How to solve this triangle problem.

[feihong47]

Homework Statement

The altitude perpendicular to the hypotenuse of a right triangle is 12 cm. Express the length of the hypotenuse as a function of the perimeter.

Homework Equations

The Attempt at a Solution

So I have to express the hypotenuse (h) in terms of the perimeter. I tried to find the area of the triangle using two different equations

1) A = (1/2)*12*h = 6h
2) A = (1/2)*xy, let x and y be the other sides of the triangle

I can set them to each other, but I get 6h = 0.5xy

The perimeter (P) is x+y+h.

I have no idea how to express h in terms of P since now I have too many variables.

I tried substitution, and the best I got (without using x and y) was some complicated polynomial with P and h that I couldn't just simply solve for h.

Any tips are appreciated.

 

[ehild]

Think of Pythagoras' Theorem.

ehild

 

[feihong47]

Think of Pythagoras' Theorem.

ehild

I certainly did. But I still have to assign a variable to one of the sides, ie x, and then the other side would be \sqrt{h^{2} - x^{2}}

It doesn't help me in putting h in terms of P however, because now

P = x + h + \sqrt{h^{2} - x^{2}}

How can I express h only in terms of P?

 

[Curious3141]

I certainly did. But I still have to assign a variable to one of the sides, ie x, and then the other side would be \sqrt{h^{2} - x^{2}}

It doesn't help me in putting h in terms of P however, because now

P = x + h + \sqrt{h^{2} - x^{2}}

How can I express h only in terms of P?

You basically have to solve for x in terms of h.

Use x\sqrt{h^2 - x^2} = 12h, which you've already figured out.

Square both sides to get a quartic in x that can be reduced to a quadratic by a sub like z = x^2, then solve for x in terms of h.

The algebra can be a little messy, though.

 

[feihong47]

Ok. If I only have to express it in terms of x then I can do it. Glad to know that it's physically impossible to express it in P. Thanks!

 

[Curious3141]

Ok. If I only have to express it in terms of x then I can do it. Glad to know that it's physically impossible to express it in P. Thanks!

NO! You *solve* for x in terms of h, then replace all the x terms in P = x + h + \sqrt{h^2 - x^2} with this expression. Finally, rearrange/solve to get h in terms of P.

As I said, it's messy. But completely possible.

 

[feihong47]

LOL. I like your emphatic NO. But yea, that sounds very messy indeed... I see some x^{2}h^{2} - x^{4} = 144h^{2}... Just this alone makes me not want to do it...

 

[Curious3141]

LOL. I like your emphatic NO. But yea, that sounds very messy indeed... I see some x^{2}h^{2} - x^{4} = 144h^{2}... Just this alone makes me not want to do it...

I just did it. Yes, it's tedious. But once you work through the algebra, you get a nice simple expression for h in terms of P.

Some things to note. You'll get two solutions for the quadratic in x^2. Keep both of them. You'll find a nice symmetry in the expression for \sqrt{h^2 - x^2} when you use each of the values of x that'll simplify your work (so you don't have to do everything twice!). Also keep the identity (a + b)(a - b) = a^2 - b^2 in mind because you'll find that a great timesaver.

Other than that, you just have to roll up your sleeves and get your hands dirty. I can't give you any more help without giving you the complete solution.

(I was hoping there'd be a simpler way with similar triangles, but no such luck).

 

[ehild]

From Pythagoras' Theorem, x²+y²=h², and you know that xy=12h.
If you add 2xy to both sides to Pythagoras' equation you get the square of x+y on he left side:

(x+y)²=h²+2xy=h²+24h.
x+y=P-h. Substitute for x+y, expand, simplify. The solution is two more lines (h² cancels).

ehild

 

[ehild]

Also keep the identity (a + b)(a - b) = a^2 - b^2 in mind because you'll find that a great timesaver.

Also keep the identity (a+b)²=a²+b²+2ab
in mind because you'll find that a great timesaver. in problems where sum and product of two variables are involved.

ehild

 

[Curious3141]

Also keep the identity (a+b)²=a²+b²+2ab
in mind because you'll find that a great timesaver. in problems where sum and product of two variables are involved.

ehild

Well, I thought that one went without saying.

 

[Curious3141]

From Pythagoras' Theorem, x²+y²=h², and you know that xy=12h.
If you add 2xy to both sides to Pythagoras' equation you get the square of x+y on he left side:

(x+y)²=h²+2xy=h²+24h.
x+y=P-h. Substitute for x+y, expand, simplify. The solution is two more lines (h² cancels).

ehild

Wow, this is really, really neat!

Mine was like then , finally ending up .

But I'm I didn't see your way from the start. I blame the fever I'm having!

 

[feihong47]

Thanks guys, that was good!

 

[ehild]

But I'm I didn't see your way from the start. I blame the fever I'm having!

Get well soon!

ehild