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Master | Next Rank: 500 Posts
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by tanyajoseph » Thu Aug 23, 2007 11:59 pm
The size of a television screen is given as the length of the screens 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 square.

A) 2
B) 4
C) 16
D) 38
E) 40
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by erdnah » Fri Aug 24, 2007 12:45 am
a x sqrt(2) = 21
b x sqrt(2) = 19

sqrt(2) = 21/a
2 = (21/a)² = 441/a^2
a^2 = 220.5

sqrt(2) = 19/b
2 = (19/b)² = 361/b^2
b^2 = 180.5

220.5-180.5 = 40
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by agps » Fri Aug 24, 2007 12:49 am
Given that for a square the diagonal is 2 times the side s squared (2s^2)
and the area of the square being s^2

for the 21 inch screen (21 inch diagonal) s^2=21^2/2 = 441/2

for the 19 inch screen s^2=19^2/2 = 361/2

so the diference between the two would be (441-361)/2 or 80/2 = 40

E is the answer
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