MGMAT Triangle properties

Expert replies

ru2008: Senior | Next Rank: 100 Posts; Posts: 58; Joined: Tue Aug 26, 2008 9:42 am

by ru2008 » Fri Sep 24, 2010 1:29 pm

Can we not assume that AD is a perpendicular bisector of BC?

If AD is 6, and ADC is a right angle, what is the area of triangular region ABC?

(1) Angle ABD = 60Â°

(2) AC = 12

Like this post

Source: Beat The GMAT — Data Sufficiency | Fri Sep 24, 2010 1:29 pm

this_time_i_will: Master | Next Rank: 500 Posts; Posts: 268; Joined: Wed Mar 17, 2010 2:32 am; Thanked: 17 times

by this_time_i_will » Fri Sep 24, 2010 7:26 pm

Yes, we can assume. But we sud understand the reason.
Both the traingles ADB and ADC are cogruent. (RHS property, one angle is right angle and one side of same length. In this case the ommon side AD.). SInce both the traingles are congruent, BD = DC and hence AD is perpendicular bisector.

Answer sud be D.

Last edited by this_time_i_will on Fri Sep 24, 2010 7:27 pm, edited 1 time in total.

My Blog:
https://www.iwantutopia.blocked/

Like this post

tlt2372: Senior | Next Rank: 100 Posts; Posts: 48; Joined: Sun Jul 20, 2008 4:45 pm; Location: Tallahassee, FL; Thanked: 2 times

by tlt2372 » Fri Sep 24, 2010 7:27 pm

I get A.

If you know a side and an angle on 30-60-90 right triangle, you can get the other two sides. Statement 1 tells us that this is a 30-60-90. Thus, you can find the other two sides using identities. Dont solve all the way through - just know that you can!

Statement 2 doesnt give us a special right triangle relationship. You cant trust the figure - it may not be drawn to scale.

Like this post

debmalya_dutta: Master | Next Rank: 500 Posts; Posts: 385; Joined: Sun Jul 12, 2009 10:16 pmFollowed by: 2 members; Thanked: 29 times; GMAT Score:710

by debmalya_dutta » Fri Sep 24, 2010 9:24 pm

Can we not assume that AD is a perpendicular bisector of BC?

We ca only assume that provided ABD and ADC are congruent... Unless and untill you can establish that , AD is not the bisector of BC

Statement 1 is insufficient because we can only establish values for AB and BD from this ... For area we need the value of DC and we have no clue how long CD is
Statement 2 is insufficient again because from this we can find the value of CD ..Nothing beyond that ..
Hence answer is C

@Deb

Like this post

saurabhmahajan: Master | Next Rank: 500 Posts; Posts: 250; Joined: Thu Jan 07, 2010 2:21 am; Thanked: 10 times

by saurabhmahajan » Sat Sep 25, 2010 1:48 am

stmt 1:
abd=60 degree
so by 30-60-90 rule we can find sides as we know ad=6

stmt 2:
ac=12
here we dont know the angle of other sides it can be 45-45-90 or 30-60-90 and so we cant find sides.

so IMO: A

Thanks and regards,
Saurabh Mahajan

I can understand you not winning,but i will not forgive you for not trying.

Like this post

hi.itz.mani: Senior | Next Rank: 100 Posts; Posts: 53; Joined: Sun May 16, 2010 9:38 pm; Thanked: 6 times

by hi.itz.mani » Tue Sep 28, 2010 4:17 am

IMO C

Whats the OA?

AD cannot be assumed as a perpendicular bisector as it hasn't been mentioned that the triangle is isosceles or equilateral ( Perpendicular from the vertix to the base bisects the base only in case the triangle is Isosceles or Equilateral)

Statement 1: Not sufficient as with the angle ABD = 60 , we can find the value of AB and BD , which is not sufficient to answer

Statement 2: Not sufficient either, citing the same reason as above

Statement 1 and Statement 2 can together give us the area , Hence C

Like this post

GadgetAddicted: Newbie | Next Rank: 10 Posts; Posts: 2; Joined: Thu Sep 23, 2010 11:34 am

by GadgetAddicted » Tue Sep 28, 2010 7:10 am

Won't the answer be D? Each statement by itself is sufficient?

Area of a triangle is 1/2BH.

H is given as 6 (AD). ABD angle is 60 so this is a 30-60-90 triangle. We can find Base by finding BD and then multiplying by 2. Therefore Statement 1 is sufficient.

For statement 2, can't we use Pythagoras Theorem? For a right angle triangle, we are given two sides.
I.E. 12^2=6^2+x^2 and then solve for x where x is DC? Multiply DC by 2 and you have the Base.

I believe the answer should be D. What do you guys think?

Like this post

kmittal82: Master | Next Rank: 500 Posts; Posts: 324; Joined: Mon Jul 05, 2010 6:44 amFollowed by: 3 members; Location: London; Thanked: 70 times

by kmittal82 » Tue Sep 28, 2010 7:43 am

I'm in the "I think its (C) camp"

We cannot assume AD to be the bisector. You can easily draw a triangle where AB and BD would be much greater than AC and CD respectively.

Stm1 helps us get BD (using 30-60-90)

Stm2 helps us get CD (using 30-60-90)

Combining 1 and 2 gives us BC, which in turn gives us the area.

