- #1
synkk
- 216
- 0
A Right-Angled Triangle has area Acm^2 and perimeter Pcm. A side other than the hypotenuse has length has length xcm. Form a quadratic equation in x in each of the following cases:
a) a=6 p=12
let the other side be y, and the hypotenuse be h
x + y + h = 12
0.5*y*x = 6, y= 12/x
x + 12/x + h = 12
x^2 + x(h-12) + 12 = 0
is there anyway to find an expression for h while keeping a quadratic equation in x? I mean I could use pythagoras but h^2 = 144/x^2 + x^2 will turn into a quartic equation no?
a) a=6 p=12
let the other side be y, and the hypotenuse be h
x + y + h = 12
0.5*y*x = 6, y= 12/x
x + 12/x + h = 12
x^2 + x(h-12) + 12 = 0
is there anyway to find an expression for h while keeping a quadratic equation in x? I mean I could use pythagoras but h^2 = 144/x^2 + x^2 will turn into a quartic equation no?