Pole trojkata o podstawie x jest rowne x^2-xy. Jaka dlugosc ma wysokosc tego trojkata.
P=x²-xy
a=x
h=??
P=½*a*h
x²-xy=½*x*h|*2
x*h=2(x²-xy)|:x
h=
podstawa Δ =x
pole PΔ=x²-xy
wzor na pole Δ P=½xh
podstawiamy :
½xh=x²-xy
xh/2 =x²-xy /·2
xh=2x²-2xy
xh=2x(x-y)
h=2x(x-y)/x
h=2(x-y)
h=2x-2y
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P=x²-xy
a=x
h=??
P=½*a*h
x²-xy=½*x*h|*2
x*h=2(x²-xy)|:x
h=![h=\frac{2(x^2-xy)}{x}=\frac{2x^2-2xy}{x}=\frac{2x(x-y)}{x}=2x-2y h=\frac{2(x^2-xy)}{x}=\frac{2x^2-2xy}{x}=\frac{2x(x-y)}{x}=2x-2y](https://tex.z-dn.net/?f=h%3D%5Cfrac%7B2%28x%5E2-xy%29%7D%7Bx%7D%3D%5Cfrac%7B2x%5E2-2xy%7D%7Bx%7D%3D%5Cfrac%7B2x%28x-y%29%7D%7Bx%7D%3D2x-2y)
podstawa Δ =x
pole PΔ=x²-xy
wzor na pole Δ P=½xh
podstawiamy :
½xh=x²-xy
xh/2 =x²-xy /·2
xh=2x²-2xy
xh=2x(x-y)
h=2x(x-y)/x
h=2(x-y)
h=2x-2y