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The size of a television screen is given as the length of

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The size of a television screen is given as the length of [#permalink] New post 27 Sep 2010, 22:44
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A
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E

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Question Stats:

76% (02:32) correct 24% (01:36) wrong based on 70 sessions
The size of a television screen is given as the length of the screen’s diagonal. If the screens were flat, then the area of a square 21-inch screen would be how many square inches greater than the area of a square 19-inch screen?

A. 2
B. 4
C. 16
D. 38
E. 40
[Reveal] Spoiler: OA
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Re: did not understand the logic [#permalink] New post 27 Sep 2010, 22:49
Expert's post
vanidhar wrote:
the square of a television screen is given as a length of the diagoal. If the screens were flat, then the area of the 21in screen would be how many square inches greater than the area of a 19in screen ?

2
4
16
38
40


d_1=21 and d_2=19 --> area_{square}=\frac{d^2}{2} --> area_1-area_2=\frac{(d_1)^2}{2}-\frac{(d_2)^2}{2}=\frac{21^2-19^2}{2}=\frac{(21-19)(21+19)}{2}=\frac{2*40}{2}=40

Answer: E.
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Re: did not understand the logic [#permalink] New post 27 Sep 2010, 22:50
vanidhar wrote:
the square of a television screen is given as a length of the diagoal. If the screens were flat, then the area of the 21in screen would be how many square inches greater than the area of a 19in screen ?

2
4
16
38
40


Answer = Area(square with diagnol 21) - Area(square with diagnol 19)

If diagnol = x. Then side = x/root(2) & area = x^2/2

So answer = \frac{21^2-19^2}{2} = \frac{40*2}{2} = 40

Answer E
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The size of a television screen [#permalink] New post 25 Dec 2010, 07:07
The size of a television screen is given as the length
of the screen’s diagonal. If the screens were flat, then
the area of a square 21-inch screen would be how
many square inches greater than the area of a square
19-inch screen?
(A) 2
(B) 4
(C) 16
(D) 38
(E) 40

not sure about the ans
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Re: The size of a television screen [#permalink] New post 25 Dec 2010, 07:15
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the diagonal of a square is always side*\sqrt{2}
and the side of a square is vice versa always \frac{diagonal}{\sqrt{2}}

Therefore:
area of the bigger one is (\frac{21}{\sqrt{2}})^2
area of the smaller one is (\frac{19}{\sqrt{2}})^2
and the difference is = 40
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Re: The size of a television screen [#permalink] New post 25 Dec 2010, 07:21
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Re: did not understand the logic [#permalink] New post 25 Dec 2010, 16:40
This question supposes that the TV-screen has the shape of a square?

Can this be solved for the rectangle as well?
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Re: did not understand the logic   [#permalink] 25 Dec 2010, 16:40
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