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30 Oct 2020, 22:37
Kudos
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
68% (02:40) correct
32%
(01:59)
wrong
based on 22
sessions
History
Date
Time
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Not Attempted Yet
What is the area of the triangle in the figure shown above?
A) 9+9√2
B) 9+9√3
C) 9√2+9√3
D) 18+18√2
E) 18 +18√3
Attachment:
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30 Oct 2020, 23:10
Kudos
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GMATBusters
Dropping a perpendicular AD as shown in the figure, we get 2 triangles.
Triangle ADB is a 45 - 45 - 90 triangle, where the ratio of sides opposite 45 - 45 - 90 = 1 : 1 : \(\sqrt{2}\)
AD = BD
\(\frac{AD}{AB} = \frac{1}{\sqrt{2}}\)
\(\frac{AD}{6} = \frac{1}{\sqrt{2}}\)
\(AD = \frac{6}{\sqrt{2}} = 3\sqrt{2}\)
Triangle ADC is a 30 - 60 - 90 triangle, where the ratio of sides opposite 30 - 60 - 90 = 1 : \(\sqrt{3}\) : 2
\(\frac{AD}{DC} = \frac{1}{\sqrt{3}}\)
\(\frac{3\sqrt{2}}{DC} = \frac{1}{\sqrt{3}}\)
\(DC = 3\sqrt{6}\)
The Base = BD + DC = \(3\sqrt{2} + 3\sqrt{6} = 3\sqrt{2} (1 + \sqrt{3})\)
Area = 1/2 * Base * Height = \(\frac{1}{2} * 3\sqrt{2}(1 + \sqrt{3}) * 3\sqrt{2} = \frac{1}{2} * 18 * (1 + \sqrt{3}) = 9(1 + \sqrt{3}) = 9 + 9\sqrt{3}\)
Option B
Arun Kumar
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