Question

If the length of a rectangle is decreased by 4 cm and the width is increased by 5 cm, the result will be a square, the area of which will be 40 cm2 greater than the area of the rectangle. Find the area of the rectangle.

Answer 1

Let x represent the side length of the square. Then
  (x +4)*(x -5) +40 = x^2
  x^2 -x -20 +40 = x^2
  20 = x

The rectangle is 15 cm by 24 cm, and its area is 360 cm^2.

Answer 2

The dimensions of the rectangle can be obtained by using the quadratic

formula.

The correct response;

• The area of the rectangle is (20 - 2·√(10)) cm²

Method by which the above response is obtained:

Let x represent the length of the rectangle, and y represent the width of

the rectangle.

x - 4 = y + 5 (sides of a square)

(x - 4) × (y + 5) = 40 cm²

Which gives;

(y + 5) × (y + 5) = 40 cm²

y² + 10·y + 25 = 40

y² + 10·y + 25 - 40 = 0

y² + 10·y - 15 = 0

Solving by using the quadratic formula gives;

$\displaystyle y = \frac{-10 \pm\sqrt{10^2 - 4 \times 1 \times (-15)} }{2 \times 1} = -5 \pm \frac{\sqrt{40} }{2} = \mathbf{-5 \pm 2\cdot \sqrt{10}}$

Using the positive values for the width, we have;

y = -5 + 2·√(10)

Which gives;

x - 4 = 5 + -5 + 2·√(10) = 2·√(10)

x = 4 + 2·√(10)

The area of the rectangle, A = x × y

Therefore;

A = (-5 + 2·√(10)) × (4 + 2·√(10)) = 20 - 2·√(10)

• The area of the rectangle, A = 20 - 2·√(10)

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