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DS Geometry (m08q22) [#permalink]
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23 Apr 2006, 09:25
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This topic is locked. If you want to discuss this question please repost 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 \(\frac{sqrt3}{2}\) Source: GMAT Club Tests  hardest GMAT questions Please explain your answer. Thank You.



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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
The answer is D



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1 is sufficiient only if ABC is equilateral triangle.. and thats not said..
was wondering if the area says that is equilateral..



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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.



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Answer is D. The midpoints essentially divide the triangle into 4 equal aread triangles.



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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.



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h = a*sqrt(3)/2
area of equi = sqrt(3)a*a/4



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Re: DS Geometry [#permalink]
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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.



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Re: DS Geometry [#permalink]
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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.



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Re: DS Geometry [#permalink]
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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



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Re: DS Geometry [#permalink]
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15 Feb 2009, 04:37
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: \(S_{MON}=S_{ABC}  (S_{MAO}+S_{MBN}+S_{NCO}) = S_{ABC}  (\frac14*S_{ABC}+\frac14*S_{ABC}+\frac14*S_{ABC}) =\frac14*S_{ABC}\)
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Re: DS Geometry [#permalink]
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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.



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Re: DS Geometry [#permalink]
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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.htmlMid point therom says that a triangle made from connecting midpoints of the sides of a given triangle is 1/4 of the original triangle. In that case, it is not required to be equilateral.
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Re: DS Geometry [#permalink]
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15 Feb 2009, 22:56
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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 coordinate 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



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Re: DS Geometry [#permalink]
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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.
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Re: DS Geometry (m08q22) [#permalink]
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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



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Re: DS Geometry (m08q22) [#permalink]
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22 Apr 2012, 06:18
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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 \(\frac{sqrt3}{2}\) Source: GMAT Club Tests  hardest GMAT questions Please explain your answer. Thank You. 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. Answer: D.
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Re: DS Geometry (m08q22) [#permalink]
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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. Answer is PS: Notice that I did not actually do any calculations in this problem. It was all conceptual.



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Re: DS Geometry (m08q22) [#permalink]
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28 Aug 2012, 23:09
I got answer but was able to solve,as both equations tells the same things answer D
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Re: DS Geometry (m08q22)
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