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In the diagram above, figure ABCD is a square with an area of 4.5 in^2
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30 May 2021, 23:23
30 May 2021, 23:23
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Question Stats:
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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. \(\frac{\sqrt{10}}{2}\)
B. \(\frac{\sqrt{14}}{2}\)
C. \(2\sqrt{2}\)
D. \(2\sqrt{3}\)
E. \(2\sqrt{5}\)
Show: ::
Attachment:
2015-06-02_1741.png [ 9.37 KiB | Viewed 1658 times ]
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In the diagram above, figure ABCD is a square with an area of 4.5 in^2
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03 Jun 2021, 23:49
03 Jun 2021, 23:49
1
GeminiHeat 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. \(\frac{\sqrt{10}}{2}\)
B. \(\frac{\sqrt{14}}{2}\)
C. \(2\sqrt{2}\)
D. \(2\sqrt{3}\)
E. \(2\sqrt{5}\)
Show: ::
Attachment:
The attachment 2015-06-02_1741.png is no longer available
Area of square = \(a^2 = 4.5 = \frac{9}{2}\)
i.e. \(a = \frac{3}{\sqrt{2}}\)
We know, diagonal of a square is \(\sqrt{2}a\) i.e. \(Diagonal = 3\)
We are also given that \(\frac{DQ}{QB} = \frac{1}{2}\)
DQ + 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 = \(\frac{3}{2}\)
Since, the diagonals in a square bisect at 90, point E is the centre of square
i.e. AE = BE = CE = DE = \(\frac{3}{2}\)
Also, DQ + QE = DE
i.e. 1 + QE = \(\frac{3}{2}\)
QE = \(\frac{1}{2}\)
Finally, Applying Pythagoras Theorem in triangle CEQ;
\(CE^2 + QE^2 = CQ^2\)
\((\frac{3}{2})^2 + (\frac{1}{2})^2 = x^2\)
\(x^2 = \frac{10}{4} \)
\(x = \frac{\sqrt{10}}{2}\)
Hence, option A
Attachments
what is the length of QC.png [ 21.27 KiB | Viewed 1408 times ]
General Discussion
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Re: In the diagram above, figure ABCD is a square with an area of 4.5 in^2
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05 Jun 2021, 23:36
05 Jun 2021, 23:36
1
KarunMendiratta wrote:
GeminiHeat 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. \(\frac{\sqrt{10}}{2}\)
B. \(\frac{\sqrt{14}}{2}\)
C. \(2\sqrt{2}\)
D. \(2\sqrt{3}\)
E. \(2\sqrt{5}\)
Show: ::
Attachment:
2015-06-02_1741.png
Area of square = \(a^2 = 4.5 = \frac{9}{2}\)
i.e. \(a = \frac{3}{\sqrt{2}}\)
We know, diagonal of a square is \(\sqrt{2}a\) i.e. \(Diagonal = 3\)
We are also given that \(\frac{DQ}{QB} = \frac{1}{2}\)
DQ + 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 = \(\frac{3}{2}\)
Since, the diagonals in a square bisect at 90, point E is the centre of square
i.e. AE = BE = CE = DE = \(\frac{3}{2}\)
Also, DQ + QE = DE
i.e. 1 + QE = \(\frac{3}{2}\)
QE = \(\frac{1}{2}\)
Finally, Applying Pythagoras Theorem in triangle CEQ;
\(CE^2 + DE^2 = CQ^2\)
\((\frac{3}{2})^2 + (\frac{1}{2})^2 = x^2\)
\(x^2 = \frac{10}{4} \)
\(x = \frac{\sqrt{10}}{2}\)
Hence, option A
This must be \(QE^2\)
