1. ## Pythagorean Word Problem

Hey guys,

Here's another word problem I am having trouble with. This section has to deal with the converse of/and pythagorean theorem.

I don't know how to set this problem up.

I tried setting the problem up in pythagorean formula. But failed

2. ## Re: Pythagorean Word Problem

Hi Cake,
I am going to try and explain without a picture. If you don't understand let me know. Try to draw it as i explain.
I know there is a lot of words here but try really hard to figure it out before you ask for more help.

FIRST
I'm just going to think of one side of the draw bridge.
It is 104 divided by 2 = 52 ft long.
As the bridge opens it pivots at the hinge end and its length doesn't change so it is opening in an arc. (A part of a circle with the hinge as the centre and the drawbridge is the radius)

SECOND Now I am thinking of both sides.
Now think of the open bridge as an isosceles trapezium. (this is like an isosceles triangle with the top cut off parallel to the base)
The base is the closed bridge and it is 104 feet long,
the top is the space between the two halves of the open draw bridge. It is 8 feet.
The height of the trapezium is the distance that you are trying to find.
Draw in the 2 heights, one for each draw bridge. These are at right angles to the closed bridge.

Now you have broken the trapezium up into two identical triangles (only facing in different directions) at the ends and a rectangle in the middle.
The height of the triangles is the distance that you are trying to find.
The closed bridge = the length of the rectangle + the base of the first triangle + the base of the second triangle = 104feet
The middle bit is 8 feet because it is the same as the top.
so 8 + 2 times the base of the triangle =104feet
so 2 times the base of the triangle =96feet
so The base of each right angled end triangle is 48feet (half of 96)

So now you have 2 sides of a right angled triangle and you can use pythagoras to find the other side.
If you have followed what I have said then you will know whether you are finding the hypotenuse or a short side.

3. ## Re: Pythagorean Word Problem

It's actually a bit easier than that. Consider just one half of the bridge. The length of a bridge section is 52 feet, which when raised forms the hypotenuse of a right triangle. The base of the triangle is 52 feet minus 4 feet (4 because one half of the 8-foot gap is attributed to each side), or 48 feet. Now use Pythagorus to claculate the height of the triangle.

$\displaystyle h = \sqrt{52^2 - 48^2} = 20$

.

4. ## Re: Pythagorean Word Problem

Got it guys,
thanks, appreciate it a lot.

5. ## Re: Pythagorean Word Problem

That's a much better explanation. I must have been giving Cake a lesson in following directions! lol
I know I was giving me a lesson on writing directions.

6. ## Re: Pythagorean Word Problem

Hi cake,
I strongly recommend drawing a picture for geometry problems. I don't mean a drawing by hand, but a drawing produced by some dynamic geometry program. (So I'm saying you need to learn how to use such a program.) When you construct the diagram with dynamic geometry, very often the act of construction will lead to a solution of the problem. Of course, this isn't a magic bullet. It doesn't always work. The second attachment refers to thread common point that I can't solve, but at least I have a good diagram.