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Inscribed Triangles

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by sureng » Wed Apr 13, 2011 7:08 am
In the figure, ABC is an equilateral triangle, and DAB is a right triangle. What is the area of the circumscribed circle?

(1) DA = 4
(2) Angle ABD = 30 degrees

Figure:
https://www.postimage.org/image.php?v=aVRSRli



Can some one pls. clarify from below link how the 2 angles ADB and ACB are equal?
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Source: — Data Sufficiency |
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by HSPA » Wed Apr 13, 2011 7:14 am
I thought I had enough math for today.. but couldnt stop

sin 30 = 4/diameter => radius = 4 and area is 16pi

I got C
First take: 640 (50M, 27V) - RC needs 300% improvement
Second take: coming soon..
Regards,
HSPA.
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by Brent@GMATPrepNow » Wed Apr 13, 2011 7:18 am
sureng wrote:
Can some one pls. clarify from below link how the 2 angles ADB and ACB are equal?
In the question, angles ADB and ACB are both "holding" the same chord, and we have a nice rule that says, "Inscribed angles holding the same chord are equal" (see attachment)

Please note that, for the rule to apply, the inscribed angles in question must be on the same side of the chord.
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by manpsingh87 » Wed Apr 13, 2011 7:20 am
sureng wrote: Can some one pls. clarify from below link how the 2 angles ADB and ACB are equal?
well its a theorem which states, all inscribed angle that subtends the same arc are equal..!!!

well as far as question is concerned i believe answer should be A.!!!
O Excellence... my search for you is on... you can be far.. but not beyond my reach!
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by Brent@GMATPrepNow » Wed Apr 13, 2011 8:05 am
This question begs a diagrammed solution:
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by sureng » Wed Apr 13, 2011 9:06 am
Thanks Brent for detailed explanation.
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by sureng » Wed Apr 13, 2011 9:08 am
HSPA wrote:I thought I had enough math for today.. but couldnt stop

sin 30 = 4/diameter => radius = 4 and area is 16pi

I got C
HSPA: sorry i didn't get you. Looks like you have quick shortcut for the solution:)
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by HSPA » Wed Apr 13, 2011 5:16 pm
sureng wrote:
HSPA wrote:I thought I had enough math for today.. but couldnt stop

sin 30 = 4/diameter => radius = 4 and area is 16pi

I got C
HSPA: sorry i didn't get you. Looks like you have quick shortcut for the solution:)
Quick, correct but cheap technique as I missed what Bernt has provided...the answer is wrong but approach is correct.

Sin A = opposite side to angle/ hypotenuse.
Here hypotenuse is diameter and oppisite side is givne as 4.
Theorm: angle subtended in semicircle is right angle so hypotenuse = diameter

why my techniqui is effective but cheap... I took option B and took the angle = 30
we dont need this... if we take bernt's another theorm above...
as we already know the angles are 60,30 in the right angle... Area is 16pi and answer option is not C but A is correct answer..
Hope this helps...
First take: 640 (50M, 27V) - RC needs 300% improvement
Second take: coming soon..
Regards,
HSPA.
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