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23 Jul 2020, 03:58
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
73% (02:31) correct
27%
(02:59)
wrong
based on 150
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If the radius of the circle above is 3 and point O is the center of the circle, what is the area of triangle ABC?
A. \(4 \sqrt{3}\)
B. \(6 \sqrt{2}\)
C. \(6 \sqrt{3}\)
D. \(12 \sqrt{2}\)
E. \(18 \sqrt{3}\)
Attachment:
1.jpg [ 18.68 KiB | Viewed 3830 times ]
Updated on: 23 Jul 2020, 18:06
Originally posted by CrackverbalGMAT on 23 Jul 2020, 04:41.
Last edited by CrackverbalGMAT on 23 Jul 2020, 18:06, edited 1 time in total.
Last edited by CrackverbalGMAT on 23 Jul 2020, 18:06, edited 1 time in total.
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Form chord CD as shown below
\(\angle{DAC}\) = \(30^o\), as the angle formed at the center by the chord CD (\(\angle{DOC}\)), will be twice the angle formed at the circumference by the same chord CD.
Triangle ABC, is a \(30^o\) - \(60^o\) - \(90^o\) triangle, with sides opposite 30-60-90 in the ration 1 : \(\sqrt{3}\) : 2
AC = 2r = 6
\(\frac{BC}{AC}\) =\(\frac{1}{\sqrt{3}}\)
BC = 6 * \(\frac{1}{\sqrt{3}}\) = 2\(\sqrt{3}\)
Area of the Triangle = \(\frac{1}{2}\) * Base * Height = = \(\frac{1}{2}\) * BC * AC = \(\frac{1}{2}\) * 2\(\sqrt{3}\) * 6 = 6\(\sqrt{3}\)
Option C
Arun Kumar
\(\angle{DAC}\) = \(30^o\), as the angle formed at the center by the chord CD (\(\angle{DOC}\)), will be twice the angle formed at the circumference by the same chord CD.
Triangle ABC, is a \(30^o\) - \(60^o\) - \(90^o\) triangle, with sides opposite 30-60-90 in the ration 1 : \(\sqrt{3}\) : 2
AC = 2r = 6
\(\frac{BC}{AC}\) =\(\frac{1}{\sqrt{3}}\)
BC = 6 * \(\frac{1}{\sqrt{3}}\) = 2\(\sqrt{3}\)
Area of the Triangle = \(\frac{1}{2}\) * Base * Height = = \(\frac{1}{2}\) * BC * AC = \(\frac{1}{2}\) * 2\(\sqrt{3}\) * 6 = 6\(\sqrt{3}\)
Option C
Arun Kumar
Attachments
Circle - Chord.jpg [ 403.66 KiB | Viewed 3633 times ]
