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Manager  Joined: 17 Mar 2010
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Question Stats: 56% (01:46) correct 44% (01:33) wrong based on 839 sessions

HideShow timer Statistics If AD is 6 and ADC is a right angle, what is the area of triangular region ABC?

(1) Angle ABD = 60°
(2) AC = 12

Attachment: Triangle.jpg [ 2.62 KiB | Viewed 19459 times ]

Originally posted by amitjash on 17 Sep 2010, 05:03.
Last edited by Bunuel on 29 Apr 2018, 23:16, edited 5 times in total.
Edited the question and added the diagram
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4 If AD is 6 and ADC is a right angle, what is the area of triangular region ABC?

Given: $$AD=6$$. Question: $$area_{ABC}=\frac{1}{2}*AD*BC=\frac{1}{2}*6*(BD+DC)=3(BD+DC)=?$$

(1) Angle ABD = 60 --> triangle ABD is 30-60-90 triangle, so the sides are in ratio $$1:\sqrt{3}:2$$ --> as AD=6 (larger leg opposite 60 degrees angle) then $$BD=\frac{6}{\sqrt{3}}$$ (smallest leg opposite 30 degrees angle). But we still don't know DC. Not sufficient.

(2) AC=12, we can find DC. But we still don't know BD. Not sufficient.

(1)+(2) We know both BD and DC, hence we can find the area. Sufficient.

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Re: If AD is 6 and ADC is a right angle, what is the area of triangular re  [#permalink]

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with statement 1 we can only figure out that the left triangle is a 30-60-90. We cannot just assume that the left triangle is also a 30-60-90. The sides could be longer or shorter than shown.
the same idea as statement 1 for 2. no assumptions can be made for the left triangle given that the side is 12.
if we combine both questions together then we can find out vital information for both triangles.
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Re: If AD is 6 and ADC is a right angle, what is the area of triangular re  [#permalink]

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but isn't triangle ABD similar to triangle ADC since AD is the perpendicular. so using the ratio of 2:1:√3 we can find all sides of triangle ABD and since AD is common using proportions we can find DC ? and thus the total area ?
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but isn't triangle ABD similar to triangle ADC since AD is the perpendicular. so using the ratio of 2:1:√3 we can find all sides of triangle ABD and since AD is common using proportions we can find DC ? and thus the total area ?

No, for (1) we cannot conclude that ABD and ADC are similar. We only know that ABD and ADC are both right triangles (one angle) and share the common side AD (one side), which is not enough to conclude that ABD and ADC are similar.

For more check Triangles chapter of Math Book: math-triangles-87197.html
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Re: If AD is 6 and ADC is a right angle, what is the area of triangular re  [#permalink]

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Hey I had the same question as the last poster. How do we know that BD and DC are of the same length? or that angle BAC has been bisected?
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Re: If AD is 6 and ADC is a right angle, what is the area of triangular re  [#permalink]

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manimgoindowndown wrote:
Hey I had the same question as the last poster. How do we know that BD and DC are of the same length? or that angle BAC has been bisected?

They are not equal: $$BD=\frac{6}{\sqrt{3}}$$ (from the first statement) and $$AD=6*\sqrt{3}$$ (from the second statement).
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GMAT 1: 670 Q39 V41 GMAT 2: 730 Q49 V41 If AD is 6 and ADC is a right angle, what is the area of triangular re  [#permalink]

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Bunuel wrote: Given: $$AD=6$$. Question: $$area_{ABC}=\frac{1}{2}*AD*BC=\frac{1}{2}*6*(BD+DC)=3(BD+DC)=?$$

(1) Angle ABD = 60 --> triangle ABD is 30-60-90 triangle, so the sides are in ratio $$1:\sqrt{3}:2$$ --> as AD=6 (larger leg opposite 60 degrees angle) then $$BD=\frac{6}{\sqrt{3}}$$ (smallest leg opposite 30 degrees angle). But we still don't know DC. Not sufficient.

