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?

Answer 1

18 cm

Explanation:

Let the width of the rectangle be x cm, then the length of the rectangle is x+5 cm.

The area of the ractangle is

`x\cdot (x+5)\ cm^2.`

The area of the square with side's length of x cm is

`x^2\ cm^2.`

The area of the square with side's length of x+5 cm is

`(x+5)^2\ cm^2.`

The area of constructed figure is

`x(x+5)+2x^2+2(x+5)^2\ cm^2.`

Since the total area of the constructed figure is 120 cm², you have

`x(x+5)+2x^2+2(x+5)^2=120.`

Solve this equation:

`x^2+5x+2x^2+2x^2+20x+50=120,\\ \\5x^2+25x-70=0,\\ \\x^2+5x-14=0,\\ \\D=5^2-4\cdot (-14)=25+56=81,\\ \\x_(1,2)=(-5\pm√(81))/(2)=-7,\ 2.`

The width of the rectangle cannot be negative, so x=2 cm and x+5=7 cm and the perimeter of the rectangle is

`2x+2(x+5)=2\cdot 2+2\cdot 7=4+14=18\ cm.`