# #25 Grid-in

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from October 11, 2003 SAT I: Reasoning Test.

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Here's the question:

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ABCD is a rectangle

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Thanks for the response

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Do those marks through the figure indicate sides that are the same? I forget stupid notations like that.

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Yep. So x=CD.

I can't see why x can't have a wide range of values, i.e. 0<x<9.

[edited stupid inequalities]

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Since both sides of the triangle on the bottom left = x and theta=90

each other angle is 45 degrees.

So the angle in the middle is 180-90-45 = 45

If we draw a horizontal line from the point above X (O) to line AC, and say the intersection is point P,

the hypoteneuse is still sqrt(2x^2), each side is x

Now, AC = AP + PC

and we know PC = x

We now have a triangle APO, with a right angle and side x

sin(:lctheta:) = AP / x

AP = x*sin(:lctheta:)

AC = x+x*sin(theta)

So the area is x*(x+xsin(:lctheta:))

Now :lctheta: is some angle between 0 and 90 (because we already have 2 45 degree angles along that line)

Between 0 and 90, which means sin(:lctheta:) is between 0 and 1.

Solving for 8 and 18, we find that if we have the case of the minumum area of 8,

x=2sqrt(2)/sqrt(sin(:lctheta:) + 1)

and that if the maximum area of 18 is the case,

x=3sqrt(2)/sqrt(sin(:lctheta:) + 1)

Playing around with those numbers, you eventually get that x is going to be between 2 and 4.3.

In any case, I personally think that question would be thrown out because of how time consuming the correct answer set is.

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