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Difference of areas of 2 TVs

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Difference of areas of 2 TVs

by pareekbharat86 » Mon Nov 04, 2013 11:28 pm
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

OA is E.
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by theCodeToGMAT » Mon Nov 04, 2013 11:34 pm
Size = Diagnol = side * sqrt(2)

Area of 21 inch - Area of 19 Inch

(21/sqrt(2))^2 - (19/sqrt(2))^2

1/2 (21+19)(21-19) = 1/2 * 40 * 2 = 40

Answer [spoiler]{E}[/spoiler]
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by ganeshrkamath » Tue Nov 05, 2013 12:41 am
pareekbharat86 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

OA is E.
The diagonal of a square with sides of length a = a sqrt(2)
Area of the square = a^2 = (diagonal^2)/2

Area of 21-inch screen = 1/2 * 21^2
Area of 19-inch screen = 1/2 * 19^2

Difference = 1/2 * (21^2 - 19^2)
= 1/2 * (21 + 19) * (21 - 19)
= 1/2 * (40) * (2)
= 40 sq. inches

Choose E

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by [email protected] » Tue Nov 05, 2013 12:52 am
Hi pareekbharat86,

Rahul has properly explained the math behind this question. Here's the pattern that you need to know:

GMAT geometry questions emphasize certain formulas and relationships. Here, it's worth noting that EVERY square can be broken down into 2 45/45/90 triangles.

45/45/90 triangles have a relationship among their sides: X/X/X(root2)

Be on the lookout for this rule, as it's likely to be something that you're tested on during your Official GMAT.

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by GMATGuruNY » Tue Nov 05, 2013 1:50 am
pareekbharat86 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
We can save a little time if we know the following:

The area of a rhombus with diagonals d� and d₂ = (d�d₂)/2.
A square is a rhombus with 4 equal angles.
In a square, the two diagonals are equal.
Thus, the area of a square with diagonal d = d²/2.

a² - b² = (a+b)(a-b).

Thus:
Big TV - Little TV = 21²/2 - 19²/2 = (1/2)(21² - 19²) = (1/2)(21+19)(21-19) = 40.

The correct answer is E.
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by pareekbharat86 » Tue Nov 05, 2013 5:06 am
Thanks all of you.

My brain's got so fried with so much practicing that I kept getting 20 as the answer. I was using 1/2*base*height formula for a square!!! My bad.
Thanks,
Bharat.
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by rairavig » Tue Nov 05, 2013 11:04 pm
in above solutions it is clear the the difference between both areas can be calculated as:

Difference = 1/2 * (21^2 - 19^2) ..... (A)

here i wana advise a general way to calculate the difference between to squares.
i.e
(n+1)^2 - n^2 = 2n+1
(n+2)^2 - n^2 = 2(2n+2)
(n+3)^2 - n^2 = 3(2n+3)
.....

so here we may consider 19 as "n" so "n+2" is 21.
and difference in area is
Difference = 1/2 * (21^2 - 19^2):
Difference = 1/2 * (19+2)^2 - 19^2 = 1/2 * 2(2*19+2)= 40

this is more efficient way to find difference between big squares.
hope you like.
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by vipulgoyal » Wed Nov 06, 2013 12:55 am
awesome, did you drive it with this PS ??
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