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the area of a playground is 216 yd^2. the width of the playground is 6 yd longer than its length find the length and width.
A. length=18, width=12
B. length=12, width=18
C. length=18, width=24
D.length=24, width=18

Respuesta :

nobillionaireNobley
Let the length of the playground be x, then the width is 6 + x.
Area = length * width = x * (6 + x) = 6x + x^2 = 216

Solving the quadratic equation x^2 + 6x - 216 = 0, we have x = 12 or -18

i.e length = 12 and width = 6 + 18 = 18
JeanaShupp

Answer: B. length=12, width=18

Step-by-step explanation:

Let x = Length of the playground then width = x+6

Area of a rectangular playground = Length x width

Since , area of playground = [tex]216\ yd^2.[/tex]

Then , we have

[tex]216=(x)\times(x+6)\\\\\Rightarrow\ x^2+6x\\\\\Rightarrow\ x^2+6x-216=0\\\\\Rightarrow\ x^2+18x-12x-216=0\\\\\Rightarrow\ x(x+18)-12(x+18)=0\\\\\Rightarrow\ (x-12)(x+18)=0\\\\\Rightarrow\ x=12\text{ or }x=-18[/tex]

But length cannot be negative , so reject x= -18 .

Thus , the length of playground = 12 yd

Then , width = 12+6 = 18 yd

Hence, the correct answer is B. length=12, width=18