"Parameter" = Perimeter.
Look at the picture.
We have the perimeter = 50 m.
The perimeter is 2l + 2w (l - length, w - width). Therefore
2l + 2w = 50 divide both sides by 2
l + w = 25 subtract w from both sides
l = 25 - w.
Use the Pythagorean theorem:
[tex]l^2+w^2=20^2\to(25-w)^2+w^2=20^2[/tex]
Use (a - b)² = a² - 2ab + b²
[tex]25^2-2(25)(w)+w^2+w^2=400\\\\625-50w+2w^2=400\qquad\text{subtract 400 from both sides}\\\\225-50w+2w^2=0\\\\2w^2-50w+225=0[/tex]
Use quadratic formula:
[tex]ax^2+bx+c=0\\\\\Delta=b^2-4ac\\\\x_1=\dfrac{-b-\sqrt\Delta}{2a};\ x_2=\dfrac{-b+\sqrt\Delta}{2a}[/tex]
We have:
[tex]a=2,\ b=-50,\ c=225[/tex]
Substitute:
[tex]\Delta=(-50)^2-4(2)(225)=2500-1000=1500\\\\\sqrt\Delta=\sqrt{1500}=\sqrt{100\cdot15}=\sqrt{100}\cdot\sqrt{15}=10\sqrt{15}\\\\w_1=\dfrac{-(-50)-10\sqrt{15}}{2(2)}=\dfrac{50-10\sqrt{15}}{4}=\dfrac{25-5\sqrt{15}}{2}\\\\w_2=\dfrac{-(-50)+10\sqrt{15}}{2(2)}=\dfrac{50+10\sqrt{15}}{4}=\dfrac{25+5\sqrt{15}}{2}[/tex]
[tex]l_1=25-w_1\\\\l_1=25-\dfrac{25-5\sqrt{15}}{2}=\dfrac{50}{2}-\dfrac{25-5\sqrt{15}}{2}=\dfrac{50-25+5\sqrt{15}}{2}=\dfrac{25+5\sqrt{15}}{2}\\\\l_2=25-w_2\\\\l_2=25-\dfrac{25+5\sqrt{15}}{2}=\dfrac{50}{2}-\dfrac{25+5\sqrt{15}}{2}=\dfrac{50-25-5\sqrt{15}}{2}=\dfrac{25-5\sqrt{15}}{2}[/tex]
[tex]Answer:\ \boxed{\dfrac{25+5\sqrt{15}}{2}\ m\times\dfrac{25-5\sqrt{15}}{2}\ m}[/tex]
