Kwadrat
a - dana przekątna
x - długość boku
z Tw. Pitagorasa obliczę długość boku
[tex]x^{2} +x^{2} =a^{2} \\\\2x^{2} =a^{2} ~~\mid \div 2\\\\x^{2} =\dfrac{a^{2} }{2} ~~\land ~~x > 0 ~~\Rightarrow ~~x=\dfrac{a}{\sqrt{2} } =\dfrac{a\sqrt{2} }{2} \\\\P_{kwadrat} =x^{2} ~~\land ~~x=\dfrac{a\sqrt{2} }{2}~~\Rightarrow ~~P_{kwadrat} =\dfrac{1}{2} a^{2}[/tex]
Prostokąt
a, 1/2a - długości boków
[tex]P_{prostokat} =a\cdot \dfrac{1}{2} a=\dfrac{1}{2} a^{2} \\\\Przekatna~~prostokata~~oblicze~~z~~Tw.~~Pitagorasa\\\\y^{2} =a^{2} +(\frac{1}{2} a)^{2} \\\\y^{2} =a^{2} +\dfrac{1}{4} a^{2} \\\\y^{2} =\dfrac{5}{4} a^{2} ~~\land~~y > 0~~\Rightarrow ~~y=\dfrac{\sqrt{5} }{2} a[/tex]
WNIOSKI:
[tex]P_{kwadrat} =P_{prostokat} =\dfrac{1}{2} a^{2} \\\\Odp_{1} :~~Pola~~maja~~rowne~~\Rightarrow~~P\\\\x=\dfrac{\sqrt{2} }{2} a~~\land~~\dfrac{\sqrt{2} }{2} \approx0,705 ~~\Rightarrow~~x\approx 0,705 a\\\\y=\dfrac{\sqrt{5} }{2} a~~\land~~\dfrac{\sqrt{5} }{2} \approx 1,12 ~~\Rightarrow~~y\approx 1,12a\\\\x < y\\\\Odp_{2}:~~Przekatna~~prostokata~~jest~~dluzsza~~od~~boku~~kwadrata~~\Rightarrow ~~F[/tex]
