d=a+2\sqrt3
Obl. Ob i P
d=a\sqrt2
a+2\sqrt3=a\sqrt2
a-a\sqrt2=-2\sqrt3
a(1-\sqrt2)=-2\sqrt3
a=\frac{-2\sqrt3}{1-\sqrt2}
a=\frac{-2\sqrt3}{1-\sqrt2}\cdot\frac{1+\sqrt2}{1+\sqrt2}
a=\frac{-2\sqrt3(1+\sqrt2)}{1-2}
a=\frac{-2\sqrt3(1+\sqrt2)}{-1}
a=2\sqrt3(1+\sqrt2)
Ob=4a=4\cdot 2\sqrt3(1+\sqrt2)=8\sqrt3(1+\sqrt2) \ [j]
\approx 33,5 \ j
P=[2\sqrt3(1+\sqrt2)]^2=(2\sqrt3)^2\cdot (1+2\sqrt2)^2=4\cdot 3(1+2\sqrt2+2)=12(3+2\sqrt2) \ [j^2]
\approx 69,9\ j^2
Odpowiedź:
Obwód kwadratu jest równy 8\sqrt3(1+\sqrt2) \ j, a pole 12(3+2\sqrt2) \ j^2.