GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 23 Jul 2018, 03:09

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# What is the length of segment BC?

Author Message
TAGS:

### Hide Tags

SVP
Joined: 29 Aug 2007
Posts: 2425
What is the length of segment BC?  [#permalink]

### Show Tags

Updated on: 11 Oct 2013, 12:49
6
15
00:00

Difficulty:

(N/A)

Question Stats:

49% (00:43) correct 51% (00:41) wrong based on 867 sessions

### HideShow timer Statistics

Attachment:

Triangle.jpg [ 28.62 KiB | Viewed 25455 times ]
What is the length of segment BC?

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

_________________

Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html

GT

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.
Math Expert
Joined: 02 Sep 2009
Posts: 47206
Re: What is the length of segment BC?  [#permalink]

### Show Tags

18 Jul 2012, 05:11
5
2
Responding to a pm.

What is the length of segment BC?

(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 [ 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.

Hope it's clear.
_________________
Director
Joined: 14 Oct 2007
Posts: 737
Location: Oxford
Schools: Oxford'10

### Show Tags

10 May 2008, 07:53
4
1
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
Attachments

trig.jpg [ 14.29 KiB | Viewed 21150 times ]

##### General Discussion
Senior Manager
Joined: 24 Feb 2008
Posts: 348
Schools: UCSD ($) , UCLA, USC ($), Stanford

### Show Tags

08 May 2008, 21:32
1
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.
_________________

Best AWA guide here: http://gmatclub.com/forum/how-to-get-6-0-awa-my-guide-64327.html

CEO
Joined: 17 May 2007
Posts: 2895

### Show Tags

08 May 2008, 21:34
4
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.
SVP
Joined: 29 Aug 2007
Posts: 2425

### Show Tags

08 May 2008, 21:45
1
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
_________________

Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html

GT

CEO
Joined: 17 May 2007
Posts: 2895

### Show Tags

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 ?
SVP
Joined: 29 Aug 2007
Posts: 2425

### Show Tags

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....
_________________

Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html

GT

SVP
Joined: 29 Aug 2007
Posts: 2425

### Show Tags

Updated on: 08 May 2008, 22:15
1
bsd_lover wrote:
"should be A" or IS A ?

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

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.
_________________

Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html

GT

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.
Manager
Joined: 11 Apr 2008
Posts: 147
Schools: Kellogg(A), Wharton(W), Columbia(D)

### Show Tags

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

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
Current Student
Joined: 28 Dec 2004
Posts: 3287
Location: New York City
Schools: Wharton'11 HBS'12

### Show Tags

09 May 2008, 05:18
1
i concur A should be the answer..

i can draw the triangle where BC>12 and where BC=12..same area..
SVP
Joined: 29 Aug 2007
Posts: 2425

### Show Tags

10 May 2008, 11:51
1
seems you missed the posts above yours. Read them again as D is incorrect..

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

_________________

Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html

GT

Director
Joined: 06 Jan 2008
Posts: 522

### Show Tags

10 May 2008, 16:59
1
GMAT TIGER wrote:
seems you missed the posts above yours. Read them again as D is incorrect..

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?
Intern
Joined: 02 Apr 2008
Posts: 37

### Show Tags

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

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.
Manager
Joined: 16 Sep 2007
Posts: 211

### Show Tags

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?
CEO
Joined: 17 Nov 2007
Posts: 3484
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40

### Show Tags

12 May 2008, 00:37
5
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.
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

Manager
Joined: 11 Apr 2008
Posts: 147
Schools: Kellogg(A), Wharton(W), Columbia(D)

### Show Tags

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.
SVP
Joined: 29 Aug 2007
Posts: 2425

### Show Tags

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?
_________________

Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html

GT

CEO
Joined: 17 Nov 2007
Posts: 3484
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40

### Show Tags

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
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

Director
Joined: 06 Jan 2008
Posts: 522

### Show Tags

15 May 2008, 09:14
2
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
Attachments

traingle.doc [39.5 KiB]

Re: DS: Triangle &nbs [#permalink] 15 May 2008, 09:14

Go to page    1   2   3    Next  [ 42 posts ]

Display posts from previous: Sort by

# Events & Promotions

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.