A garden measuring 9 meters by 4 meters is going to have a walkway constructed all around the perimeter, increasing the total area to 150 square meters. What will be the width of the pathway? (The pathway will be the same width around the entire garden).

Respuesta :

Answer:

3 metres

Step-by-step explanation:

Let the width of the pathway be x

Area: (9+2x)(4+2x)

36 + 8x + 18x + 4x²

4x² + 26x + 36

Increased area is 150:

4x² + 26x + 36 = 150

4x² + 26x - 114 = 0

2x² + 13x - 57 = 0

2x² + 19x - 6x - 57 = 0

x(2x + 19) - 3(2x + 19) = 0

(x - 3)(2x + 19) = 0

x = 3, x = -19/2

The width of the considered pathway after increasing the total area to 150 square meters is 3 meters.

How to find the area and the perimeter of a rectangle?

For a rectangle with length and width L and W units, we get:

  • Area of the rectangle = [tex]L \times W \: \rm unit^2[/tex]
  • Perimeter of the rectangle = [tex]2(L + W) \: \rm unit[/tex]

How to find the roots of a quadratic equation?

Suppose that the given quadratic equation is

[tex]ax^2 + bx + c = 0[/tex]

Then its roots are given as:

[tex]x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]

For this case, we're given that:

  • The dimensions of the rectangular garden is 9 m by 4 m (assume its rectangular shaped)
  • Pathway is constructed around the garden
  • The area of the total garden + pathway = 150 sq. meters.
  • Pathway is of same width on all sides of the garden.

Consider the diagram given below.

The pathway is rectangular.

Let its width be x.

One dimension of the pathway = width on either side parallel to that garden's one dimension + that dimension of garden =  x + x + 9 = 2x + 9 meters (as width on all the sides is 'x' m)

Other dimension of the pathway = width on either side parallel to that garden's other dimension + that dimension of garden = x + x + 4 = 2x + 4 meters

Thus, the dimensions of the rectangular pathway are 2x + 9 m and 2x + 4 m.

The area of the rectangle of dimension 2x +9 and 2x + 4 is covering area of the pathway and area of the garden.

This summed area is 150 sq. meters.

This area is [tex](2x+9)(2x+4)[/tex] sq. meters.

Thus, we get:

[tex](2x + 9)(2x +4) = 150\\4x^2 + 26x + 36 = 150\\4x^2 + 26x -114 = 0\\[/tex]

Using the formula for solving quadratic equations, we get:

[tex]x = \dfrac{-26 \pm \sqrt{26^2 - 4(4)(-114)}}{2(4)} = \dfrac{-26 \pm \sqrt{2500}}{8}\\\\x = \dfrac{-26 \pm 50}{8}\\\\x = 3, -76/8[/tex]

As x represents width (a measurement of how long a thing) cannot be negative, therefore, we take only x = 3 (in meters)

This is the width of the pathway.

Thus, the width of the considered pathway after increasing the total area to 150 square meters is 3 meters.

Learn more about finding the solutions of a quadratic equation here:

https://brainly.com/question/3358603

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