Question

Length of a rectangle is 5 cm longer than the width. Four squares are constructed outside the rectangle such that each of the squares share one side with the rectangle. The total area of the constructed figure is 120 cm2. What is the perimeter of the rectangle? Thanks, lots of points!

Answer 1

The perimeter of rectangle is
`18\ cm`

Explanation:

Let

x------> the length of rectangle

y----> the width of rectangle

we know that

The area of the constructed figure is equal to

`A=xy+2x^(2) +2y^(2)`

`A=120\ cm^(2)`

so

`120=xy+2x^(2) +2y^(2)` -----> equation A

`x=y+5` -----> equation B

substitute equation B in equation A

`120=y(y+5)+2(y+5)^(2) +2y^(2)`
`120=y^(2)+5y+2(y^(2)+10y+25)+2y^(2)`
`120=y^(2)+5y+2y^(2)+20y+50+2y^(2)`

`5y^(2)+25y+50-120=0`

`5y^(2)+25y-70=0`

using a graphing tool to resolve the quadratic equation

the solution is

`y=2\ cm`

Find the value of x
`x=2+5=7\ cm`

Find the perimeter of rectangle

The perimeter of rectangle is equal to

`P=2(x+y)`

`P=2(7+2)=18\ cm`