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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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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:
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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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If AD is 6 and ADC is a right angle, what is the area of triangular re [#permalink]
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17 Sep 2010, 05:25




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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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15 Apr 2012, 15:11
with statement 1 we can only figure out that the left triangle is a 306090. We cannot just assume that the left triangle is also a 306090. 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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29 Jun 2012, 04:44
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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Re: If AD is 6 and ADC is a right angle, what is the area of triangular re [#permalink]
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29 Jun 2012, 05:05



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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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04 Apr 2013, 11:00
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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04 Apr 2013, 11:25



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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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26 Oct 2013, 11:54
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 306090 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. Answer: C. 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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If AD is 6 and ADC is a right angle, what is the area of triangular re [#permalink]
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27 Oct 2013, 07:03
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 306090 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. Answer: C. 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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If AD is 6 and ADC is a right angle, what is the area of triangular re [#permalink]
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15 Dec 2013, 16:05
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 306090 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. Answer: C. 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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If AD is 6 and ADC is a right angle, what is the area of triangular re [#permalink]
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15 Dec 2013, 23:21
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 3857 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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If AD is 6 and ADC is a right angle, what is the area of triangular re [#permalink]
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15 Dec 2013, 23:50
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. Hence Answer is C Pushpinder Gill
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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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06 Feb 2014, 11:28
I think answer is E, since it is not given that BDC points are collinear.



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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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07 Feb 2014, 05:27
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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18 Mar 2014, 01:11
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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18 Mar 2014, 08:51
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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Re: If AD is 6 and ADC is a right angle, what is the area of triangular re [#permalink]
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12 Apr 2015, 17:23
Main take away from this question is: Figures are not drawn to scale.rest explanations are already given.
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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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25 Oct 2016, 06:46
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 306090 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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If AD is 6 and ADC is a right angle, what is the area of triangular re [#permalink]
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15 Oct 2017, 18:25
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 306090 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. Answer: C. Hi Bunuel Cant we assume that that AD bisects BC into two? Please clear my doubt thanks



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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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15 Oct 2017, 21:36
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 306090 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. Answer: C. Hi Bunuel Cant we assume that that AD bisects BC into two? Please clear my doubt thanks 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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