Question 891556
The diagonal of a rectangle is 
{{{L^2+W^2=26^2}}}
Also,
{{{L=W+14}}}
Substitute,
{{{(W+14)^2+W^2=26^2}}}
{{{W^2+28W+196+W^2=676}}}
{{{2W^2+28W-480=0}}}
{{{W^2+14W-240=0}}}
{{{(W-10)(W+24)=0}}}
Two solutions but a negative results doesn't make sense here.
{{{W-10=0}}}
{{{W=10}}}
Then,
{{{L=10+14}}}
{{{L=24}}}
So then,
{{{A=L*W}}}
{{{A=24*10}}}
{{{A=240}}}