Question

A pool measuring 20 meters by 24 meters is surrounded by a path of uniform

Answer 1

Answer:

8

Step-by-step explanation:

If the width of the path is x, then the dimensions of the combined area would be (20 + 2x) × (24 + 2x). Since the combined area is given as 1440, we can equate this with the area of a rectangle with these dimensions.

1440 = (20 + 2x)(24 + 2x)

Expand the brackets

1440 = 480 + 88x + 4x^2

Rearrange the equation, moving the 1440 to the other side so that it is equal to zero

0 = 4x^2 + 88x - 960

Divide by 4 as it is a common factor

0 = x^2 + 22x - 240

Use the quadratic formula

For an equation of the form 0 = ax^2 + bx + c, the quadratic formula is:

$x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac } }{2a}$

$x = \frac{ - 22 \pm \sqrt{ {22}^{2} - 4 (1)( - 240)} }{2(1)}$

$x = \frac{ - 22 \pm \sqrt{484 + 960} }{2}$

$x = \frac{ - 22 \pm \sqrt{1444} }{2}$

$x = \frac{ - 22 \pm38}{2}$

$x = - 11 \pm19$

$x = 8$