Question

A rectangular parking lot has a length that is 5 yards greater than the width the area of the parking lot is 300 yd.² find the length and width

Answer 1

Answer:

width=5yd and length=10yd

Step-by-step explanation:

To solve this problem, you need to start by asking what you do and don't know (always where to start with word problems)

We know that Area (A) = Length (L) * Width (W)

You know Area (A) = 50 sq. yds.

You don't know anything about the width, so let's just say Width (W) = x

And you know that the length is 5 yds longer than the width, so L = W + 5yd and, since W = x, we can say L = x + 5yd

Now, we recall that A = L * W, and, since A = 50, we can say 50 sq yd = L * W.

Well, now just plug in. What does W equal and L equal?

From above we plug in and can say that 50 sq yd = x * (x + 5 yd).

Multiply it out and you get 50 = x^2 + 5x (removing units for a moment so they don't clutter the lines).

Now there are a number of ways to solve this. You can tell (because of the "x^2") that it is quadratic. So let's get it equal to zero:

0 = x^2 + 5x - 50

Now you can factor it, use your quadratic formula, or (if you hated yourself) do it as a completing the square problem. I'm going to factor.

I can see from going through a list in my head that the factors 10 and -5 multiply to -50 and add to 5, so I know that

0 = x^2 + 5x - 50 = (x + 10)(x - 5)

And so you know that x = -10 or 5.

Well, you know the parking lot cannot be -10 yd long. So we must say that W = x = 5, and, since L = W + 5, then L = 5+5 = 10. So your width is 5 yd and length is 10 yd.

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