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15 Jun 2020, 05:12
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
15% (02:47) correct
85%
(01:51)
wrong
based on 73
sessions
History
Date
Time
Result
Not Attempted Yet
In the diagram above, what is the value of x?
A. 6
B. 8
C. \(\sqrt{71}\)
D. \(\sqrt{80}\)
E. 9
Are You Up For the Challenge: 700 Level Questions
Attachment:
2020-06-15_1607.png [ 48.2 KiB | Viewed 5250 times ]
firas92
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Joined: 16 Jan 2019
Last visit: 24 Jul 2026
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Given Kudos: 142
Location: India
Concentration: General Management
Schools: Tepper '23 (M$) Foster '23 (D)
GMAT 1: 740 Q50 V40
WE:Sales (Other)
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15 Jun 2020, 13:49
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Extend the diagram to make an 8x4 rectangle as shown
So the diagonal of this rectangle BC is \(\sqrt{8^2+4^2} = \sqrt{80}\)
Now from right triangle ABC we have
\(x=\sqrt{(\sqrt{80})^2-3^2}=\sqrt{80-9}=\sqrt{71}\)
Answer is (C)
So the diagonal of this rectangle BC is \(\sqrt{8^2+4^2} = \sqrt{80}\)
Now from right triangle ABC we have
\(x=\sqrt{(\sqrt{80})^2-3^2}=\sqrt{80-9}=\sqrt{71}\)
Answer is (C)
Attachments
2020-06-15_1607.png [ 60.34 KiB | Viewed 4682 times ]
02 Nov 2021, 01:24
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Bunuel
beeblebrox, of course almost all answers above are incorrect, but it is a simple question if you draw that 'all important additional line'.
Join BD as shown.
WHY? Because we will be able to convert x as a side of a right angled triangle.
Mark the remaining point as shown in the attached figure.
Take \(\triangle ABC\) and \(\triangle EDC\)
1) \(\angle ACB\) = \(\angle ECD\) - Opposite angles
2) \(\angle CAB\) = \(\angle CED\) - 90
3) Remaining angle has to be equal to each other, both being 90-a or b here
Thus, \(\triangle ABC\) is similar to \(\triangle EDC\), so sides are in the same ratio.
As sides AB:DE = 3:5, AC and CE will also be in the ratio 3:5.
So, AC = \(4*\frac{3}{3+5}*\frac{12}{8}=\frac{3}{2}\) and EC = \(4*\frac{5}{3+5}*\frac{20}{8}=\frac{5}{2}\)
Thus, from \(\triangle ABC\)
\(BC =\sqrt{ AC^2+AB^2}=\sqrt{3^2+\frac{3}{2}^2}=\sqrt{\frac{45}{4}}=\frac{3\sqrt{5}}{2}\)
Also, from \(\triangle DEC\)
\(DC =\sqrt{ DE^2+CE^2}=\sqrt{5^2+\frac{5}{2}^2}=\sqrt{\frac{125}{4}}=\frac{5\sqrt{5}}{2}\)
Therefore, BD = BC + CD = \(\frac{3\sqrt{5}}{2}+\frac{5\sqrt{5}}{2}=\frac{8\sqrt{5}}{2}=4\sqrt{5}\)
Now, take \(\triangle BDF\)
DF = x = \(BD^2-FD^2=\sqrt{4\sqrt{5}^2-3^2}=\sqrt{80-9}=\sqrt{71}\)
C
Attachments
2020-06-15_1607.png [ 56.58 KiB | Viewed 3355 times ]
General Discussion
