Question 1195773

a diagonal {{{d=13}}} centimeters 

if the length of the rectangle {{{L}}} is to be {{{3}}} centimeters more than {{{triple }}} the width {{{W}}}, we have
 
{{{L=3W+3}}}.....eq1

using Pythagorean theorem we know that

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

substituting {{{d=13}}} and {{{L=3W+3}}} we have

{{{13^2=(3W+3)^2+W^2}}} 

{{{169=9W^2 + 18W + 9+W^2}}} 

{{{10W^2 + 18W + 9-169=0}}} 

{{{10W^2 + 18W -160=0}}} .....factor

{{{2 (W + 5) (5 W - 16) = 0}}} 

solutions;

{{{W + 5=0}}}  -> {{{W =- 5}}}  -> disregard negative solution

{{{(5 W - 16) = 0}}}  ->{{{W =16/5}}}->exact solution

now find the length:

{{{L=3(16/5)+3}}}.....eq1

{{{L=63/5}}}->exact solution


 dimensions of the rectangle : {{{ 63/5}}} centimeter by {{{ 16/5}}}centimeter