see In the picture given below, one can see rectangular in-ground swimming pool installed by a family in their backyard. There is a concrete sidewalk around the pool of width x m. The outside edges of the sidewalk measure 7 m and 12 m. The area of the pool is 36 sq.m.
(a) Based on the information given above, form a quadratic equation in terms of x.
(b) Find the width of the sidewalk around the pool.
(a) Based on the information given above, form a quadratic equation in terms of x.
(b) Find the width of the sidewalk around the pool.
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Answer:
The correct answers for the given problems are found to be:
(i) The required quadratic expression in terms of [tex]x[/tex] is found to be [tex]4x^2-38x+48=0[/tex].
(ii) the width of the sidewalk is found to be either of the two following values: (1) 8m or (2) 1.5m or [tex]\frac{3}{2}[/tex] m.
Step-by-step explanation:
The outside edges of the sidewalk are given to us as 7 meters and 12 meters respectively with the width of the sidewalk being given as [tex]x[/tex] meters.
Now, the edges of the swimming pool will be: [tex]7-2x[/tex] meters and [tex]12-2x[/tex] meters.
The area of the pool is given to us as: [tex]36 m^2[/tex]
We know that the area of a rectangular surface is: [tex]A=length\times breadth[/tex]
Substituting the edges of the pool along with its area in the expression for area, we get:
[tex]36=(7-2x)\times (12-2x)[/tex]
Simplifying this, we get:
[tex]36=4x^2-24x-14x+84[/tex]
or we can say:
[tex]4x^2-24x-14x+84-36=0\\4x^2-38x+48=0\\[/tex]
Thus, the required quadratic expression in terms of [tex]x[/tex] is found to be [tex]4x^2-38x+48=0[/tex].
Now, solving for [tex]x[/tex], we get:
[tex]4(x-8)(x-\frac{3}{2})=0\\x=8,\frac{3}{2}[/tex]
Thus, the width of the sidewalk is found to be either of the two following values:
(i) 8m or (ii) 1.5m or [tex]\frac{3}{2}[/tex] m