Question 1125600


let dimensions be {{{L}}} and {{{W}}}


given:

The area of a rectangle is {{{45}}}, so we have 

{{{L*W=45}}}....solve or {{{W}}}

{{{W=45/L}}}......eq.1



its perimeter is {{{28}}}, so we have

 {{{2(L+W)=28}}}

{{{L+W=14}}} ...solve or {{{W}}}

{{{W=14-L}}}...eq.2


from eq.1 and eq.2 we have


{{{45/L=14-L}}}...solve or {{{L}}}

{{{45=14L-L*L}}}

{{{45=14L-L^2}}}

{{{L^2-14L+45=0}}}

{{{(L - 9) (L - 5) = 0}}}

solutions:

{{{(L - 9)  = 0}}}-> {{{L=9}}}
or
{{{(L - 5)  = 0}}}-> {{{L=5}}}


go to eq.2, substitute {{{L=9}}}

{{{W=14-L}}}...eq.2

{{{W=14-9}}}

{{{W=5}}}


so, we will choose the length to be {{{highlight(L=9)}}} and the width to be {{{highlight(W=5)}}}-> dimensions


and we can ind diagonal {{{d}}} using Pythagorean theorem since diagonal cuts rectangle into two right angle triangles where diagonal is actually hypotenuse


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

{{{d^2=9^2+5^2}}}

{{{d^2=81+25}}}

{{{d^2=106}}}

{{{d=sqrt(106)}}}

{{{d=10.295630140987}}}

approximately

{{{highlight(d=10.3)}}}