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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 Updated on: 08 Jul 2012, 05:06
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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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Originally posted by Stiv on 08 Jul 2012, 04:45.
Last edited by Bunuel on 08 Jul 2012, 05:06, edited 1 time in total.
Edited the question.
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Re: The size of a television screen is given as the length of [#permalink]

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New post 08 Jul 2012, 05:07
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Stiv 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


\(d_1=21\) 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: The size of a television screen is given as the length of [#permalink]

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New post 08 Jul 2012, 10:05
Bunuel wrote:
Stiv 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


\(d_1=21\) 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.



Just to refresh my understanding, how is the \(area_{square}=\frac{d^2}{2}\)

The side x side is simple enough. Given that a square is essentially 2 isoceles triangles, I was going with 21=\(x\sqrt{2}\)

That's where I got stuck. But I still don't get your formula...
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Re: The size of a television screen is given as the length of [#permalink]

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New post 08 Jul 2012, 10:08
1
manavecplan wrote:
Bunuel wrote:
Stiv 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


\(d_1=21\) 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.



Just to refresh my understanding, how is the \(area_{square}=\frac{d^2}{2}\)

The side x side is simple enough. Given that a square is essentially 2 isoceles triangles, I was going with 21=\(x\sqrt{2}\)

That's where I got stuck. But I still don't get your formula...


The area of a square is \(side^2\), but since here the length of the diagonal is given, then it's better to use another formula \(area_{square}=\frac{diagonal^2}{2}\).
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Re: The size of a television screen is given as the length of [#permalink]

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New post 12 Dec 2013, 12:43
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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

We're dealing with squares here. When you take the diagonal of a square it creates two isosceles triangles with a ratio of x: x: x√2. Because the hypotenuse is 21 inches it is equal to x√2. To find the length of the legs, we set 21 = x√2. x = 21/√2
To find the area we square 21/√2 which equals 441/2. Similarly, for the 19 inch screen we follow the same steps. We find that the legs of the triangle (or the sides of the square screen) are 19/√2 inches. To find the area we square this to get 361/2. Finally, subtract 361/2 from 441/2 to get 80/2 = 40.

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The size of a television screen is given as the length of [#permalink]

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New post 13 Nov 2017, 23:16
manavecplan wrote:
Bunuel wrote:
Stiv 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


\(d_1=21\) 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.



Just to refresh my understanding, how is the \(area_{square}=\frac{d^2}{2}\)

The side x side is simple enough. Given that a square is essentially 2 isoceles triangles, I was going with 21=\(x\sqrt{2}\)

That's where I got stuck. But I still don't get your formula...


I had to try it out myself to understand the formula. As you mentioned with the diagonal, we calculate the length of each side = \(\frac{21}{\sqrt{2}}\) = x
Area of a Square is \((Side^2) = (\frac{21}{\sqrt{2}} * \frac{21}{\sqrt{2}}) = \frac{21^2}{2} = \frac{diagonal^2}{2}\)
The size of a television screen is given as the length of   [#permalink] 13 Nov 2017, 23:16
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