Question 1169125

Length ={{{L}}}
Width ={{{W}}}
perimeter=>{{{P=2(L+W)}}}
area=> {{{A=LW}}}

A rectangle has {{{perimeter}}} {{{56}}} feet we have

{{{56ft=2(L+W)}}}...solve for {{{L}}}
{{{56ft/2=L+W}}}
{{{28ft-W=L}}}....eq.1


and it’s area is {{{180}}} sq feet find the dimensions of this rectangle, we have

{{{180=LW}}}..solve for {{{L}}}
{{{180/W=L}}}...........eq.2

from eq.1 and eq.2 we have

{{{28ft-W=180/W}}}....solve for {{{W}}}
{{{28W-W^2=180}}}
{{{0=W^2-28W+180}}}....factor
{{{0=W^2-10W-18W+180}}}
{{{0=(W^2-10W)-(18W-180)}}}
{{{0=W(W-10)-18(W-10)}}
{{{0 = (W - 18) (W - 10)}}}

=>{{{W=10ft}}} or {{{W=18ft}}}

{{{28ft-W=L}}}....eq.1, substitute {{{W=10ft}}}
{{{28ft-10ft=L}}}
{{{L=18ft}}}

Length ={{{18ft}}}
Width ={{{10ft}}}