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

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17 Sep 2010, 05:03
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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 04 Apr 2013, 11:21, edited 4 times in total.
Edited the question and added the diagram
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Re: Area of triangular region [#permalink]

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17 Sep 2010, 05:22
I am not 100% sure of my answer but I think it's correct:

(1) it tells us that we have a triangle 90-60-30 so now, we are able to place the points on the triangle (where is A, where is D and where is C). So we know that A is the right angle, D the 60 degree angle and C the 30 degree angle.

That is not sufficient though...

(2) AD = 6, AC = 12 but which side is the hypothenuse ?????? we can't compute the area !
Insufficient !

(1) and (2)

We know that AC is the hypothenuse, AD the height and AC the basis.

Area = AD * AC / 2 so BOTH TOGETHER

ANS : C.
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Re: Area of triangular region [#permalink]

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17 Sep 2010, 05:25
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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

Can someone explain???

There should be a diagram attached:
Attachment:

2cr0lsz.jpg [ 7.64 KiB | Viewed 12941 times ]

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.

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Re: Area of triangular region [#permalink]

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17 Sep 2010, 05:38
What if the figure is something like i have attached??? How can we assume that it is an equilateral triangle???
Attachments

Full page fax print.pdf [198.24 KiB]

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Re: Area of triangular region [#permalink]

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17 Sep 2010, 05:43
amitjash wrote:
What if the figure is something like i have attached??? How can we assume that it is an equilateral triangle???

I didn't draw it, the diagram is attached to the original question. I guess you took it from the source which didn't have it.
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Re: Area of triangular region [#permalink]

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17 Sep 2010, 06:31
Yes, you are right bunuel thanks for the information...
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15 Apr 2012, 15:11
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 [#permalink]

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29 Jun 2012, 04:44
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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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Re: If AD is 6 and ADC is a right angle, what is the area of [#permalink]

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29 Jun 2012, 05:05
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 [#permalink]

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23 Oct 2012, 06:19
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Hello Bunuel ..

there is a rule that ..perpendicular line bisect the angle..

so cant we here assume it that BD=DC??..

i think its A..if we knw its 30 60 90 triangle..and we can get BD..BD=CD..we can get the area..

where m i wrong Bunuel..?
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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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23 Oct 2012, 06:25
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Expert's post
sanjoo wrote:
Hello Bunuel ..

there is a rule that ..perpendicular line bisect the angle..

so cant we here assume it that BD=DC??..

i think its A..if we knw its 30 60 90 triangle..and we can get BD..BD=CD..we can get the area..

where m i wrong Bunuel..?

There is no such rule. It's not even clear what does it mean.
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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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04 Apr 2013, 11:00
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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 [#permalink]

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04 Apr 2013, 11:25
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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Re: If AD is 6 and ADC is a right angle, what is the area of [#permalink]

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27 Apr 2013, 14:23
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).

What if this was a problem solving and you could not verify using the way in which you did above?
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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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28 Apr 2013, 03:30
jmuduke08 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).

What if this was a problem solving and you could not verify using the way in which you did above?

Sorry, don't understand what you mean.

We are NOT given that BD=DC and we are NOT assuming that when solving, so what should we verify and why?
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Re: Area of triangular region [#permalink]

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26 Oct 2013, 11:54
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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.

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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Re: Area of triangular region [#permalink]

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27 Oct 2013, 07:03
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.

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?

Of course not. Are you saying that the height and the median in any triangle coincide? That's not true.
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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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28 Oct 2013, 15:05
The triangles are similar if you take both statements together.
(1) gives you all angles of the left triangle.
(2) gives you the length of two sides of a triangle, and it's given that it's a right triangle
From the above you know that the triangles are similar and equal to each other (because they share the same
side that's relative to the same angle)...
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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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11 Dec 2013, 09:36
If AD is 6 and ADC is a right angle, what is the area of triangular region ABC?

(1) Angle ABD = 60°

If ABD is 60 then we know triangle ABD = 30:60:90 triangle. Coupled with the one measurement we are given we can find the length of the other two sides of this triangle. However, we know nothing about AC or DC, only that triangle ADC is right - we don't know the measurements of the angles so we cannot find their leg lengths. For example, DC could be six inches long or 20. We don't know and therefore cannot find the area of this triangle. Insufficient.

(2) AC = 12
We have a leg and a hypotenuse of a right triangle so we can find the length of the second leg but we know nothing about triangle ABD, except for one leg length. Insufficient.

1+2) 1 Gives us the length of all 3 sides of ABD. 2 gives us the length of all three sides of ADC. Sufficient.

C
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Re: Area of triangular region [#permalink]

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15 Dec 2013, 16:05
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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.

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?
Re: Area of triangular region   [#permalink] 15 Dec 2013, 16:05

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