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24 Mar 2020, 01:09
Kudos
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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:
25%
(medium)
Question Stats:
82% (02:41) correct
18%
(02:49)
wrong
based on 17
sessions
History
Date
Time
Result
Not Attempted Yet
Attachment:
12.jpg [ 8.79 KiB | Viewed 1526 times ]
In the figure above, BC || AD. What is the area of quadrilateral ABCD ?
(A) 5\(\sqrt{2}\)
(B) \(\frac{9\sqrt{3}}{2}\)
(C) \(\frac{27\sqrt{3}}{4}\)
(D) \(\frac{27}{2}\)
(E) 16
26 Mar 2020, 01:46
Kudos
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SajjadAhmad
Solution:
- • ABCD is a trapezium.
- o BC||AD
o \(∠BCD + ∠ADC = 180\)
o \(∠BCD = 180-120 = 60\).
- o In triangle DMC
o \(∠DMC = 90\)
- DMC is a 30-60-90 triangle.
\(CM:DM:DC = 1: \sqrt{3}: 2\)
- o \(Area of ABCD = \frac{1}{2} *DM*(AD+BC)= \frac{1}{ 2}* 3 \frac{\sqrt{3}}{2} *(5+4) =\frac{27 \sqrt{3}}{4}\)
