[tex]|AB|=|BC|=|CD|=|AD|=x\\|KL|=|LM|=|MN|=|KN|=y\\\\y^2=(\frac45x)^2+(\frac15x)^2\\y^2=\frac{16}{25}x^2+\frac1{25}x^2\\\\y^2=\frac{17}{25}x^2\\\\P_1=x^2=100\%\\P_2=y^2=\frac{17}{25}x^2=\frac{68}{100}x^2=68\%[/tex]
Odp. Pole kwadratu KLMN stanowi 68% kwadratu ABCD