GeminiHeat wrote:
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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.
√102B.
√142C.
2√2D.
2√3E.
2√5Attachment:
The attachment 2015-06-02_1741.png is no longer available
Area of square =
a2=4.5=92i.e.
a=3√2We know, diagonal of a square is
√2a i.e.
Diagonal=3We are also given that
DQQB=12DQ + QB = DB (diagonal)
x + 2x = 3
x = 1
i.e. DQ = 1 and QB = 2
Now, drop a perpendicular CE on DB (refer to the figure below)
CE will bisect DB (as BCD is an isosceles right triangle)
i.e. DE = BE =
32Since, the diagonals in a square bisect at 90, point E is the centre of square
i.e. AE = BE = CE = DE =
32Also, DQ + QE = DE
i.e. 1 + QE =
32QE =
12Finally, Applying Pythagoras Theorem in triangle CEQ;
CE2+QE2=CQ2(32)2+(12)2=x2x2=104x=√102Hence, option A
Attachments
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