(2) AC=12, we can find DC. But we still don't know BD. Not sufficient.

(1)+(2) We know both BD and DC, hence we can find area. Sufficient.

I thought if you dropped something that was perpindicular to the base of a triangle then it was automatically a bisector; so if you found BD using (1) BD MUST equal dc, no?
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AccipiterQ wrote: Given: $$AD=6$$. Question: $$area_{ABC}=\frac{1}{2}*AD*BC=\frac{1}{2}*6*(BD+DC)=3(BD+DC)=?$$

(1) Angle ABD = 60 --> triangle ABD is 30-60-90 triangle, so the sides are in ratio $$1:\sqrt{3}:2$$ --> as AD=6 (larger leg opposite 60 degrees angle) then $$BD=\frac{6}{\sqrt{3}}$$ (smallest leg opposite 30 degrees angle). But we still don't know DC. Not sufficient.

(2) AC=12, we can find DC. But we still don't know BD. Not sufficient.

(1)+(2) We know both BD and DC, hence we can find area. Sufficient.

I thought if you dropped something that was perpindicular to the base of a triangle then it was automatically a bisector; so if you found BD using (1) BD MUST equal dc, no?[/quote]

Of course not. Are you saying that the height and the median in any triangle coincide? That's not true.
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GMAT 1: 670 Q39 V41 GMAT 2: 730 Q49 V41 If AD is 6 and ADC is a right angle, what is the area of triangular re  [#permalink]

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Bunuel wrote: Given: $$AD=6$$. Question: $$area_{ABC}=\frac{1}{2}*AD*BC=\frac{1}{2}*6*(BD+DC)=3(BD+DC)=?$$

(1) Angle ABD = 60 --> triangle ABD is 30-60-90 triangle, so the sides are in ratio $$1:\sqrt{3}:2$$ --> as AD=6 (larger leg opposite 60 degrees angle) then $$BD=\frac{6}{\sqrt{3}}$$ (smallest leg opposite 30 degrees angle). But we still don't know DC. Not sufficient.

(2) AC=12, we can find DC. But we still don't know BD. Not sufficient.

(1)+(2) We know both BD and DC, hence we can find area. Sufficient.

I thought if you dropped a line down from a triangle vertex and it formed a right angle on the opposite side then that line bisected the side? So in this case if you know what BD is then you know what DC is?
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AccipiterQ wrote: I thought if you dropped a line down from a triangle vertex and it formed a right angle on the opposite side then that line bisected the side? So in this case if you know what BD is then you know what DC is?

To figure out whether it holds, why don't you try drawing some extreme figures, say, something like this:
Attachment: Ques3.jpg [ 3.67 KiB | Viewed 6481 times ]

Will this be true in this case?
When will it be true? When the triangle is equilateral, sure. Also when the triangle is isosceles if the equal sides form the angle from which the altitude is dropped.

Don't put your faith in the figure given. It may be just one of the many possibilities or may be somewhat misleading.
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GMAT 1: 780 Q51 V46 If AD is 6 and ADC is a right angle, what is the area of triangular re  [#permalink]

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amitjash wrote: If AD is 6 and ADC is a right angle, what is the area of triangular region ABC?

(1) Angle ABD = 60°
(2) AC = 12

Statement I is insufficient

The left triangle becomes a 90 30 60 triangle however we don't know anything about the right triangle

Statement II is insufficient

The right triangle becomes a 90 30 60 triangle however we don't know anything about the left triangle

Combining is sufficient
We know both the triangles are 90 30 60 with same sides.

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I think answer is E, since it is not given that BDC points are collinear.
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virendrasd wrote:
I think answer is E, since it is not given that BDC points are collinear.

That's not correct.

OG13, page 272:
A figure accompanying a data sufficiency problem will conform to the information given in the question but will not necessarily conform to the additional information given in statements (1) and (2).
Lines shown as straight can be assumed to be straight and lines that appear jagged can also be assumed to be straight.
You may assume that the positions of points, angles, regions, and so forth exist in the order shown and that angle measures are greater than zero degrees.
All figures lie in a plane unless otherwise indicated.

