Question 1137122

let  rectangle’s base be {{{b}}} and height {{{h}}}

if a rectangle’s base is {{{7cm}}} longer than its height, we have 

 {{{b=h+7cm}}}........eq.1

 
if the rectangle’s area is {{{144cm^2}}}, we have

 {{{b*h=144cm^2}}}...substitute {{{b}}} from eq.1

{{{(h+7cm)*h=144cm^2}}}

{{{h^2+7cm*h=144cm^2}}}

{{{h^2+7cm*h-144cm^2=0}}}....factor completely

{{{h^2+16cm*h-9cm*h-144cm^2=0}}}

{{{(h^2+16cm*h)-(9cm*h+144cm^2)=0}}}

{{{h(h+16cm)-9cm(h+16cm)=0}}}

{{{(h - 9cm) (h + 16cm) = 0}}}

solutions: {{{h=9cm}}} or {{{h=-16cm}}}->disregard negative solution or height

so,{{{h=9cm}}}

go to {{{b=h+7cm}}}........eq.1, substitute {{{h}}}

{{{b=9cm+7cm}}}

{{{b=16cm}}}


answer: rectangle’s base is {{{16cm}}} and height is {{{9cm}}}