z twierdzenia Pitagorasa mamy
(\frac{1}{2}e)^2+(\frac{1}{2}f)^2=4^2
\frac{1}{2}e=x
\frac{1}{2}f=y
x^2+y^2=4^2
2x+2y=10
…
x^2+y^2=4^2
x+y=5
…
x^2+y^2=16
x=5-y
…
(5-y)^2+y^2=16
x=5-y
…
5^2-10y+y^2+y^2=16
x=5-y
…
2y^2-10y+25-16=0
x=5-y
…
2y^2-10y+9=0
x=5-y
…
2y^2-10y+9=0
\Delta=b^2-4ac=100-4*2*9=100-72=28
\sqrt{28}=\sqrt{4*7}=2\sqrt{7}
y_1=\frac{-b-\sqrt{\Delta}}{2a}=\frac{10-2\sqrt{7}}{4}=2(\frac{5-\sqrt{7})}{4}=\frac{5-\sqrt{7}}{2}
y_2=\frac{-b+\sqrt{\Delta}}{2a}=\frac{10+2\sqrt{7}}{4}=2(\frac{5+\sqrt{7})}{4}=\frac{5+\sqrt{7}}{2}
czyli przekątne
e = 10 -f
e_1 = 10 - f_1
e_1 = 10 -(5-\sqrt{7}) = 10 -5 +\sqrt{7} = 5 +\sqrt{7}
e_2 = 10 -f_2
e_2 = 10 -(5+\sqrt{7}) = 10 -5 -\sqrt{7} = 5 -\sqrt{7}
Ponieważ wartości liczbowe obu przekątnych są zamiennie jednakowe to
e =5 -\sqrt{7}cm
f =5 + \sqrt{7}cm
P= 1/2*e*f
P = 1/2*( 5-\sqrt{7})(5 + \sqrt{7})
P =\frac{1}{2}*(5^2 - \sqrt{7}^2)
P =\frac{1}{2}* (25 - 7)
P =\frac{1}{2}*18
P = 9 cm^2
P = 9
P =a*h
a*h = 9
4*h =9
h = 9: 4
h = 2,25 cm