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by smallsorrow » Sat Sep 20, 2008 1:36 am
There are two flat screens, one with a diagonal of 21 inches and the other one with a diagonal of 19 inches. What is deifference of the areas of the 21 inch screen and the 19 inch screen.

1.2
2.4
3.30
4.35
5.40

I only can come to a solution which a huge calculation. Do you know any efficient way?

OA will follow
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by manulath » Sat Sep 20, 2008 6:01 am
is the ans 40?
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by Ian Stewart » Sat Sep 20, 2008 6:22 am
I assume the screens are square; otherwise the question is impossible. If the diagonal of a square is x, you can see by using the Pythagorean Theorem that the length of a side of the square is x/sqrt(2), and the area of the square is therefore (x^2)/2. So, in this question we need to find (21^2)/2 - (19^2)/2. I'd find this easiest to calculate using the difference of squares factorization:

21^2/2 - 19^2/2
= (1/2)(21^2 - 19^2)
= (1/2)(21+19)(21-19)
= (1/2)(40)(2)
= 40
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by smallsorrow » Sat Sep 20, 2008 9:10 am
Hi Ian,

no, the screen is rectangular... as far as I can remember...
... I will look for the original problem tonite and post it..

Thank you.

The OA is 40 btw
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by stop@800 » Sat Sep 20, 2008 9:15 am
Use a^2 - b^2 = (a-b)(a+b)
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by Stuart@KaplanGMAT » Sat Sep 20, 2008 10:35 am
smallsorrow wrote:Hi Ian,

no, the screen is rectangular... as far as I can remember...
... I will look for the original problem tonite and post it..

Thank you.

The OA is 40 btw
If the screens are not square, there's no way to answer the question from the information given, since you cannot use the diagonal of a rectangle to calcuate the area.

As Ian pointed out, there are two different formulas for the area of a square:

area = side^2

and

area = (diagonal^2)/2

Most people are only familiar with the first formula, but the second is great on questions in which you're given the length of the diagonal instead of the length of a side.
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