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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

User Mahasam
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2 Answers

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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
User Enomad
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Answer: B. length=12, width=18

Explanation:

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

Area of a rectangular playground = Length x width

Since , area of playground =
216\ yd^2.

Then , we have


216=(x)*(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

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

User Resonance
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