Question

the perimeter of a rectangle is 34yd , and the area of the rectangle is 72yd^2 . Find the dimensions of the rectangle.

Answer 1

\[34=2l+2w\]
and

Solve the first equation for w (or l).
\[34-2l=2w\]
\[w=17-l\]
Now substitute this result into the Area equation.
\[72=(17-l)l\]
\[72=17l-l^2\]
\[l^2-17l+72=0\]
Now factor that.
\[(l-9)(l-8)=0\]
Thus l = 8 or l= 9 . Either way the dimensions of the rectangle was 8 x 9.
This would be the standard algebraic method of solving this equation.
One could have solved it by guess and check with the hope that the answer was integer values.

Answer 2

thank you