The area of a rectangular sign is 36 square meters. If the length of the sign is one meter more than twice the width, then what is the width of the sign?



Answer :

[tex]l-length\\w-width\\\\2w=l-1\ \ \ \ \ /:2\\\\w=\frac{l-1}{2}\\\\Area:A=w\times l;\ A=36\ m^2\\\\w\times l=36\\\\put\ w=\frac{l-1}{2}\ into\ an\ equation\ w\times l=36:\\\\\frac{l-1}{2}\times l=36\ \ \ \ \ \ /\times2[/tex]

[tex](l-1)\times l=72\\\\l^2-l-72=0\\\\l^2-9l+8l-72=0\\\\l(l-9)+8(l-9)=0\\\\(l-9)(l+8)=0\iff l-9=0\ or\ l+8=0\\\\l=9\ or\ l=-8 < 0\\\\l=9\ m\ then\ w=\frac{9-1}{2}=\frac{8}{2}=4\ (m)\\\\Answer:The\ width\ of\ the\ sign\ equal\ 4m.[/tex]

Other Questions