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  Discussion about math, puzzles, games and fun.   Useful symbols: √ ∞ ≠ ≤ ≥ ≈ ⇒ ∈ Δ θ ∴ ∑ ∫ π -




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Topic review (newest first)

2006-01-26 21:14:24

h = (2a +  p - 2ap) / 2(p-a) = (p^2 - (2)(17)(p) + (2)17^2)/((2)(p - 17)) when a=17 upup

2006-01-26 20:48:21

hrm . that looks real good . nope no domain needed .. well the online hw is overdued and i was able to see the answer.. supposely the correct answer.

(p^2 - (2)(17)(p) + (2)17^2)/((2)(p - 17))

any justifications? or should i just blame the online hw system for screwin me up haha


thx a lot by the way everyone . you guys are awesome

2006-01-26 09:46:08

do we need to say about the domain?

2006-01-26 00:14:56

I think MathsIsFun summed up that solution perfectly:

MathsIsFun wrote:


It's right, as far as I can tell, but it just annoys me when things can't be simplified and you have to leave them in a mess like that. The best alternative I could come up with is:

h = (p-a)/2 - a²/2(p-a)

But that's not really any simpler. Maybe a little bit.

2006-01-25 19:23:21

The triangle has the usual three sides smile

We know one of them is 17 cm, let's call that side "a"
Another side can be called "b"
And the hypotenuse is already called "h"

The perimeter is a+b+h

And we also know that a²+b²=h² (Pythagoras Theorem)

So, we have two formulas:

Lets start with Pythagoras:

And the perimeter formula can be used to find b
p=a+b+h ==> b = p-a-h

h² = a²+b² = a²+(p-a-h)² = a²+(p-a-h)²

It is looking like it is going to very complicated!

Expanding: (p-a-h)² = p² - 2ap - 2hp + 2ha +a² + h²

So:  h² = a² +  p² - 2ap - 2hp + 2ha +a² + h²

Simplifying: 0 = a² +  p² - 2ap - 2hp + 2ha +a²
Put "h" terms on left: 2hp - 2ha = a² +  p² - 2ap  +a²
Simplify: 2h(p-a) = 2a² +  p² - 2ap

And last: h = (2a² +  p² - 2ap) / 2(p-a)

up (I hope!)

(You can put in a value of 17 for a if you want)

2006-01-25 18:40:03

hey all. i was doin hw and ran into this problem. seems pretty impossible to me. any help would be nice. thx

"The altitude of a right triangle is 17 cm. Let h be the length of the hypotenuse and let p be the perimeter of the triangle. Express h as a function of p ."


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