OG13, page 150:
Figures: A figure accompanying a problem solving question is intended to provide information useful in solving the problem. Figures are drawn as accurately as possible. Exceptions will be clearly noted. Lines shown as straight are straight, and lines that appear jagged are also straight. The positions of points, angles, regions, etc., exist in the order shown, and angle measures are greater than zero. All figures lie in a plane unless otherwise indicated.

Hope it helps.
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Re: If AD is 6 and ADC is a right angle, what is the area of triangular re  [#permalink]

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what if question wud have been like this..

If AD is 6, and ADC is a right angle, what is the area of equilateral triangle ABC?
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Re: If AD is 6 and ADC is a right angle, what is the area of triangular re  [#permalink]

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sanjoo wrote:
what if question wud have been like this..

If AD is 6, and ADC is a right angle, what is the area of equilateral triangle ABC?

Then statement 1 would have been irrelevant and statement 2 would have been incorrect. If side of an equilateral triangle is 12, the altitude would be $$(\sqrt{3}/2)*12 = 6*\sqrt{3}$$
But AD is given to be 6.
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Main take away from this question is:
Figures are not drawn to scale.

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If AD is 6 and ADC is a right angle, what is the area of triangular re  [#permalink]

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amitjash wrote: If AD is 6 and ADC is a right angle, what is the area of triangular region ABC?

(1) Angle ABD = 60°
(2) AC = 12

I tried not to solve everything as it is a DS question...
Area is equal to base x height / 2
AD is height, BC is the base. We need to find the value of BC.
1. since B is 60 degree, A in ABD is 30.
we have 1 leg, and we know the property of 30-60-90 right triangle
we can find BD.
but since we don't know whether AD divides BC in two equal parts, then we can't tell exactly whether we can find BC.

2. AC = 12. We can find DC. we are faced with the same problem - can't find BD.

1+2
we have BD and we have DC. we can find the answer.

C
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Bunuel wrote: Given: $$AD=6$$. Question: $$area_{ABC}=\frac{1}{2}*AD*BC=\frac{1}{2}*6*(BD+DC)=3(BD+DC)=?$$

(1) Angle ABD = 60 --> triangle ABD is 30-60-90 triangle, so the sides are in ratio $$1:\sqrt{3}:2$$ --> as AD=6 (larger leg opposite 60 degrees angle) then $$BD=\frac{6}{\sqrt{3}}$$ (smallest leg opposite 30 degrees angle). But we still don't know DC. Not sufficient.

(2) AC=12, we can find DC. But we still don't know BD. Not sufficient.

(1)+(2) We know both BD and DC, hence we can find area. Sufficient.

Hi Bunuel

Cant we assume that that AD bisects BC into two?

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If AD is 6 and ADC is a right angle, what is the area of triangular re  [#permalink]

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zanaik89 wrote:
Bunuel wrote: Given: $$AD=6$$. Question: $$area_{ABC}=\frac{1}{2}*AD*BC=\frac{1}{2}*6*(BD+DC)=3(BD+DC)=?$$

(1) Angle ABD = 60 --> triangle ABD is 30-60-90 triangle, so the sides are in ratio $$1:\sqrt{3}:2$$ --> as AD=6 (larger leg opposite 60 degrees angle) then $$BD=\frac{6}{\sqrt{3}}$$ (smallest leg opposite 30 degrees angle). But we still don't know DC. Not sufficient.

(2) AC=12, we can find DC. But we still don't know BD. Not sufficient.

(1)+(2) We know both BD and DC, hence we can find area. Sufficient.

Hi Bunuel

Cant we assume that that AD bisects BC into two?

No. The height and the median coincide only in isosceles/equilateral triangle, when the height is dropped from the vertex formed by equal sides to the base.
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