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If AD is 6 and ADC is a right angle, what is the area of

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Re: Area of triangular region [#permalink]

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New post 25 Jan 2014, 03:39
dasikasuneel wrote:
Hi Bunuel,

I have a question.

Can I not imagine that, in a right angled triangle if one of the side of a triangle bears x square root 3 as its length can I not imagine that it's a 30-60-90? If not why.

Please help.
Thanks
Suneel


What do you mean by "imagine"? How does the length of one side define the angles?
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Re: If AD is 6 and ADC is a right angle, what is the area of [#permalink]

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New post 06 Feb 2014, 11:28
Bunuel wrote:
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).



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

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

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New post 07 Feb 2014, 05:27
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virendrasd wrote:
Bunuel wrote:
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).



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 [#permalink]

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New post 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 [#permalink]

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New post 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 [#permalink]

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

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New post 25 Oct 2016, 06:46
amitjash wrote:
Attachment:
Triangle.jpg
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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Re: If AD is 6 and ADC is a right angle, what is the area of [#permalink]

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New post 14 Aug 2017, 00:28
AccipiterQ wrote:
Bunuel wrote:
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

Can someone explain???


There should be a diagram attached:
Attachment:
2cr0lsz.jpg


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.

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?


I also assumed a similar thing . But I believe a perpendicular only bisects the angle at the vertex IF the perpendicular bisects the opposite side .. and vice versa . So we cannot assume the perpendicular is an angle / side bisector unless it was explicitly mentioned in the stem

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

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New post 15 Oct 2017, 18:25
Bunuel wrote:
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

Can someone explain???


There should be a diagram attached:
Attachment:
2cr0lsz.jpg


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.

Answer: C.


Hi Bunuel

Cant we assume that that AD bisects BC into two?

Please clear my doubt thanks

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

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New post 15 Oct 2017, 21:36
zanaik89 wrote:
Bunuel wrote:
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

Can someone explain???


There should be a diagram attached:
Attachment:
2cr0lsz.jpg


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.

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

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New post 16 Oct 2017, 02:46
zanaik89 wrote:
Bunuel wrote:
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

Can someone explain???


There should be a diagram attached:
Attachment:
2cr0lsz.jpg


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.

Answer: C.


Hi Bunuel

Cant we assume that that AD bisects BC into two?

Please clear my doubt thanks



So u mean that since it isnt given that triangle ABC is an isoceles triangle or equilateral triangle, we can not consider AD as a perpendicular bisector right?

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

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New post 16 Oct 2017, 02:53
zanaik89 wrote:

So u mean that since it isnt given that triangle ABC is an isoceles triangle or equilateral triangle, we can not consider AD as a perpendicular bisector right?


Yes, we know that AD is perpendicular to BC but we don't know whether it bisects BC.
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Re: If AD is 6 and ADC is a right angle, what is the area of   [#permalink] 16 Oct 2017, 02:53

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