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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]

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New post 27 Sep 2010, 23:44
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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
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Re: did not understand the logic [#permalink]

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New post 27 Sep 2010, 23:49
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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]

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New post 27 Sep 2010, 23:50
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]

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New post 25 Dec 2010, 08:07
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]

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New post 25 Dec 2010, 08:15
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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]

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New post 25 Dec 2010, 08:21

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Re: did not understand the logic [#permalink]

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New post 25 Dec 2010, 17:40
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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New post 17 Sep 2017, 22:59
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Re: The size of a television screen is given as the length of [#permalink]

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New post 22 Sep 2017, 07:45
vanidhar wrote:
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


Let’s determine the side of the square 21-inch screen (i.e., the diagonal of the screen is 21 inches). Recall that the diagonal of a square is equal to side√2.

21 = side√2

21/√2 = side

Since area is side^2, the area of the 21-inch screen is (21/√2)^2 = 441/2.

Let’s determine the side of the square 19-inch screen:

19 = side√2

19/√2 = side

The area of the 19-inch screen is (19/√2)^2 = 361/2.

Thus, the difference is 441/2 - 361/2 = 80/2 = 40.

Alternate solution:

We are given two square TV screens with diagonals 21 and 19, respectively. We have to determine the difference between the areas of the screens. Recall that the area of a square, given its diagonal d, is A = d^2/2. Thus, the area of the 21-inch screen is 21^2/2 = 441/2 and the area of the 19-inch screen is 19^2/2 = 361/2. Therefore, the difference in areas is 441/2 - 361/2 = 80/2 = 40.

Answer: E
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Re: The size of a television screen is given as the length of   [#permalink] 22 Sep 2017, 07:45
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