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=
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