MGMAT: area of a triangle

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Nermal: Senior | Next Rank: 100 Posts; Posts: 77; Joined: Sun Jun 21, 2009 10:25 am; Location: Germany; Thanked: 7 times

by Nermal » Sat Sep 12, 2009 1:35 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

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Source: Beat The GMAT — Data Sufficiency | Sat Sep 12, 2009 1:35 am

dtweah: Master | Next Rank: 500 Posts; Posts: 487; Joined: Fri Mar 27, 2009 5:49 am; Thanked: 36 times

Re: MGMAT: area of a triangle

by dtweah » Sat Sep 12, 2009 2:06 am

Nermal wrote: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

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To find the area of a circle we need only the radius. There is no way we can deduce the radius from this circle from any of its chords or figures. Clearly 1 is not suff.

Under 2, knowing any of the angles doesn't help also because we still don't know the Diameter, ie No CENTER.
Combining is futile.

Choose E.

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rohan_vus: Master | Next Rank: 500 Posts; Posts: 189; Joined: Thu Apr 03, 2008 2:03 pm; Location: USA; Thanked: 21 times

by rohan_vus » Sat Sep 12, 2009 4:35 am

IMO A . First you can deduce radius . You know DA = 4 and the DAB is right angle triangle and inside a circle DB must be the diameter as any chord lesser than diameter will not be able to suspend angle 90 degree inside a circle.

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nauruz: Newbie | Next Rank: 10 Posts; Posts: 6; Joined: Sun Aug 02, 2009 1:25 am

by nauruz » Sat Sep 12, 2009 4:55 am

rohan_vus wrote:IMO A . First you can deduce radius . You know DA = 4 and the DAB is right angle triangle and inside a circle DB must be the diameter as any chord lesser than diameter will not be able to suspend angle 90 degree inside a circle.

ok, but in a right triangle knowing only one side, how would you find the other?

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rohan_vus: Master | Next Rank: 500 Posts; Posts: 189; Joined: Thu Apr 03, 2008 2:03 pm; Location: USA; Thanked: 21 times

by rohan_vus » Sat Sep 12, 2009 5:13 am

This is how u get . angle ADB = 60 degree . based on priciple of equal chord suspend equal angle on the circle. Angle ADB = Angle ACB = 60 degree .

Now you know angle ADB = 60 then diameter DB is easy to get as DB * cos 60 degree = DA (right angle theory) .....and then thus u get DB and hence the area

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glorydefined: Senior | Next Rank: 100 Posts; Posts: 43; Joined: Tue Mar 31, 2009 9:54 am; Thanked: 2 times

by glorydefined » Sat Sep 12, 2009 7:38 am

Answer A.

Given : DB = diameter (since angle(DAB)=90), triangle ADB and ACB are supplementary as angle(ADB)=angle(ACB), triangle ACB = equilateral triangle and each angle is 60 degree in equilateral triangle.

From statement 1 : we know DA=4. angle(ADB)=angle(ACB)=60. And so trianlge ADB is of 30-60-90 format and the sides are of a,(sqrt)3 a,2a respectively. since DA=4, AD=(sqrt)3 *4, and DB=8. We know the diameter and we can find the radius and the area. Sufficient.

so answer is between AD.

From statement 2: we know angle(ABD)=30, with this we only know that the trianlge ADB is of 30-60-90 format and the sides are a,(sqrt)3 a,2a respectively. But we dont know the sides (consider only statement B). So not enough. Not sufficient.

so Answer A

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hariharakarthi: Master | Next Rank: 500 Posts; Posts: 131; Joined: Tue Apr 22, 2008 1:54 pm; Thanked: 3 times; GMAT Score:550

by hariharakarthi » Sat Sep 12, 2009 7:55 pm

@glorydefined
Ans A is correct. But, we can not say DB is a Diameter. As center of the the circle is not given.

1. DA =4
Angle of ADB =60. Since, ABC is a equilaeral triangle and hence each angle is 60. Angle made by a chord at the circumfernece is equal. Hence, ADB =60. Now, we know two angles of triangle DAB and thrid angle is 30. We know side DA=4. We can calculated (DB)one side of triangle.

Now, circle is passing through the vertices of triangle.
radius of circle = side/sqrt(3).
From, we can calculate the area. Hence it is Suff.

2. Angle ABD = 30. But, this we already calculated using statement 1. Hence NS.

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hariharakarthi: Master | Next Rank: 500 Posts; Posts: 131; Joined: Tue Apr 22, 2008 1:54 pm; Thanked: 3 times; GMAT Score:550

by hariharakarthi » Sat Sep 12, 2009 7:59 pm

@glorydefined

You are correct. DB is diameter. The circle passes through the vertices of triangle DAB. In this case, hypotnuse of the triangle is diameter of the circle.

Thanks,
hhk

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