Question 1133678


The diagonal of a rectangle field is {{{y^2+16}}}cm and one side is {{{8y}}} cm. For what value of y the perimeter of the rectangle will be {{{62 }}}cm

To find the width, multiply the length that you have been given by{{{ 2}}}, and subtract the result from the perimeter. You now have the total length for the remaining {{{2}}} sides. This number divided by {{{2}}} is the width.

{{{2W= 62-2L}}}

{{{2W= 62-2*8y}}}

{{{W= 62/2-2*8y/2}}}

{{{W= 31-8y}}}

diagonal cuts rectangle into two right triangles, so if

{{{d=y^2+16}}}, {{{L=8y}}}, and {{{W= 31-8y}}} we have 

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

{{{d=sqrt(L^2+W^2)}}}

{{{y^2+16=sqrt((8y)^2+(31-8y)^2)}}}

{{{y^2+16=sqrt(64y^2+64y^2 - 496 y + 961)}}}

{{{y^2+16=sqrt(128y^2 - 496 y + 961)}}}

{{{(y^2+16)^2=(sqrt(128y^2 - 496 y + 961))^2}}}

{{{y^4 + 32 y^2 + 256= 128y^2 - 496 y + 961 }}}

{{{128y^2-y^4  - 496 y- 32 y^2+ 961 -256=0}}}

{{{-y^4 + 96 y^2 - 496 y + 705 = 0}}}

{{{-(y - 5) (y - 3) (y (y + 8) - 47) = 0}}}


if {{{-(y - 5)=0 }}}=>{{{-y+5=0}}}=>{{{5=y}}}....one solution to use

if {{{(y - 3)=0}}}=>{{{y=3}}}....another solution to use

{{{(y (y + 8) - 47) = 0}}}=>{{{y^2 + 8y - 47= 0}}}=>{{{y = -4 - 3 sqrt(7)}}} , {{{y = 3 sqrt(7) - 4}}}.....disregard these solutions

go with:

=>{{{y=5}}}
{{{L=8*5}}} 
{{{L=40cm}}} 
and
 {{{W= 31-8*5}}}
{{{W= 31-40}}}
{{{W= -9cm}}}=> disregard negative solution


go with:

=>{{{y=3}}}
{{{L=8*3}}} 
{{{L=24cm}}} 
and
 {{{W= 31-8*3}}}
{{{W= 31-24}}}
{{{W= 7cm}}}

check the perimeter:

{{{2*24cm+2*7cm=62cm}}}
{{{48cm+14cm=62cm}}}
{{{62cm=62cm}}}

diagonal:
{{{d=y^2+16}}}
{{{d=3^2+16}}}
{{{d=25}}}

{{{d=sqrt(L^2+W^2)}}}
{{{25=sqrt(24^2+7^2)}}}
{{{25=sqrt(625)}}}
{{{25=25}}}


so, your answer is: I. {{{3cm}}}