# Homework Help: Doing math the wrong way(TM)

1. May 16, 2012

### KiwiKid

Ok, so this was the simple exercise problem I was trying to solve:
"One of the legs of a right triangle has length 4. Express the length of the altitude perpendicular to the hypothenuse as a function of the length of the hypothenuse."

Let's assume for a moment that h = the length of the hypothenuse, a = the length of the altitude perpendicular, x = 4, one of the legs of the triangle, and y is unknown.

Because I didn't know any corners, I was trying very hard to find the solution by doing things with only the Pythagorean theorem. This turned out not to work, so I tried the law of sines, the law of cosines, and what have you. None of it worked.

In the end, the solution turned out to be this: the area of the triangle A = a*h/2 = 4y/2. Therefore, a = 4y/h. Use the Pythagorean theorem to substitute y for h, and then we get a = 4*sqrt(h^2-16)/h. Easy as pie.

So why do I post this? Because it was so easy, and I was unable to solve it without looking at the solutions manual. The reason for this is simple: I totally forgot about the area formula and was too focused on solving things with the Pythagorean theorem and sines and cosines. How do you avoid these kind of mistakes? Is it a simple matter of doing lots of these problems, and in time you will make less of these stupid mistakes, or do you use a certain method to make sure you're using all the 'tools' you have?

2. May 16, 2012

### micromass

The problem IS possible to solve with only Pythagoras' theorem.

3. May 16, 2012

### KiwiKid

*blinks* Ok. Now I feel even more stupid. :P

4. May 16, 2012

### micromass

Don't feel stupid. It's pretty tricky.

It's easy to see in a picture, but I'll describe it mathematically:

Take a right triangle A,B,C where C is the hypothenuse. We know |A|=4. We wish to express everything with |C|. By Pythagoras theorem, you can find an expression for |B| in function of |C|.

Take the altitude H. This divides the hypothenus in two parts E and D. We know that |D|=|C|-|E|.
Apply Pythagoras theorem on the triangle A,H,E and B,H,D.
You can express |H|² thus twice in function of |A| and |E| and again in function of |B| and |D|. Equating everything eventually gives you that you can express everything in function of |C|.

OK, this was a horrible explanation. Try to do as much as you can, if you get stuck, then I'll guide you further.

5. May 16, 2012

### HallsofIvy

I would be inclined to use "similar triangles". The altitude from the right angle to the hypotenuse divides the right triangle into two smaller triangles, both of which are similar to the original triangle.

6. May 16, 2012

### micromass

Aah yes, that's also a nice thing to do.

Kiwi, as you see, there are multiple solutions to this problem. The solution using areas that you provided is very nice. So don't feel bad about it.

7. May 18, 2012

### Alshia

Pardon, but can someone explain to me why the area of triangle is equal to:

A = (1/2) x hypothenuse x altitude

I intuitively understand it....but I won't be able to prove it. I can represent it graphically though. The idea is the same as the area of trapezium where you 'cut' a rectangle in half in a certain way to get the area, correct?

Also, the altitude is also called the height right?

8. May 18, 2012

### Mentallic

Can you show why a right-angled triangle with sides a and b (not the hypotenuse) has an area of $\frac{1}{2}ab$? If you can do that, then you can answer your own question because it's basically the same idea except you'll have two triangles (the splits up the whole triangle into two smaller triangles).

9. May 18, 2012

### Alshia

The OP's post on the answer confused me because of this:

A = a*h/2

...where a is the hypothenuse. I am just verifying whether it would be the same. Yes I know the answer already.

10. May 18, 2012

### HallsofIvy

A triangle has three altitudes so this doesn't make sense until you have specified which altitude.

It is true in any triangle that the area is (1/2)bh where b is the length of any one side and h is altitude measured perpendicular to that side. IF the altitude you are talking about is measured perpendicular to the hypotenuse then your formula is just the general formula for area of a triangle. If it is one of the other two altitudes (which happen to be the legs of a right triangle) this is not true. Because two of the altitudes of a right triangle are the legs, it is also true that the area of a right triangle is given by A= (1/2)ab where a and b are the lengths of the legs.