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What is the length of segment BC?

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What is the length of segment BC?  [#permalink]

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New post Updated on: 11 Oct 2013, 12:49
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What is the length of segment BC?

(1) Angle ABC is 90 degrees.
(2) The area of the triangle is 30.

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Originally posted by GMAT TIGER on 08 May 2008, 21:28.
Last edited by Bunuel on 11 Oct 2013, 12:49, edited 2 times in total.
Edited the question and added the OA.
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Re: What is the length of segment BC?  [#permalink]

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New post 18 Jul 2012, 05:11
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Responding to a pm.

What is the length of segment BC?

Image

(1) Angle ABC is 90 degrees. \(BC=\sqrt{13^2-5^2}\). Sufficient.

(2) The area of the triangle is 30. Consider the diagram below:
Attachment:
ABC.png
ABC.png [ 6.06 KiB | Viewed 11360 times ]
Notice that segment \(AB_2\) is the mirror reflection of segment \(AB_1\) around the vertical line passing through point A. Now, if the height of triangles \(ACB_1\) and \(ACB_2\) is \(\frac{60}{13}\), then the area of both triangles is \(area=\frac{1}{2}*base*height=\frac{1}{2}*13*\frac{60}{13}=30\). So, as you can see we can have different lengths of segment CB (\(CB_1\) and \(CB_2\)). Not sufficient.

Answer: A.

Hope it's clear.
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Re: DS: Triangle  [#permalink]

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New post 10 May 2008, 07:53
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The answer is infact D, here is how

1) this is sufficient (as agreed by all)

2) refer to my diagram. Given the area, we can calculate the height h since we know the base (13). Therefore now we have two 90 right triangles. We can find out x using pythagoras ( i am not solving it here since this is a DS problem). We can then find out y by doing 13 - x. Now that we know y and h, we can find out z (or AB), hence this info is sufficient
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Re: DS: Triangle  [#permalink]

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New post 08 May 2008, 21:32
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A

1. Suff since it's a 5-12-13 triangle
2. Insuff because the height to side 5, which can be calculated to be 12 may not necessarily be the leg.
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Re: DS: Triangle  [#permalink]

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New post 08 May 2008, 21:34
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I'd say D.

There is a forumla to derive the area of a triangle if you know all three sides so B alone is sufficient.

A is obviously sufficient on its own.
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Re: DS: Triangle  [#permalink]

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New post 08 May 2008, 21:45
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bsd_lover wrote:
I'd say D.

There is a forumla to derive the area of a triangle if you know all three sides so B alone is sufficient.

A is obviously sufficient on its own.


is the following?

Sqrt [S(S-a)(S-b)(S-c)]

Where,

S = (a + b + c)/3
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Re: DS: Triangle  [#permalink]

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New post 08 May 2008, 22:03
I guess we'll end up with a pretty complex equation - one with \(x^3\) in it. But do we need to solve ? for the purpose of DS all we need to know is that there is a method to solve.

There was a problem in one of the challenges where I knew we could solve and find out an answer using trigonometry, but trigonometry is not required knowledge for GMAT so does that mean that it can't be solved ??? I guess I dont know what is correct here..

What is the OA ?
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Re: DS: Triangle  [#permalink]

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New post 08 May 2008, 22:06
bsd_lover wrote:
I guess we'll end up with a pretty complex equation - one with \(x^3\) in it. But do we need to solve ? for the purpose of DS all we need to know is that there is a method to solve.

There was a problem in one of the challenges where I knew we could solve and find out an answer using trigonometry, but trigonometry is not required knowledge for GMAT so does that mean that it can't be solved ??? I guess I dont know what is correct here..

What is the OA ?


OA should be A.

BC could be 12 or > 12. So only A is suff....
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Re: DS: Triangle  [#permalink]

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New post Updated on: 08 May 2008, 22:15
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bsd_lover wrote:
"should be A" or IS A ?


I do not have official answer but I concur with A. so should be A. :wink:

if we draw a perpendicular from b to base ac, then bc becomes 12.

also we can draw the trangle with same area, same height, same measures for ab and ac. in this case bc is > 12.
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Originally posted by GMAT TIGER on 08 May 2008, 22:11.
Last edited by GMAT TIGER on 08 May 2008, 22:15, edited 1 time in total.
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Re: DS: Triangle  [#permalink]

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New post 08 May 2008, 23:50
Quote:
I do not have official answer but I concur with A. so should be A. :wink:

if we draw a perpendicular from b to base ac, then bc becomes 12.

also we can draw the trangle with same area, same height, same measures for ab and ac. in this case bc is > 12.


Are we not restricted by the fact that the third side IS "13", a fixed value.
For this reason, IMO=> D
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Re: DS: Triangle  [#permalink]

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New post 09 May 2008, 05:18
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i concur A should be the answer..

i can draw the triangle where BC>12 and where BC=12..same area..
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Re: DS: Triangle  [#permalink]

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New post 10 May 2008, 11:51
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seems you missed the posts above yours. Read them again as D is incorrect..

buffdaddy wrote:
The answer is infact D, here is how

1) this is sufficient (as agreed by all)

2) refer to my diagram. Given the area, we can calculate the height h since we know the base (13). Therefore now we have two 90 right triangles. We can find out x using pythagoras ( i am not solving it here since this is a DS problem). We can then find out y by doing 13 - x. Now that we know y and h, we can find out z (or AB), hence this info is sufficient

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Re: DS: Triangle  [#permalink]

