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If angle ABC is 30 degrees, what is the area of triangle BCE? [#permalink]
27 Aug 2011, 09:19

1

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

44% (02:49) correct
56% (02:20) wrong based on 80 sessions

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

Re: If angle ABC is 30 degrees, what is the area of triangle BCE? [#permalink]
27 Aug 2011, 09:23

Not sure how to add this in a hide mode, this is the OA

#

1. Since EC = 2AC, EA = CA, EC = 2(6) = 12 and line AB is an angle bisector of angle EBC. This means that angle ABC = angle ABE. Since we know that angle ABC = 30, we know that angle ABE = 30. Further, since lines L and M are parallel, we know that line AB is perpendicular to line EC, meaning angle BAC is 90. 2. Since all the interior angles of a triangle must sum to 180: angle ABC + angle BCA + angle BAC = 180 30 + angle BCA + 90 = 180 angle BCA = 60 3. Since all the interior angles of a triangle must sum to 180: angle BCA + angle ABC + angle ABE + angle AEB = 180 60 + 30 + 30 + angle AEB = 180 angle AEB = 60 4. This means that triangle BCA is an equilateral triangle. 5. To find the area of triangle BCE, we need the base (= 12 from above) and the height, i.e., line AB. Since we know BC and AC and triangle ABC is a right triangle, we can use the Pythagorean theorem on triangle ABC to find the length of AB. 62 + (AB)2 = 122 AB2 = 144 - 36 = 108 AB = 1081/2 6. Area = .5bh Area = .5(12)(1081/2) = 6*1081/2 7. Statement (1) is SUFFICIENT

# Evaluate Statement (2) alone.

1. The sum of the interior angles of any triangle must be 180 degrees. DCG + GDC + CGD = 180 60 + 30 + CGD = 180 CGD = 90 Triangle CGD is a right triangle. 2. Using the Pythagorean theorem, DG = 1081/2 (CG)2 + (DG)2 = (CD)2 62 + (DG)2 = 122 DG = 1081/2 3. At this point, it may be tempting to use DG = 1081/2 as the height of the triangle BCE, assuming that lines AB and DG are parallel and therefore AB = 1081/2 is the height of triangle BCE. However, we must show two things before we can use AB = 1081/2 as the height of triangle BCE: (1) lines L and M are parallel and (2) AB is the height of triangle BCE (i.e., angle BAC is 90 degrees). 4. Lines L and M must be parallel since angles FDG and CGD are equal and these two angles are alternate interior angles formed by cutting two lines with a transversal. If two alternate interior angles are equal, we know that the two lines that form the angles (lines L and M) when cut by a transversal (line DG) must be parallel. 5. Since lines L and M are parallel, DG = the height of triangle BCE = 1081/2. Note that it is not essential to know whether AB is the height of triangle BCE. It is sufficient to know that the height is 1081/2. To reiterate, we know that the height is 8 since the height of BCE is parallel to line DG, which is 1081/2. 6. Since we know both the height (1081/2) and the base (CE = 12) of triangle BCE, we know that the area is: .5*12*1081/2 = 6*1081/2 7. Statement (2) alone is SUFFICIENT.

# Since Statement (1) alone is SUFFICIENT and Statement (2) alone is SUFFICIENT, answer D is correct.

Re: If angle ABC is 30 degrees, what is the area of triangle BCE? [#permalink]
27 Aug 2011, 20:49

I see more assumptions than statements in the following part

and line AB is an angle bisector of angle EBC. This means that angle ABC = angle ABE. Since we know that angle ABC = 30, we know that angle ABE = 30. Further, since lines L and M are parallel, we know that line AB is perpendicular to line EC, meaning angle BAC is 90. _________________

Re: If angle ABC is 30 degrees, what is the area of triangle BCE? [#permalink]
27 Aug 2011, 22:20

DeepGagan wrote:

I see more assumptions than statements in the following part

and line AB is an angle bisector of angle EBC. This means that angle ABC = angle ABE. Since we know that angle ABC = 30, we know that angle ABE = 30. Further, since lines L and M are parallel, we know that line AB is perpendicular to line EC, meaning angle BAC is 90.

Agreed. Point B can easily be shifted to the left and still comply with all the data, but the angle will not be 90. _________________

Re: If angle ABC is 30 degrees, what is the area of triangle BCE? [#permalink]
14 Jun 2014, 11:13

2

This post was BOOKMARKED

Attachment:

DS.gif [ 3.64 KiB | Viewed 1001 times ]

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

Last edited by Bunuel on 14 Jun 2014, 11:34, edited 2 times in total.

Re: If angle ABC is 30 degrees, what is the area of triangle BCE? [#permalink]
20 Mar 2015, 16:58

gaurav245 wrote:

Attachment:

DS.gif

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

The answer I am getting does not match the OA. Please let me know what I am doing wrong. Here's my approach to this problem.

From Stem, angle ABC = 30º

From (1), AC = 6 & BC = 12

Using Law of Sines, sin 30º/AC = sin BAC/BC So, Sin BAC = 1 => angle BAC = 90º => AB is the altitude and by applying pythagorean theorem, we have, AB = \(6\sqrt{3}\). Since, we have AB and EC, it is possible to find area of triangle BCE. Thus, this is SUFFICIENT.

From (2), angle FDG = 90º, GC = 6, CD = 12, angle DCG = 60º and angle CDG = 30º. Thus, triangle CDG is a right triangle with angle DGC = 90º. Hence, we can conclude that L || M.

EC = 12, however, we can't assume that AB is perpendicular to EC, nor can we prove this based on the information provided. Hence, this information is INSUFFICIENT.

Re: If angle ABC is 30 degrees, what is the area of triangle BCE? [#permalink]
20 Mar 2015, 19:41

rdoshi wrote:

gaurav245 wrote:

Attachment:

DS.gif

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

The answer I am getting does not match the OA. Please let me know what I am doing wrong. Here's my approach to this problem.

From Stem, angle ABC = 30º

From (1), AC = 6 & BC = 12

Using Law of Sines, sin 30º/AC = sin BAC/BC So, Sin BAC = 1 => angle BAC = 90º => AB is the altitude and by applying pythagorean theorem, we have, AB = \(6\sqrt{3}\). Since, we have AB and EC, it is possible to find area of triangle BCE. Thus, this is SUFFICIENT.

From (2), angle FDG = 90º, GC = 6, CD = 12, angle DCG = 60º and angle CDG = 30º. Thus, triangle CDG is a right triangle with angle DGC = 90º. Hence, we can conclude that L || M.

EC = 12, however, we can't assume that AB is perpendicular to EC, nor can we prove this based on the information provided. Hence, this information is INSUFFICIENT.

Therefore, answer (according to me) is A.

(2) is sufficient as well. Let me explain.

Area of all the triangles drawn on the same base on one of the parallel lines with third vertex on the other parallel line is same.

You have already figured out that L||M so here it goes.

From the diagram above:

Area of triangle BCE = Area of triangle DCE

Area of triangle DCE = 1/2 * EC * DG

You already know EC and using pythagorean theoram can find DG. Hence, you can get area of triangle DCE and also BCE. _________________

Regards, J -------------------------------------------------- Consider Kudos if I helped in some way!!!

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