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

Perimeter of rectangle = 38 cm

Explanation:

It is given that, Length of a rectangle is 5 cm longer than the width.

Let 'x' be the width then length = x + 5

Area of of rectangle = x(x + 5)

One side of square = ( x+ 5)/4

Area of all squares = [(x +5)/4]² * 4 = (x + 5)²/4

To find the value of x

Total area = 120 cm²

Area of of rectangle + Area of all squares = 120

x(x + 5) + (x + 5)²/4 = 120

x² + 5x + (x² + 10x + 25)/4 = 120

4x² + 20x + x² + 10x + 25 = 120 * 4

x² + 6x + 5 = 24 * 4

x² + 6x - 96 =0

x = 7

To find the perimeter of rectangle

width = x = 7

Length = x + 5 = 7 + 5 = 12

Perimeter = 2(length + width) = 2(7 + 12) = 38 cm

Answer 2

The perimeter of rectangle is
`18\ cm`

Explanation:

Let

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

y----> the width of the rectangle

we know that

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

`120=xy+2x^(2)+2y^(2)` ---> equation B (area of the constructed figure)

substitute the equation A in equation B

`120=(y+5)y+2(y+5)^(2)+2y^(2)\\ 120=(y+5)y+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)\\120=5y^(2)+25y+50\\5y^(2)+25y-70=0`

using a graphing calculator -----> solve the quadratic equation

The solution is

`y=2\ cm`

Find the value of x

`x=y+5 ----> x=2+5=7\ cm`

Find the perimeter of rectangle

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