Like this post

GMATMadeEasy: Legendary Member; Posts: 768; Joined: Mon Nov 30, 2009 3:46 amFollowed by: 7 members; Thanked: 21 times

by GMATMadeEasy » Tue Sep 28, 2010 10:44 am

ru2008 wrote:Can we not assume that AD is a perpendicular bisector of BC?

If AD is 6, and ADC is a right angle, what is the area of triangular region ABC?

(1) Angle ABD = 60Â°

(2) AC = 12

1.You definitely can not assume that it is bisector . Why would you do so ? There are many traiangles possible with perpendicular and not being bisector .
2 . OA should be A

Explanation :
1.
ABD = 60 ; that makes left hand triangle 3Ã ,60 and 90 traingle.If you know one side (which we know here) , you get the left hand side traingle area. RIght hand triangle is also 30,60 and 90 triangle so you get the area of this triangle too. Hence you can get total area.

2. Statement 2 gives you the right hand triangle sides but you have no way to know left hand side. Because sides of right hand traingle do not give any special relationship either.

Like this post

uwhusky: Legendary Member; Posts: 1172; Joined: Wed Apr 28, 2010 6:20 pmFollowed by: 4 members; Thanked: 74 times

by uwhusky » Tue Sep 28, 2010 10:59 am

GMATMadeEasy wrote:
ru2008 wrote:Can we not assume that AD is a perpendicular bisector of BC?

If AD is 6, and ADC is a right angle, what is the area of triangular region ABC?

(1) Angle ABD = 60Â°

(2) AC = 12

1.You definitely can not assume that it is bisector . Why would you do so ? There are many traiangles possible with perpendicular and not being bisector .
2 . OA should be A

Explanation :
1.
ABD = 60 ; that makes left hand triangle 30, 60 and 90 triangle. If you know one side (which we know here) , you get the left hand side traingle area. RIght hand triangle is also 30, 60 and 90 triangle so you get the area of this triangle too. Hence you can get total area.

But you just did when you make the assumption that the left side triangle is equal to the right side triangle.

There's nothing in the stimulus that says this is an "equilateral triangle", so there is no evidence that AD bisects BC. So I would think the answer should be C as well, unless this is a test of some obscure triangle knowledge, in which if left side triangle is 30/60/90, then the right must also be the same, and that this triangle must be an equilateral triangle.

Yep.

Like this post

uwhusky: Legendary Member; Posts: 1172; Joined: Wed Apr 28, 2010 6:20 pmFollowed by: 4 members; Thanked: 74 times

by uwhusky » Tue Sep 28, 2010 11:16 am

Btw, if you punched in the number, you can see that this is definitely not an equilateral triangle.

30/60/90 = x/xâˆš3/2x

So if xâˆš3 = 6, then 2x must be 4âˆš3.

so AB is not the same as AC, which is 12.

Also, if your reasoning is that A is sufficient because left side triangle is the same as right side triangle, then you must assume all measure is the same on both sides. Therefore if AC = 12, and AD is 6, we can easily find DC, and since both triangles are the same, the same assumption of (1), then we could do the same for ABD.

So I think the answer would either be D or C depending on your reasoning, but it shouldn't be A or B by itself.

Yep.

Like this post

GMATMadeEasy: Legendary Member; Posts: 768; Joined: Mon Nov 30, 2009 3:46 amFollowed by: 7 members; Thanked: 21 times

by GMATMadeEasy » Tue Sep 28, 2010 11:30 am

Agreed , I made an error. It should be C .

I ,by mistake,assumed one of the angles as 30 in right side triangle .. sorry.. I didnt mean equilateral triangle at all.

Like this post

uwhusky: Legendary Member; Posts: 1172; Joined: Wed Apr 28, 2010 6:20 pmFollowed by: 4 members; Thanked: 74 times

by uwhusky » Tue Sep 28, 2010 11:37 am

Don't be sorry man, it's a discussion! We're all here to learn and to improve.

Yep.

Like this post

GadgetAddicted: Newbie | Next Rank: 10 Posts; Posts: 2; Joined: Thu Sep 23, 2010 11:34 am

by GadgetAddicted » Tue Sep 28, 2010 11:42 am

Any way to find out the actual answer to the question??

Like this post

uwhusky: Legendary Member; Posts: 1172; Joined: Wed Apr 28, 2010 6:20 pmFollowed by: 4 members; Thanked: 74 times

by uwhusky » Tue Sep 28, 2010 11:43 am

magic of Google: https://www.manhattangmat.com/forums/post3967.html

Lol, Harish Dorai drew the triangle I was going to draw and post here.

Yep.

Like this post

Post new topic Post Reply

• Page 1 of 1

GMAT Course Reviews

560 reviews

1168 reviews

431 reviews

359 reviews

147 reviews

5/5

5 Star (560 Reviews)

"Target Test Prep is the closest to the official version of the GMAT exam, about 99% accuracy in terms of the quality and quantity of information. The course has excellently created singular sets of focused lessons and tests for every possible topic that one could come across in the official GMAT exam."

"The TTP course maximizes the efficiency of the time you spend studying. It will take time and effort but I could almost guarantee that if you complete the course exactly as it is laid out you will get an amazing score. They also have a very responsive team willing to help with any questions you might have."

"TTP has two things that I think no other test prep company offers: A teaching approach that reinforces understanding and an attitude that will give you the mental preparedness needed to succeed on the test. TTP gives you a deep understanding of the concept you need to know while teaching you how to think."

Write a Review