Question 1006429


let the length be {{{L}}}, the width {{{W}}}, and the diagonal {{{d}}}
if the length of a rectangle is {{{5}}}in longer than its width, we have
 {{{L=W+5}}}
if the diagonal is {{{5}}} in. shorter than twice the width, we have 

{{{d=2W-5}}}

we know that {{{d^2=L^2+W^2}}}...substitute {{{d}}} and {{{L}}} from above

{{{(2W-5)^2=(W+5)^2+W^2}}}........solve for {{{W}}}

{{{4W^2-20W+25=W^2+10W+25+W^2}}}

{{{4W^2-20W+25=2W^2+10W+25}}}

{{{4W^2-20W+25-2W^2-10W-25=0}}}

{{{2W^2-30W=0}}}....simplify, divide by {{{2}}}

{{{W^2-15W=0}}}

{{{(W-15)W=0}}}

solutions:

{{{W=0}}}....since {{{W}}} represents the width, disregard this solution
{{{highlight(W=15)}}}=> your solution

now find the length
{{{L=15+5}}}

{{{highlight(L=20)}}}

and diagonal

{{{d=2*15-5}}}

{{{d=30-5}}}

{{{highlight(d=25)}}}

check the result:


{{{d^2=L^2+W^2}}}

{{{25^2=20^2+15^2}}}

{{{625=400+225}}}

{{{625=625}}}