It is currently 18 Oct 2017, 08:19

Live Now:

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

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

DS Geometry (m08q22)

Author Message
Intern
Joined: 12 Feb 2006
Posts: 26

Kudos [?]: 12 [2], given: 0

Show Tags

23 Apr 2006, 09:25
2
KUDOS
5
This post was
BOOKMARKED
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If $$M$$ , $$N$$ , and $$O$$ are midpoints of sides $$AB$$ , $$BC$$ , and $$AC$$ of triangle $$ABC$$ . What is the area of triangle $$MON$$ ?

1. The area of $$ABC$$ is $$\frac{\sqrt{3}}{4}$$
2. $$ABC$$ is an equilateral triangle with height $$\frac{sqrt3}{2}$$

[Reveal] Spoiler: OA
D

Source: GMAT Club Tests - hardest GMAT questions

Kudos [?]: 12 [2], given: 0

Intern
Joined: 20 Feb 2006
Posts: 44

Kudos [?]: 2 [1], given: 0

Show Tags

23 Apr 2006, 10:17
1
KUDOS
Since M, N, and O are midpoints of sides AB, BC, and AC of triangle ABC,

Area of AMO = Area of BMN = Area of CNO = Area of MNO = Area of ABC/4

Area of MNO = sqrt(3)/16

(1) is sufficient

Again (2) also sufficient as we know area of an equilateral triangle = sqrt(3)*a^2/4 where a is the side and height is sqrt(3)*a/2

hence a = 1 and Area of ABC = sqrt(3)/4

Kudos [?]: 2 [1], given: 0

Director
Joined: 04 Jan 2006
Posts: 922

Kudos [?]: 37 [0], given: 0

Show Tags

23 Apr 2006, 11:25
1 is sufficiient only if ABC is equilateral triangle.. and thats not said..
was wondering if the area says that is equilateral..

Kudos [?]: 37 [0], given: 0

Director
Joined: 04 Jan 2006
Posts: 922

Kudos [?]: 37 [0], given: 0

Show Tags

23 Apr 2006, 11:27
I would go with B..

if ABC is equilateral, then area of mno is 1/4th of abc..

1) says area is sqrt3/4... which is base *height = sqrt3/2.. which gives many possibilites for base.

Kudos [?]: 37 [0], given: 0

VP
Joined: 29 Apr 2003
Posts: 1403

Kudos [?]: 30 [0], given: 0

Show Tags

23 Apr 2006, 14:40
Answer is D. The midpoints essentially divide the triangle into 4 equal aread triangles.

Kudos [?]: 30 [0], given: 0

Senior Manager
Joined: 08 Jun 2004
Posts: 494

Kudos [?]: 93 [0], given: 0

Location: Europe

Show Tags

24 Apr 2006, 05:08
AgreeD.

Guys would you please remaind me how to find the area of the equilateral (or any triangle) triangle knowing only the height? Thank you.

Kudos [?]: 93 [0], given: 0

Director
Joined: 04 Jan 2006
Posts: 922

Kudos [?]: 37 [0], given: 0

Show Tags

24 Apr 2006, 17:44
h = a*sqrt(3)/2
area of equi = sqrt(3)a*a/4

Kudos [?]: 37 [0], given: 0

Retired Moderator
Joined: 18 Jul 2008
Posts: 960

Kudos [?]: 294 [0], given: 5

Show Tags

28 Nov 2008, 15:52
Does anyone have a better explanation for this?

My 2 questions are:

1) how do we prove that triangle ABC is divided into equal 4 smaller triangles by knowing the midpoints.

2) How do we find the area of MON if we only know the height of ABC.

Kudos [?]: 294 [0], given: 5

Manager
Joined: 27 May 2008
Posts: 200

Kudos [?]: 45 [0], given: 0

Show Tags

28 Nov 2008, 21:13
Hi Yach,

1 - cant be proved by just knowing midpoint. we need height and sides of the triangle. With Area we cant prove unless it is an equilateral triangle.

2) we know h = V3/2 * Side for equilateral triangle
so side = 1 in our case since h = V3/2

Area of MON = 1/4 of Area ABC. since equilateral.