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New post 10 May 2008, 16:59
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GMAT TIGER wrote:
seems you missed the posts above yours. Read them again as D is incorrect..

buffdaddy wrote:
The answer is infact D, here is how

1) this is sufficient (as agreed by all)

2) refer to my diagram. Given the area, we can calculate the height h since we know the base (13). Therefore now we have two 90 right triangles. We can find out x using pythagoras ( i am not solving it here since this is a DS problem). We can then find out y by doing 13 - x. Now that we know y and h, we can find out z (or AB), hence this info is sufficient

I disagree. The answer must be D.
S2: Sqrt [S(S-a)(S-b)(S-c)] =30
Where S=(a+b+c)/3.
@GMATTIGER, so what you are saying is that you can draw other triangle with two sides 5, and 13, with area = 30. And you are saying you can draw other triangle with the third side greater than 12 but can keep the area and other two sides constant.I think it is not possible.
Can you draw a triangle with sides 5, 13, 1million, with area 30? AND
at the same time can you draw other triangle with sides 5,13, 20 with area 30?
Answer should be D.Please help me if I missed anything.
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Re: DS: Triangle  [#permalink]

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New post Updated on: 11 May 2008, 22:26
GMAT TIGER wrote:
seems you missed the posts above yours. Read them again as D is incorrect..

buffdaddy wrote:
The answer is infact D, here is how

1) this is sufficient (as agreed by all)

2) refer to my diagram. Given the area, we can calculate the height h since we know the base (13). Therefore now we have two 90 right triangles. We can find out x using pythagoras ( i am not solving it here since this is a DS problem). We can then find out y by doing 13 - x. Now that we know y and h, we can find out z (or AB), hence this info is sufficient


Sorry, answer has to be D.

GMAT TIGER wrote:
also we can draw the trangle with same area, same height, same measures for ab and ac. in this case bc is > 12.


i guess, drawing a perpendicular to a base AB (if that's what you meant) will in fact lead you to the same result while drawing a perpendicular to base AC will tell you nothing.

As to BC=12 or >12 in different instances, again differing results may be due to rounding. Try calculating it in excel. Under buffdaddy's approach you ll get 12.

Ill be happy to be proven wrong.

Cheers

Originally posted by NightAlum on 11 May 2008, 19:21.
Last edited by NightAlum on 11 May 2008, 22:26, edited 1 time in total.
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Re: DS: Triangle  [#permalink]

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New post 11 May 2008, 19:50
For everyone doubting the OA, check with Heron's formula. 2*sqrt(61) and 12 will both work for line segment BC. The more interesting question is why ASS is sufficient for triangle congruence. If I had not recognized the triangle, I would've said A is insufficient. Is 90 degrees a special case?
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Re: DS: Triangle  [#permalink]

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New post 12 May 2008, 00:37
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Hi all!

My 2 cents....

Let imagine AB and AC as sticks bounded in point A by a hinge. So, we can move AB relatively to AC and change angle BAC form 0 to 180. The area of ABC triangle will change form 0 (angle BAC=0) through maximum value 12*5*13=32.5 (angle BAC=90) to 0 (angle BAC=180). If you imagine that, It will be obvious that for 0<area<32.5 we will get two possible value of BC and for area=32.5 we will get only one value of BC. In our case area=30<32.5. Therefore, the second condition is insufficient.
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Re: DS: Triangle  [#permalink]

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New post 13 May 2008, 02:47
Thanks all of you guys for all yor explanations. But I still can;t get this straight. It would be helpful if someone could draw any sample triangle that is not 5, 12, 13 and meets condition II. Thanks for your help.
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Re: DS: Triangle  [#permalink]

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New post 14 May 2008, 19:57
walker wrote:
Hi all!

My 2 cents....

Let imagine AB and AC as sticks bounded in point A by a hinge. So, we can move AB relatively to AC and change angle BAC form 0 to 180. The area of ABC triangle will change form 0 (angle BAC=0) through maximum value 12*5*13=32.5 (angle BAC=90) to 0 (angle BAC=180). If you imagine that, It will be obvious that for 0<area<32.5 we will get two possible value of BC and for area=32.5 we will get only one value of BC. In our case area=30<32.5. Therefore, the second condition is insufficient.


walker,
you said "maximum value 12*5*13=32.5" that i did not get. if the sides are 12, 5, and 13, then it is a right angle triangle and should have 30 as area. how did you get 32.5 for a triangle with sides 12, 5, and 13?
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Re: DS: Triangle  [#permalink]

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New post 14 May 2008, 20:44
GMAT TIGER wrote:
walker,
you said "maximum value 12*5*13=32.5" that i did not get. if the sides are 12, 5, and 13, then it is a right angle triangle and should have 30 as area. how did you get 32.5 for a triangle with sides 12, 5, and 13?


Hi, Tiger!

Sorry for typo. It should be 1/2*5*13=32.5 instead of 12*5*13=32.5.

I used formula for area: area=1/2*a*h. if a=5, the maximum height will be h=13
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Re: DS: Triangle  [#permalink]

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New post 15 May 2008, 09:14
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What a bliss! I was being jerk!
For those who did not understand Walker's expln, check out this link.
(Interactive triangle)
http://www.mathopenref.com/heronsformula.html
For there are 2 triangles with the area 30.
1: 5,13,12 (For acute angle)
2: 5,13,15.5... (Not 15.5) ( For obtuse angle)
see the attachment. Thank you all :-D
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Re: DS: Triangle &nbs [#permalink] 15 May 2008, 09:14

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