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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cans
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d=a*root(2)
area = (d^2)/2.
thus diff = (21^2 - 19^2)/2 = 40
IMO E
area = (d^2)/2.
thus diff = (21^2 - 19^2)/2 = 40
IMO E
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Cans!!
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Cans!!
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pemdas
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in squares we rule sqroot(2)*a=c where a is a side of the square and c is diagonal (from right triangles/ half of the square "45`-45` angle property rule")
21=sqroot(2)*a and 19=sqroot(2)*b where a and b are two different sides of two square screens
Hence a=21/sqroot(2) and b=19/sqroot(2), a^2-b^2=(21^2)/2 - 19^2/2= (21-19)(21+19)/2= (2*40)/2=40
e
hope this is shortcut too
21=sqroot(2)*a and 19=sqroot(2)*b where a and b are two different sides of two square screens
Hence a=21/sqroot(2) and b=19/sqroot(2), a^2-b^2=(21^2)/2 - 19^2/2= (21-19)(21+19)/2= (2*40)/2=40
e
hope this is shortcut too
abuc0112 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
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navami
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Consider the length of the diagonal of the larger screen = A = 21
Length of the larger television screen = A/ Root 2 = (21) / Root 2
Consider the length of the diagonal of the smaller screen = a = 19
Length of the smaller television screen = a /Root 2 = 19 /Root 2
now area of the larger screen = (21 / root 2)^2
now area of the smaller screen =(19 / root 2)^2
(21^2 - 19^2 )/ 2 = ((21 + 19 ) * (21 - 19 )) / 2
40
hence E
Length of the larger television screen = A/ Root 2 = (21) / Root 2
Consider the length of the diagonal of the smaller screen = a = 19
Length of the smaller television screen = a /Root 2 = 19 /Root 2
now area of the larger screen = (21 / root 2)^2
now area of the smaller screen =(19 / root 2)^2
(21^2 - 19^2 )/ 2 = ((21 + 19 ) * (21 - 19 )) / 2
40
hence E
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Navami
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Brent@GMATPrepNow
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Here's my approach (without skipping any stepsabuc0112 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 x be the length (and width) of the square screen with diagonal 21
The area of the large screen will be x^2
Let y be the length (and width) of the square screen with diagonal 19
The area of the small screen will be y^2
Our goal is to find the value of x^2 - y^2
Large TV: If we examine the right triangle created by 2 sides (both with length x) and the diagonal, we can apply the Pythagorean Theorem to get x^2 + x^2 = 21^2
When we simplify this, we get 2x^2 = 441, which means x^2 = 441/2
Small TV: If we examine the right triangle created by 2 sides (both with length y) and the diagonal, we can apply the Pythagorean Theorem to get y^2 + y^2 = 19^2
When we simplify this, we get 2y^2 = 361,, which means y^2 = 361/2
We can now find the value of x^2 - y^2
We get x^2 - y^2 = 441/2 - 361/2 = 80/2 = 40 = E
Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