Kudos [?]: 45 [0], given: 0

Director
Joined: 29 Aug 2005
Posts: 855

Kudos [?]: 487 [0], given: 7

Show Tags

15 Feb 2009, 03:52
I agree that the answer should be D.

In the link below, I found explanation showing that a triangle formed by connecting midpoints of the triangle divides the area of this bigger triangle into 4 equal areas:
http://mathworld.wolfram.com/MedialTriangle.html

Kudos [?]: 487 [0], given: 7

CEO
Joined: 17 Nov 2007
Posts: 3584

Kudos [?]: 4580 [1], given: 360

Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40

Show Tags

15 Feb 2009, 04:37
1
KUDOS
Expert's post
yach wrote:
M, N, and O are midpoints of sides AB, BC, and AC of triangle ABC. What is the area of triangle MON?

The main idea here is realizing that $$S_{MON}=\frac14*S_{ABC}$$

1. Let's consider vertex A: M and O are midpoints of AB and AC. In other words, all linear sizes of MAO triangle is smaller by 2 times than all linear sizes of BAC. Therefore,$$S_{MAO}=\frac14*S_{ABC}$$

2. Applying the same reasoning for each vertex we will get:
$$S_{MON}=S_{ABC} - (S_{MAO}+S_{MBN}+S_{NCO}) = S_{ABC} - (\frac14*S_{ABC}+\frac14*S_{ABC}+\frac14*S_{ABC}) =\frac14*S_{ABC}$$
_________________

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

Kudos [?]: 4580 [1], given: 360

Manager
Joined: 02 Aug 2007
Posts: 223

Kudos [?]: 74 [0], given: 0

Schools: Life

Show Tags

15 Feb 2009, 16:24
walker wrote:
yach wrote:
M, N, and O are midpoints of sides AB, BC, and AC of triangle ABC. What is the area of triangle MON?

The main idea here is realizing that $$S_{MON}=\frac14*S_{ABC}$$

But how do you arrive at this conclusion, were are not told that either ABC or MNO are equilateral.
AB, BC, and OC can each be a different length, as can MN, ON, OM.

Kudos [?]: 74 [0], given: 0

SVP
Joined: 29 Aug 2007
Posts: 2472

Kudos [?]: 842 [0], given: 19

Show Tags

15 Feb 2009, 22:40
xALIx wrote:
walker wrote:
yach wrote:
M, N, and O are midpoints of sides AB, BC, and AC of triangle ABC. What is the area of triangle MON?

The main idea here is realizing that $$S_{MON}=\frac14*S_{ABC}$$

But how do you arrive at this conclusion, were are not told that either ABC or MNO are equilateral.
AB, BC, and OC can each be a different length, as can MN, ON, OM.

This is good reference: http://mathworld.wolfram.com/MedialTriangle.html

Mid point therom says that a triangle made from connecting mid-points of the sides of a given triangle is 1/4 of the original triangle. In that case, it is not required to be equilateral.
_________________

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

GT

Kudos [?]: 842 [0], given: 19

Director
Joined: 27 May 2008
Posts: 541

Kudos [?]: 363 [2], given: 0

Show Tags

15 Feb 2009, 22:56
2
KUDOS
for any triangle, ABC if you join mid points and divide it into 4 parts, the area will be divided into 4 triangles with each having an area equal to 1/4 of triangle ABC.

Lets prove is with simple method, visualize the triangle ABC on co-ordinate plane

A = (0,0)
B = (x,0)
C = (a,b)

Area = xb/2 (Note that area does not depend on a)

M = (x/2, 0)
N = ((x+a)/2, b/2)
O = (a/2, b/2)

base of triangle, NO = x/2
Height = b/2
Area of MNO = xb/8

Similarly triangle AMO
base AM = x/2
height = b/2
Area = xb/8

Kudos [?]: 363 [2], given: 0

CEO
Joined: 17 Nov 2007
Posts: 3584

Kudos [?]: 4580 [0], given: 360

Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40

Show Tags

16 Feb 2009, 00:01
xALIx wrote:
But how do you arrive at this conclusion, were are not told that either ABC or MNO are equilateral.
AB, BC, and OC can each be a different length, as can MN, ON, OM.

This problem test similarity. ABC and MAO (as other small triangles) are similar triangles: the same angle A and the same relation between AB/AC=AM/AO.
_________________

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

Kudos [?]: 4580 [0], given: 360

Intern
Joined: 22 Dec 2009
Posts: 22

Kudos [?]: 71 [0], given: 1

Show Tags

03 Sep 2010, 10:34
Please explain how 2 is sufficient. I know how to get the area of an equilateral but a bit puzzled as we are only given the height

Kudos [?]: 71 [0], given: 1

Director
Joined: 28 Jun 2011
Posts: 879

Kudos [?]: 244 [0], given: 57

Show Tags

30 Nov 2011, 09:41
willget800 wrote:
I would go with B..

if ABC is equilateral, then area of mno is 1/4th of abc..

