from October 11, 2003 SAT I: Reasoning Test.

Here's the question:

ABCD is a rectangle

 

 

Thanks for the response

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

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]

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.