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Geometry - television screens

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Geometry - television screens

by Xeb » Sat Oct 08, 2011 4:35 am
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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by rohanberi » Sat Oct 08, 2011 4:56 am
Xeb 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
I think there is a typo in options.
Because, for first tv screen, diagonal = 21 inches, side of a square screen = sqrt2 * 21
Hence, area = 441/2 sq inches.

Similarly, for second tv screen, diagonal = 19 inches, side of a square screen = sqrt2 * 19
Hence, area = 381/2 sq inches.

Subtracting these two, 441/2 - 361/2 = 80/2 = 40 sq inches. ( E is the answer.)
Last edited by rohanberi on Wed Apr 10, 2013 10:50 pm, edited 1 time in total.
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by gmatclubmember » Sat Oct 08, 2011 4:58 am
Xeb 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
The answer would be 21^2/2-19^2/2=(21+19)(21-19)/2=2*40/2=40
E should be the answer
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by gmatclubmember » Sat Oct 08, 2011 4:59 am
rohanberi wrote:
Xeb 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
I think there is a typo in options.
Because, for first tv screen, diagonal = 21 inches, side of a square screen = sqrt2 * 21
Hence, area = 441/2 sq inches.

Similarly, for second tv screen, diagonal = 19 inches, side of a square screen = sqrt2 * 19
Hence, area = 381/2 sq inches.

Subtracting these two, 441/2 - 381/2 = 60/2 = 30 sq inches. ( which is not an option)
19^2 is 361.
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by Abhishek009 » Sat Oct 08, 2011 5:02 am
Xeb 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

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


So the 21 inch and 19 inch represents diagonals....

Diagonal of a square is sqrt(2)*a

sqrt(2)*a = 21

a = 21/1.414 = 14.85

Area of the square = 220.57



Again -

19 inch diagonal screen

sqrt(2)*a = 19

a = 19/1.414 = 13.44

Area of the square = 180.55


Difference in Area is 220.57-180.55 = 40.02


So IMO : E
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by rohanberi » Sat Oct 08, 2011 5:08 am
Oops. My bad.
gmatclubmember wrote:
rohanberi wrote:
Xeb 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
I think there is a typo in options.
Because, for first tv screen, diagonal = 21 inches, side of a square screen = sqrt2 * 21
Hence, area = 441/2 sq inches.

Similarly, for second tv screen, diagonal = 19 inches, side of a square screen = sqrt2 * 19
Hence, area = 381/2 sq inches.

Subtracting these two, 441/2 - 381/2 = 60/2 = 30 sq inches. ( which is not an option)
19^2 is 361.
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by Bek » Sat Oct 08, 2011 6:55 am
there is another way:

hypotenuse: a^2 + b^2 = c^2

1) c = 21
c^2 = 441
441/2 + 441/2 = 441
sqrt(441/2) + sqrt(441/2) = 21
Area: sqrt(441/2) x sqrt(441/2) = 441/2 = 220.5

2) c = 19
c^2 = 361
361/2 + 361/2 = 361
sqrt(361/2) + sqrt(361/2) = 19
Area: sqrt(361/2) x sqrt(361/2) = 361/2 = 180.5

3) the difference between areas is: 220.5 - 180.5 = 40
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by parul9 » Sat Oct 08, 2011 7:01 am
E!
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by rafael.amado » Mon Jun 27, 2016 4:27 pm
Another way to do this:

Area of a square = side^2 ; Pythagoras=> Diag^2 = side^2 + side^2 = 2*side^2, then:
Area=(Diag^2)/2

So, the difference between the area of a 21-inches diagonal square and a 19-inches diagonal square would be:
(21^2)/2 - (19^2)/2 = 1/2*[(21^2)-(19^2)]

If (a^2-b^2) = (a+b)*(a-b), we have:
1/2*(21+19)*(21-19)=1/2*(40)*(2)=40

So, with some properties in mind, one would not have to use difficult divisions or big squares.

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by Scott@TargetTestPrep » Thu Jan 04, 2018 10:05 am
Xeb 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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