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

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

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New post 28 Sep 2017, 22:57
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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
[Reveal] Spoiler: OA

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Re: The size of a television screen is given as the length of the screen's [#permalink]

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New post 28 Sep 2017, 23:04
Bunuel 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



area of a square 21-inch screen = (21/\sqrt{2})^2 = 441/2
area of a square 19-inch screen = (19/\sqrt{2})^2 = 361/2

Diff = 80/2 = 40
E
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The size of a television screen is given as the length of the screen's [#permalink]

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New post 29 Sep 2017, 00:42
side of first television = 21/sqrt 2
side of second television = 19/sqrt 2
area difference =
(21/sqrt 2)^2 - (19/sqrt 2) ^2
take 1/2 out side then
= 1/2* ( 21^2 - 19^2)
=1/2* ((20+1)^2-(20-1)^2) e.g ((a+b)^2-(a-b)^2)
=1/2*(20^2+1^2+40-20^2-1^2+40)
=1/2*80
=40

Answer E

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

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New post 29 Sep 2017, 12:43
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Bunuel 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

If the size of a square television screen is given by its diagonal, we need side lengths to calculate area.

The relationship between the a square's side and its diagonal, d, is given by

\(s\sqrt{2} = d\)
\(s = \frac{d}{\sqrt{2}}\)

The side of the 21-inch size television (d = 21), therefore, is

\(\frac{21}{\sqrt{2}}\). Square that to find area:

\((\frac{21}{\sqrt{2}}\) * \(\frac{21}{\sqrt{2}})\) = \(\frac{21*21}{2}\) = \(\frac{441}{2}\)

The side of the 19-inch size television (d = 19) is

\(\frac{19}{\sqrt{2}}\). Square that to find area:

\((\frac{19}{\sqrt{2}}\) * \(\frac{19}{\sqrt{2}})\) = \(\frac{19*19}{2}\) = \(\frac{361}{2}\)

Difference in area between larger and smaller, in square inches:

\((\frac{441}{2} - \frac{361}{2}) =\frac{80}{2} = 40\)

Answer E

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Re: The size of a television screen is given as the length of the screen's [#permalink]

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New post 03 Oct 2017, 15:36
Bunuel 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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Kudos [?]: 991 [0], given: 5

Re: The size of a television screen is given as the length of the screen's   [#permalink] 03 Oct 2017, 15:36
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