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705-805 Level|   Geometry|      
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If angle ABC is 30 degrees, what is the area of triangle BCE? 05 Nov 2019, 06:00
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If angle ABC is 30 degrees, what is the area of triangle BCE?

(1) Angle CDF is 120 degrees, lines L and M are parallel, and AC = 6, BC = 12, and EC = 2AC
(2) Angle DCG is 60 degrees, angle CDG is 30 degrees, angle FDG = 90, and GC = 6, CD = 12 and EC = 12


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D
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If angle ABC is 30 degrees, what is the area of triangle BCE? 05 Nov 2019, 08:49
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If angle ABC is 30 degrees, what is the area of triangle BCE?

STATEMENT (1) Angle CDF is 120 degrees, lines L and M are parallel, and AC = 6, BC = 12, and EC = 2AC

in triangle BCE
BC = 12 and EC = 12 (this proves that BCE is isosceles triangle)
now AB bisects side EC in two equal parts
AB is the median = altitude = angle bisector (since BCE is an isosceles triangle)
we can find the altitude AB
now we can find the area of triangle BCE
SUFFICIENT

STATEMENT (2) Angle DCG is 60 degrees, angle CDG is 30 degrees, angle FDG = 90, and GC = 6, CD = 12 and EC = 12

in triangle BCE
EC = 12 (base) we need to know the height of the triangle

angle DGC = 90
height of triangle BCE = DGC that is = 6
now we know the base of triangle BCE = 12
height of BCE = 6
We can find the area
SUFFICIENT

D is the answer
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If angle ABC is 30 degrees, what is the area of triangle BCE? 05 Nov 2019, 22:24
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D is the answer

As

Option one we can easily find the area of triangle being equilateral as we know the angle.

So we AD/BCE one is sufficient

AD

Now in second point the two lines are parallel meaning which both trainable are same

So D

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If angle ABC is 30 degrees, what is the area of triangle BCE? 30 Dec 2019, 10:14
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shridhar786

height of triangle BCE = DGC that is = 6
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I am being picky here, but the height is actually \(6\sqrt{2}\)

Your solution is flawless otherwise.
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If angle ABC is 30 degrees, what is the area of triangle BCE? 17 Jan 2020, 01:16
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Very good question...worth trying !!!
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If angle ABC is 30 degrees, what is the area of triangle BCE? 15 Feb 2020, 11:59
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shridhar786
If angle ABC is 30 degrees, what is the area of triangle BCE?

STATEMENT (1) Angle CDF is 120 degrees, lines L and M are parallel, and AC = 6, BC = 12, and EC = 2AC

in triangle BCE
BC = 12 and EC = 12 (this proves that BCE is isosceles triangle)
now AB bisects side EC in two equal parts
AB is the median = altitude = angle bisector (since BCE is an isosceles triangle)
we can find the altitude AB
now we can find the area of triangle BCE
SUFFICIENT

STATEMENT (2) Angle DCG is 60 degrees, angle CDG is 30 degrees, angle FDG = 90, and GC = 6, CD = 12 and EC = 12

in triangle BCE
EC = 12 (base) we need to know the height of the triangle

angle DGC = 90
height of triangle BCE = DGC that is = 6
now we know the base of triangle BCE = 12
height of BCE = 6
We can find the area
SUFFICIENT

D is the answer
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For Highlighted part, For an Isosceles Triangle, doesn't the mentioned rule only apply to the unequal side? Side EB is the unequal side...If so, how else do we solve this?
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If angle ABC is 30 degrees, what is the area of triangle BCE? 12 Mar 2021, 08:42
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Can anyone please explain how is AB the angle bisector? We only know that it is median.

Additionally, even if we say that triangle BCE is an isosceles triangle, we still cannot infer that AB is the angle bisector, since only the median that falls on the unequal side from the opposite vertex is the angle bisector. Here however that is not the case.

Bunuel
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chetan2u

Could you please help explain this? It will really help. Have searched all sources however cannot figure it out.
Thanks
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If angle ABC is 30 degrees, what is the area of triangle BCE? 13 Mar 2021, 07:27
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Can anyone please explain how is AB the angle bisector? We only know that it is median.

Additionally, even if we say that triangle BCE is an isosceles triangle, we still cannot infer that AB is the angle bisector, since only the median that falls on the unequal side from the opposite vertex is the angle bisector. Here however that is not the case.

Bunuel
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chetan2u

Could you please help explain this? It will really help. Have searched all sources however cannot figure it out.
Thanks
Show more


You are right. You require further info.
Nothing tells us that BAC is 90.

Something is to be added : Either DC is parallel to BE or AB is the shortest distance between L and M.
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If angle ABC is 30 degrees, what is the area of triangle BCE? 14 Mar 2021, 02:26
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pgwodehouse
Can anyone please explain how is AB the angle bisector? We only know that it is median.

