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Simple concept problem

by showbiz » Tue Dec 07, 2010 1:44 pm
The diagonal of a rectangular flowerbed is 15 sqrt(2) feet.  What is the area of the flowerbed, if its length and width are equal?

The answer is
225

My question is this.

If sides are equal, then we have a square. So a^2 + a^2 = (15sqrt(2))^2. So 2*a^2=225. So a^2=225/2. I'm missing something here, but can't figure out what it is. Can anyone help?
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by Night reader » Tue Dec 07, 2010 2:39 pm
showbiz wrote:The diagonal of a rectangular flowerbed is 15 sqrt(2) feet.  What is the area of the flowerbed, if its length and width are equal?

The answer is
225

My question is this.

If sides are equal, then we have a square. So a^2 + a^2 = (15sqrt(2))^2. So 2*a^2=225. So a^2=225/2. I'm missing something here, but can't figure out what it is. Can anyone help?
length=width=15

proportion in a right triangle non-hypotenuse sides a+a, hypotenuse a sqrt(2)

area of rectangle = 15^2 = 225
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by Laura GMAT Tutor » Tue Dec 07, 2010 2:41 pm
I'm having a little trouble understanding what you're doing here. But I can tell you that the easiest way to understand this is to apply a 45/45/90 right triangle.

The diagonal of a square is always equal to (side)(sqrt 2). For example, if each side of a square is 5, then the diagonal is 5sqrt2. Or if each side of a square is 12, then the diagonal is 12sqrt2.

So if the diagonal of a square is 15sqrt2, then each side is 15, easy-breezy!
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by Rahul@gurome » Tue Dec 07, 2010 8:18 pm
showbiz wrote:The diagonal of a rectangular flowerbed is 15 sqrt(2) feet.  What is the area of the flowerbed, if its length and width are equal?

The answer is
225

My question is this.

If sides are equal, then we have a square. So a^2 + a^2 = (15sqrt(2))^2. So 2*a^2=225. So a^2=225/2. I'm missing something here, but can't figure out what it is. Can anyone help?
(15√2)² = (15)²(√2)² = 225*2 = 450

Thus, (a² + a²) = (15√2)² = 450
=> 2a² = 450
=> a² = 225

Area = a² = 450
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