Anaira Mitch wrote:In the diagram above, figure ABCD is a square with an area of 4.5 in^2. If the ratio of the length of DQ to the length of QB is 1 to 2, what is the length of QC, in inches?
A. sqrt(10)/2
B. sqrt(14)/2
C. 2√2
D. 2sqrt(3)
E. 2sqrt(5)
Hi Anaira Mitch,
We have area of square ABCD = 4.5.
=> Side of the square = Sqrt(4.5) = Sqrt(9/2) = 3/Sqrt(2)
=> Diagonal = BD = [3/Sqrt(2)]*Sqrt(2) = 3
Since BD is divided in the ratio of the length of DQ to the length of QB is 1 to 2, we have DQ = 1 and QB = 2.
From the point Q, drop a perpendicular to DC. It meets DC at point P.
This forms a rightangled traingle DQP, where angle QPD =90, angle QDP = angle DQP = 45.
Since DQ = 1, QP = 1/Sqrt(2).
Since DC = side of the square = 3/Sqrt(2), we have PC = 3/Sqrt(2) - 1/Sqrt(2) = 2/Sqrt(2)
In the rightangled traingle QPC,
QC^2 = QP^2 + PC^2 = 1/2 + 4/2 = 5/2
=> QC = Sqrt(5/2) = Sqrt(10/4) = [spoiler]Sqrt(10)/2[/spoiler].
The correct answer:
A
Hope this helps!
Relevant book:
Manhattan Review GMAT Geometry Guide
-Jay
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