Answer :
[tex]x,y-\ dimensions\ of\ rectangle\\\\\
Perimeter\\\\
P=2x+2y\\
P=26\\\\
26=2x+2y\ |:2\\
13=x+y\\
x=13-y\\\\\
Area:\\\\
A=x*y\\
A=40\\
40=x*y\\\\
Substituting\ x=13-y\\\\
40=(13-y)*y\\
40=13y-y^2\\y^2-13y+40=0\\
\Delta=13^2-4*1*40=169-160=9\\ \sqrt{\Delta}=\sqrt9=3\\\\
y=\frac{13-3}{2}=\frac{10}{2}=5\ \ or y=\frac{13+3}{2}=\frac{16}{2}=8\\x=13-y\\x=13-5=8\ or\ x=13-8=5\\\\Dimensions\ are\ 5m\ and\ 8m.
[/tex]
