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 ."
The triangle has the usual three sides
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)
(You can put in a value of 17 for a if you want)
I think MathsIsFun summed up that solution perfectly:
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.
Why did the vector cross the road?
It wanted to be normal.
do we need to say about the domain?
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
h = (2a² + p² - 2ap) / 2(p-a) = (p^2 - (2)(17)(p) + (2)17^2)/((2)(p - 17)) when a=17