MATH SOLVE

3 months ago

Q:
# 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?

Accepted Solution

A:

Answer:18Step-by-step explanation:RemarkThis 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'tHowever if you don't the question gives you 2 complex answers. (answers with the sqrt( - 1) in them.SolutionLet the width = xLet the length = x + 5Area of the rectangle: L * w = x * (x + 5)Area of the smaller squares (there are 2)Area = 2*s^2x = sArea = 2 * x^2Area of the larger squares = 2 * (x+5)^2Total Areax*(x + 5) + 2x^2 + 2(x + 5)^2 = 120 Expandx^2 + 5x + 2x^2 + 2(x^2 + 10x + 25) = 120 Remove the bracketsx^2 + 5x + 2x^2 + 2x^2 + 20x + 50 = 120 collect the like terms on the left5x^2 + 25x + 50 = 120 Subtract 120 from both sides.5x^2 + 25x - 70 = 0 Divide through by 5x^2 + 5x - 14 = 0 Factor(x + 7)(x - 2) = 0 x + 7 has no meaning x - 2 = 0x = 2 PerimeterP = 2*w + 2*Lw = 2L = 2 + 5L = 7 P = 2*2 + 2 * 7P = 4 + 14P = 18