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

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If AD is 6 and ADC is a right angle, what is the area of [#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
[Reveal] Spoiler: OA

Last edited by Bunuel on 04 Apr 2013, 10:21, 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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New post 17 Sep 2010, 04:22
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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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
2cr0lsz.jpg [ 7.64 KiB | Viewed 15479 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.

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

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shradhagrover18 wrote:
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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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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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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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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New post 04 Apr 2013, 10:25
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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New post 27 Apr 2013, 13:23
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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New post 28 Apr 2013, 02:30
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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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 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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New post 27 Oct 2013, 06:03
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.

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?


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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New post 28 Oct 2013, 14:05
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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New post 11 Dec 2013, 08:36
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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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?

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Re: Area of triangular region [#permalink]

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New post 15 Dec 2013, 22:21
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
Ques3.jpg [ 3.67 KiB | Viewed 2151 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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Re: If AD is 6 and ADC is a right angle, what is the area of [#permalink]

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New post 15 Dec 2013, 22:50
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


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

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Re: Area of triangular region [#permalink]

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New post 24 Jan 2014, 21:10
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
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Re: Area of triangular region   [#permalink] 24 Jan 2014, 21:10

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

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