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

Answer:

18

Step-by-step explanation:

Remark

This is one of those questions that can throw you. The problem is that do you include the original rectangle or not. The way it is written it sounds like you shouldn't

However if you don't the question gives you 2 complex answers. (answers with the sqrt( - 1) in them.

Solution

Let the width = x

Let the length = x + 5

Area of the rectangle: L * w = x * (x + 5)

Area of the smaller squares (there are 2)

Area = 2*s^2

x = s

Area = 2 * x^2

Area of the larger squares = 2 * (x+5)^2

Total Area

x*(x + 5) + 2x^2 + 2(x + 5)^2 = 120                 Expand

x^2 + 5x + 2x^2 + 2(x^2 + 10x + 25) = 120     Remove the brackets

x^2 + 5x + 2x^2 + 2x^2 + 20x + 50 = 120      collect the like terms on the left

5x^2 + 25x +  50 = 120                                   Subtract 120 from both sides.

5x^2 + 25x - 70 = 0                                         Divide through by 5

x^2 + 5x - 14 = 0                                               Factor

(x + 7)(x - 2) = 0                                                 x + 7 has no meaning  

x - 2 =  0

x = 2                                                              

Perimeter

P = 2*w + 2*L

w = 2

L = 2 + 5

L = 7

P = 2*2 + 2 * 7

P = 4 + 14

P = 18