is utterly genius! And may be an entirely different way of solving the problem!

Last edited by mikau (2006-01-13 11:06:38)

mikau · 2006-01-13 17:32:12

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.

mikau · 2006-01-13 18:06:45

Here we are! The solution to this nightmare of a problem is really quite simple, it just requires a little trick.

As you can see, the trick is to extend the beam out just enough till it touches the uper left corner and the required lengths and angles can be found as shown. Once you have these, the solution to the problem is easily attainable.

MathsIsFun · 2006-01-13 20:43:59

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!

mikau · 2006-01-14 11:10:26

Thats a good way to do it. :-)

Math Is Fun Forum

#1 2006-01-12 08:24:20

Real life architectual problem

#2 2006-01-12 11:14:34

Re: Real life architectual problem

#3 2006-01-12 12:15:25

Re: Real life architectual problem

#4 2006-01-12 17:18:03

Re: Real life architectual problem

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Re: Real life architectual problem

#6 2006-01-12 18:10:04

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#7 2006-01-12 18:44:52

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#8 2006-01-12 18:49:32

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#9 2006-01-12 18:54:59

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#10 2006-01-12 19:06:32

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#11 2006-01-12 19:09:28

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#12 2006-01-12 21:04:17

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#13 2006-01-13 03:20:54

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#14 2006-01-13 10:00:27

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