The area of a playground is 320 yd2. The width of the playground is 4 yd longer than its length. Find the length and width of the playground.
a
length = 24 yd, width = 20 yd
b
length = 20 yd, width = 16 yd
c
length = 20 yd, width = 24 yd
d
length = 16 yd, width = 20 yd
Let's assume the length of the playground is x yd.
According to the problem, the width of the playground is 4 yd longer than its length. So the width of the playground is (x + 4) yd.
The area of the playground is given as 320 yd^2, so we can write the equation:
length * width = area
x * (x + 4) = 320
Expanding the equation:
x^2 + 4x = 320
Rearranging the equation:
x^2 + 4x - 320 = 0
We can factor this quadratic equation:
(x + 20)(x - 16) = 0
Setting each factor equal to zero:
x + 20 = 0 or x - 16 = 0
Solving each equation:
x = -20 or x = 16
Since the length cannot be negative, we discard the -20 solution. So the length of the playground is 16 yd.
The width of the playground is given as (x + 4) yd, so the width is (16 + 4) = 20 yd.
Therefore, the length of the playground is 16 yd and the width is 20 yd.
Therefore, the correct answer is: d) length = 16 yd, width = 20 yd.