A Playground
A playground is
m by
m. When 10m is added to the width and 20m to the length, the overall area is now double the size.
What was the original area of the playground?
The answer is 1200.
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1 solution
It is actually a very simple algebra problem: We are going to write: for the previous area 1: A1 = x * 3x for the modified area 2 afterwards: A2 = (x+10) * (3x+20)
As we know that A2 is two times area one, we can write: A1 * 2 = A2 (note that we have to multiply the SMALLER area by two to GET to the doubled sized A2! I confused that sometimes in secobdary school!) -> x * 3x * 2 = (x + 10) * (3x + 20)
Solve towards x: 6x^2 = 3x^2 + 30x + 20x + 200 |-3x^2 3x^2 = 50x + 200 |-3x^2 0 = -3x^2 + 50x + 200
Solve by hand or with formular for square functions: (-b (+/-) squareroot (b^2 - 4ac)) / 2a where a = -3, b = 50, c = 200 The two solutions would be x = - 10/3 and x = 20. Because x has to be positive (there are no negative lengths in this example) x is 20, whereas A1 is: A1 = 3x * x -> A1 = 3*20 * 20 -> A1 = 1200
And to be sure: A2 = (x + 10) * (3x + 20) -> (20 + 10) (3 20 + 20) = 2400
YEY! :)