Additionally, even if we say that triangle BCE is an isosceles triangle, we still cannot infer that AB is the angle bisector, since only the median that falls on the unequal side from the opposite vertex is the angle bisector. Here however that is not the case.

Bunuel
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chetan2u

Could you please help explain this? It will really help. Have searched all sources however cannot figure it out.
Thanks


You are right. You require further info.
Nothing tells us that BAC is 90.

Something is to be added : Either DC is parallel to BE or AB is the shortest distance between L and M.
Show more
Hi,

I think we can use this property: Two lines that are perpendicular to the same line are parallel. Hence, line L and line M are parallel to each other.
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If angle ABC is 30 degrees, what is the area of triangle BCE? 14 Mar 2021, 03:28
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pgwodehouse
Can anyone please explain how is AB the angle bisector? We only know that it is median.

Additionally, even if we say that triangle BCE is an isosceles triangle, we still cannot infer that AB is the angle bisector, since only the median that falls on the unequal side from the opposite vertex is the angle bisector. Here however that is not the case.

Bunuel
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chetan2u

Could you please help explain this? It will really help. Have searched all sources however cannot figure it out.
Thanks


You are right. You require further info.
Nothing tells us that BAC is 90.

Something is to be added : Either DC is parallel to BE or AB is the shortest distance between L and M.
Hi,

I think we can use this property: Two lines that are perpendicular to the same line are parallel. Hence, line L and line M are parallel to each other.
Show more

But where is it given that any line is perpendicular. Line AB may look like a perpendicular but you cannot assume it. It could be just 89 too.
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If angle ABC is 30 degrees, what is the area of triangle BCE? 14 Mar 2021, 03:39
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chetan2u

Hello :)

Statement 1 is sufficient.

Per statement 2 , angle FDG AND CDG are equal to 90 degrees. So line L is parralel to line M. Subsequently, triangle ABC gets fixed and has a definite invariable area. I maybe wrong 😂

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If angle ABC is 30 degrees, what is the area of triangle BCE? 14 Mar 2021, 04:07
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chetan2u

Hello :)

Statement 1 is sufficient.

Per statement 2 , angle FDG AND CDG are equal to 90 degrees. So line L is parralel to line M. Subsequently, triangle ABC gets fixed and has a definite invariable area. I maybe wrong 😂

Posted from my mobile device
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Hi

Statement II is sufficient, but I don’t think statement I is sufficient because nothing tells us that AB is perpendicular or DC is parallel to BE. We just know that ABC is isosceles but our solution requires it to be equilateral
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If angle ABC is 30 degrees, what is the area of triangle BCE? 15 Mar 2021, 02:42
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Bunuel

If angle ABC is 30 degrees, what is the area of triangle BCE?

(1) Angle CDF is 120 degrees, lines L and M are parallel, and AC = 6, BC = 12, and EC = 2AC
(2) Angle DCG is 60 degrees, angle CDG is 30 degrees, angle FDG = 90, and GC = 6, CD = 12 and EC = 12


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The answer here is (D). Each statement alone is sufficient. Statement 1 alone is sufficient too because of the given data. Normally, I agree it would not be sufficient.

Statement 1:

First think about this - we know that SAS is a congruency rule but SSA is not. So we cannot make a unique triangle using SSA (two sides and one non-included angle). We can make two such triangles given side-side-angle.
Here is a simple video that explains this: https://www.youtube.com/watch?v=8jaVoGOwbC0
The areas of the two triangles would be different.

Considering ACB triangle in our figure, we know AC, CB and angle CBA. So normally, it would give us two such triangles, one acute and one obtuse (as shown in the video). But the measurements given are such that it will give us a 90 degree triangle and hence only one such triangle is possible.

Using the sine law, we know x/Sin X = y/Sin Y
So AC/Sin 30 = BC/Sin BAC
6/Sin 30 = 12/Sin BAC
Since Sin 30 = 1/2, we get 12 = 12/Sin BAC
Sin BAC = 1 which means angle BAC = 90 degrees
(Normally we would get two values of angle BAC because sin Q = sin (180 - Q). So when Q = 90, we get a single value.

This makes BA the altitude of triangle BEC. We can easily find BA because BAC is a 30-60-90 triangle so BA = 6*sqrt(3)
Area of triangle BEC = (1/2) * Altitude * Base \(= \frac{1}{2} * 6\sqrt{3}*12\)

Sufficient


Statement II.
DG is the perpendicular distance between the parallel lines L and M.
Since CDG is a 30-60-90 triangle, we get that DG = 6*sqrt(3)

Since triangle BEC is between the same two parallel lines, the length of altitude will be the same i.e. 6*sqrt(3).

Area of triangle BEC = (1/2) * Altitude * Base \(= \frac{1}{2} * 6\sqrt{3}*12\)

Sufficient


Note: Sufficiency of statement 1 may not be within GMAT scope.
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If angle ABC is 30 degrees, what is the area of triangle BCE? 17 Nov 2022, 08:31
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