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

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Joined: 16 Feb 2012
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The size of a television screen is given as the length of  [#permalink]

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Updated on: 08 Jul 2012, 04:06
5
00:00

Difficulty:

45% (medium)

Question Stats:

70% (01:20) correct 30% (01:39) wrong based on 239 sessions

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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, 03:45.
Last edited by Bunuel on 08 Jul 2012, 04:06, edited 1 time in total.
Edited the question.
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Posts: 52294
Re: The size of a television screen is given as the length of  [#permalink]

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08 Jul 2012, 04:07
4
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 $$d_2=19$$ --> $$area_{square}=\frac{d^2}{2}$$ --> $$area_1-area_2=\frac{(d_1)^2}{2}-\frac{(d_2)^2}{2}=\frac{21^2-19^2}{2}=\frac{(21-19)(21+19)}{2}=\frac{2*40}{2}=40$$

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

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08 Jul 2012, 09:05
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 $$d_2=19$$ --> $$area_{square}=\frac{d^2}{2}$$ --> $$area_1-area_2=\frac{(d_1)^2}{2}-\frac{(d_2)^2}{2}=\frac{21^2-19^2}{2}=\frac{(21-19)(21+19)}{2}=\frac{2*40}{2}=40$$

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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Posts: 52294
Re: The size of a television screen is given as the length of  [#permalink]

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08 Jul 2012, 09:08
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 $$d_2=19$$ --> $$area_{square}=\frac{d^2}{2}$$ --> $$area_1-area_2=\frac{(d_1)^2}{2}-\frac{(d_2)^2}{2}=\frac{21^2-19^2}{2}=\frac{(21-19)(21+19)}{2}=\frac{2*40}{2}=40$$

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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12 Dec 2013, 11:43
1
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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Re: The size of a television screen is given as the length of the screen's  [#permalink]

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28 Sep 2017, 23:04
1
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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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

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

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29 Sep 2017, 12:43
1
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:

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

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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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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.

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

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13 Nov 2017, 22:16
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 $$d_2=19$$ --> $$area_{square}=\frac{d^2}{2}$$ --> $$area_1-area_2=\frac{(d_1)^2}{2}-\frac{(d_2)^2}{2}=\frac{21^2-19^2}{2}=\frac{(21-19)(21+19)}{2}=\frac{2*40}{2}=40$$

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 $$(Side^2) = (\frac{21}{\sqrt{2}} * \frac{21}{\sqrt{2}}) = \frac{21^2}{2} = \frac{diagonal^2}{2}$$
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02 Jan 2019, 04:26
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