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yach · **Posted:** Sun Apr 23, 2006 9:25 am


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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

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Please explain your answer. Thank You.

durgesh79 · **Posted:** Sun Feb 15, 2009 10:56 pm

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

sherinaparvin · **Posted:** Sun Apr 23, 2006 10:17 am

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

willget800 · **Posted:** Sun Apr 23, 2006 11:25 am

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

willget800 · **Posted:** Sun Apr 23, 2006 11:27 am

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.

sm176811 · **Posted:** Sun Apr 23, 2006 2:40 pm

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

M8 · **Posted:** Mon Apr 24, 2006 5:08 am

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.

willget800 · **Posted:** Mon Apr 24, 2006 5:44 pm

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

bigfernhead · **Posted:** Fri Nov 28, 2008 3:52 pm

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.

selvae · **Posted:** Fri Nov 28, 2008 9:13 pm

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.

seofah · **Posted:** Sun Feb 15, 2009 3:52 am

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

walker · 04/27/2012

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}

xALIx · **Posted:** Sun Feb 15, 2009 4:24 pm

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.

GMAT TIGER · **Posted:** Sun Feb 15, 2009 10:40 pm

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.

walker · 04/27/2012

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.

cmugeria · **Posted:** Fri Sep 03, 2010 10:34 am

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

prateekbhatt · **Posted:** Wed Nov 30, 2011 9:41 am

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

Bunuel · **Posted:** Sun Apr 22, 2012 6:18 am

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

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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