A small, rectangular park has

[arthi9487]

A small, rectangular park has a perimeter of 560 feet and a diagonal measurement of 200 feet. What is its area, in square feet?

A. 19,200
B. 19,600
C. 20,000
D. 20,400
E. 20,800

I get to the point of realizing that L+W=280 and that both sides need to be squared per the Pythagorean theorem, but after that, I am not sure what to do with resulting quadratic equation: L^2 + 2LW + W^2, since there are 2 solutions. Can someone please walk through this?

[RonPurewal]

You don't need two variables. You know that the sum of the length and the width is 280, so you should just use "x" and "280 - x".
Try that. If there are still any problems, post back.

[RonPurewal]

In general -- NEVER use more variables than you have to!
In general, if there is any sort of easy relationship between the quantities at hand, this means you should not use more than one variable.

E.g.,

"I took an hour longer than my brother to do ____"
---> He took t hours. I took t + 1 hours. One variable.

"Two children inherited a sum of money in the ratio 3:4"
--> One child inherited 3x; the other inherited 4x. One variable. (For the love of all that's good in this world don't do x/y = 3/4.)

"Jim is twice as old as Sam."
--> Jim is 2x years old; Sam is x years old. One variable.

Etc.
Why do more work than you have to?

[arthi9487]

I am still unable to solve. Can you please walk through the solution?

[RonPurewal]

The length and width of the park are "x" and "280 - x". The diagonal is 200. That's a right triangle.

x^2 + (280 - x)^2 = 200^2

From there it's routine algebra. Find x, then find the other dimension (which is 280 - x), and you're good to go.

[justine_gonsalves]

Hi Ron can you explain further by the rest is just algebra. If I work out x~2 + (280 - x)~2 = 200 Those are some large numbers to work with in 2 mins, is that the correct understanding?

[tim]

Are you asking whether the correct understanding is that those are some large numbers to work with in two minutes? Some people would consider them large, others would not. It really doesn't matter though; this is what you have to do to get through the problem, and if it takes you any longer than a minute and a half you will probably want to work on getting faster at crunching numbers and manipulating algebraic expressions.

[RonPurewal]

Hi Ron can you explain further by the rest is just algebra. If I work out x~2 + (280 - x)~2 = 200 Those are some large numbers to work with in 2 mins, is that the correct understanding?

What else are you going to do?

If the answer is "I don't know", then ... well, just start doing the work.
If you can only think of one approach ... don't talk yourself out of it! Because then you'll have zero approaches.

This whole complaint is distressingly common, really: "But... that will take too long!"

1/ No, it probably won't.
Most people are REALLY BAD at estimating how long things will take. (I've had lots of students tell me—with a straight face—that they thought they'd take several minutes to do some task... but then they just did it, and it only took 15-30 seconds. Happens all the time.)

2/ If you don't have a Plan B, then this complaint is really a non-thing, since .. well, what else are you gonna do?

[RonPurewal]

There's another way, by the way, if you're familiar with the 3-4-5 right triangle.

See, here, you have a diagonal that's a whole number (i.e., it doesn't contain the "√" sign).
That means this diagonal is a very special animal. If you take most right triangles with random integer legs, then you'll wind up with diagonal = √something.

So... well, let's just GUESS that we're dealing with a 3-4-5-ratio triangle. Yep—just toss it out there.

The diagonal is 200. So, if this is going to work, then we're going to be multiplying the template by 40—producing a triangle with sides 4x30, 4x40, and 4x50. In other words, 120, 160, and 200.
The 120 and 160 would be the sides of the rectangle, and the 200 would be the diagonal.

Well... let's see here...
Perimeter = 120 + 160 + 120 + 160
= 560
Holy whoa what, it works.
And we're done.
(Well, not totally; we still need to multiply 120 x 160. But, we snuck out of the hard part.)

This is a thing. If a right triangle has "nice" sides, then, MUCH more often than not, it will be a multiple of the 3-4-5 or 5-12-13 template.
If you get stuck with algebra, it's worth the time to guess these templates, throw them at the problem, and see whether they work.

[RonPurewal]

And one more thing. If you are a SUPER FAN of the "special" quadratics, you might recognize the opportunity to do this—although it's certainly not obvious.

Let's say the sides are L and W.
The perimeter is 2L + 2W. So 2L + 2W = 560, or, more simply, L + W = 280.
Also, √(L^2 + W^2) = 200. So, L^2 + W^2 = 40,000.

The goal is not to find L and W individually; the goal is just to find the product LW.

If you are a SUPER FAN of the "special" qudratics, then you'll recognize that, if you square L + W, you'll get
L^2 + 2LW + W^2 = 280^2 = 78,400
... but wait a minute...
The blue thing is 40,000 (see above), and the pink thing is THE GOAL OF THE WHOLE PROBLEM.

SO...
40,000 + 2LW = 78,400
2LW = 38,400
LW = 19,200
Got it.

Again, this is not the kind of thing you'll think of, unless you are a SUPER FAN of the special quadratics. But, the special quadratics have a large fan base.

[justine_gonsalves]

Thanks this is helpful, I was moreso referring if there was another solution. I like the 3-4-5 Triangle approach Ron, this is great.

[RonPurewal]

Thanks this is helpful, I was moreso referring if there was another solution. I like the 3-4-5 Triangle approach Ron, this is great.

OK, yeah, there's the 3-4-5 approach as well as the "other" algebra approach (the one in which you don't get the individual L and W values, but instead play around with "special" quadratics).

MORE IMPORTANTLY
Do not EVER think, "I don't want to do this because I think it will take too long".
Of course, if your goal is to get a lower score on the GMAT, then this thought process is the best possible strategy for achieving that goal.

[ajaym8]

This 3-4-5 & 5-12-13 is a really neat & a never-seen-before approach. Thanks Ron.

[RonPurewal]

you're welcome.