1) says area is sqrt3/4... which is base *height = sqrt3/2.. which gives many possibilites for base.

Even i selected B....but this weblink is a good solution to this mistake

http://mathworld.wolfram.com/MedialTriangle.html
_________________

Kudos [?]: 244 [0], given: 57

Math Expert
Joined: 02 Sep 2009
Posts: 41885

Kudos [?]: 128722 [3], given: 12182

Show Tags

22 Apr 2012, 06:18
3
KUDOS
Expert's post
2
This post was
BOOKMARKED
yach wrote:
If $$M$$ , $$N$$ , and $$O$$ are midpoints of sides $$AB$$ , $$BC$$ , and $$AC$$ of triangle $$ABC$$ . What is the area of triangle $$MON$$ ?

1. The area of $$ABC$$ is $$\frac{\sqrt{3}}{4}$$
2. $$ABC$$ is an equilateral triangle with height $$\frac{sqrt3}{2}$$

[Reveal] Spoiler: OA
D

Source: GMAT Club Tests - hardest GMAT questions

Look at the diagram below:

MN, NO and OM each are midsegments of triangle ABC (midsegment is a line segment joining the midpoints of two sides of a triangle). Important property of a midsegment: the midsegment is always half the length of the third side. So, $$MN=\frac{AC}{2}$$, $$NO=\frac{AB}{2}$$ and $$OM=\frac{BC}{2}$$

Next, since each side of triangle MNO is half of the side of triangle ABC then these triangles are similar (the ratio of all the sides are the same). Important property of similar triangles: if two similar triangles have sides in the ratio $$\frac{x}{y}$$, then their areas are in the ratio $$\frac{x^2}{y^2}$$.

Since the sides of two similar triangles MNO and ABC are in the ratio 1:2 then then their areas are in the ratio 1:4 --> (area of MNO)=(area of ABC)/4.

So, in order to find the area of MNO we should find the area of ABC.

(1) The area of ABC is $$\frac{\sqrt{3}}{4}$$. Sufficient.
(2) ABC is an equilateral triangle with height $$\frac{sqrt3}{2}$$ --> we can find the area of equilateral triangle with given altitude. Sufficient.

_________________

Kudos [?]: 128722 [3], given: 12182

Current Student
Joined: 07 Sep 2011
Posts: 74

Kudos [?]: 50 [0], given: 13

GMAT 1: 660 Q41 V40
GMAT 2: 720 Q49 V39
WE: Analyst (Mutual Funds and Brokerage)

Show Tags

28 Aug 2012, 10:52
Based on the prompt, We know that the each side of triangle MON is going to be exactly HALF the length of the corresponding sides of triangle ABC. Which means that these two are SIMILAR triangles, and we will be able to figure out the area of MON if we are given the area of ABC. based on the ratio s^2:s^2 (s=side).

1) SUFFICIENT.
2) if ABC is an equilateral triangle, its height is the same from any base, and also cuts the triangle ABC in half to make it 2 right triangles with sides ratio of x:x^(1/2):2x. Knowing the side of the longest leg, we are able to calculate the rest of the sides and hence the area of ABC. SUFFICIENT.

[Reveal] Spoiler:
D.

PS: Notice that I did not actually do any calculations in this problem. It was all conceptual.

Kudos [?]: 50 [0], given: 13

Intern
Status: Looking for High GMAT Score
Joined: 19 May 2012
Posts: 36

Kudos [?]: 6 [0], given: 58

Location: India
Concentration: Strategy, Marketing
WE: Marketing (Internet and New Media)

Show Tags

28 Aug 2012, 23:09
I got answer but was able to solve,as both equations tells the same things answer D
_________________

“The best time to plant a tree was 20 years ago. The second best time is now.” – Chinese Proverb

Kudos [?]: 6 [0], given: 58

Re: DS Geometry (m08q22)   [#permalink] 28 Aug 2012, 23:09

Go to page    1   2    Next  [ 21 posts ]

Display posts from previous: Sort by

DS Geometry (m08q22)

Moderator: Bunuel

 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®.