Given a triangle ABC, where G is the intersection of point of medians

Question

Given a triangle ABC, where G is the intersection of point of medians, what fraction of the triangle ABC is shaded?

A.  \frac{1}{4}

B.  \frac{1}{12}

C.  \frac{1}{6}

D.  \frac{1}{8}

E.  \frac{1}{9}

ManjariMishra · Answer

Could you please explain this

2asc · Answer

From the question, we can deduct that ADE &ABC are similar triangles, so DE=1/2BC, DE//BC
hence height of triangle ADE=1/2ABC

Since DE//BC, DGE&ABC are similar triangles
Since DE=1/2BC, then height of the two triangles (from G) also is 1:2, which makes the height of the trapezoid DECB 3
Since height of triangle ADE=1/2ABC, which makes height of ADE 3 as well. 
So, the height of ABC=3+3=6

Since 2DE=BC, then, Area of ABC = (6x2)times of DGE.

So answer, DGE:ABC = 1/12.

lotif · Answer

2asc  I dont understand how you get the height from G to base.

ManjariMishra · Answer

I have solved it like this .
We can deduce that ADE and ABC are similar triangles.(SAS property) AD/AB= 1/2 ,AE/Ac=1/2 and angle A is common .
Therefore DE/BC=1/2.
Also BC !! DE (D and E are mid points so angle EDG = angle GCB and angle DEG = angle GBC so the third angles are also same and hence both the triangles DGE are GBC are similar .For a centroid the areas of all the six trinagles are equal so area of BGC will be (1/6 of area of ABC + 1/6 of area of ABC i.e 2/6 of area of ABC .
Also as DE/BC=1/2 so the area of DBC/Area of BGC = 1/4

So the area of DBC = 1/4 area pf BGC 
 = 1/4*2/6 area of ABC
==> Area of DBC/Area of ABC = 1/12

2asc · Answer

rezalotif 
Because DE//BC, it makes DGE & BGC similar triangles. 
From earlier we established that ADE and ABC are similar triangles, where the sides of ABC is twice of ADE, which means DE=1/2BC
Now look at DE & BC in terms of triangle DGE and BGC. We established DGE&BGC are similar triangles and DE=1/2BC, then the height of the two triangles should reflect the same ratio. 
Suppose height from G to DE is a, and G to BC is b, then DE/BC=a/b=1/2. 
Suppose a is 1, then b is 2, a+b=3=half of height of triangle ABC
So height of ABC is 3x2=6 times of height of DGE

Hope this helps!

BrentGMATPrepNow · Answer

Since the question does not mention anything about the kind of triangle we have (isosceles, scalene, etc), we can assume that we can use ANY triangle. 
So, let's use a nice equilateral triangle with sides of length 2
 https://i.imgur.com/fikB9kw.png

Area of an equilateral triangle = (√3)(side²/4)
So, area = (√3)(2²/4) 
= (√3)(4/4)
=   √3

Now let's draw our medians 
 https://i.imgur.com/aQo3XFK.png

We can also add some angles
 https://i.imgur.com/NkCLglS.png

We need to find the area of the RED triangle in the middle
 https://i.imgur.com/7sLUlSi.png

So, let's flip that red triangle like so...
 https://i.imgur.com/yspVhua.png

If we draw an altitude from the top vertex, the line divides the triangle into two 30-60-90 special right triangles. 
 https://i.imgur.com/ajr1HP9.png 
When we compare the RED 30-60-90 triangles with the base 30-60-90 triangle we can calculate the height (h) 
Since we're comparing SIMILAR triangles, we can write:  0.5 /  √3   =  h /  1  
Cross multiply to get: (0.5)(1) = (√3)(h)
Simplify: 1/2 = (√3)(h)
Solve: h = 1/(2√3)

We have the following
 https://i.imgur.com/sPDrusA.png 
Area of triangle = (base)(height)/2
= (1) /2
=   1/(4√3)

What fraction of the triangle ABC is shaded?  
Fraction =   1/(4√3)  / √3 
=      ( 1/√3 )
= 1/12

Answer: B

Cheers, 
Brent

BrentGMATPrepNow · Answer

https://i.imgur.com/e1JwGyX.png

IMPORTANT: the diagrams in GMAT Problem Solving questions (like the one above) are  DRAWN TO SCALE unless stated otherwise .
So, even if we have no idea where to begin, we can certainly eliminate 2 or 3 answers by simply "eyeballing" the diagram.

RELATED VIDEO FROM OUR COURSE
 https://www.youtube.com/watch?v=MaPuJh3m3Pk

paulbro4 · Answer

I don't understand why we can just call it an equilateral triangle.

EDIT:

To anyone else confused. The fact the that the question is only asking what  fraction  of the triangle is shaded means we can pic any triangle we want because the proportions created by this scenario remain the same for any triangle. Choosing an equilateral triangle then just makes the math much simpler.

Fdambro294 · Answer

IF you know the concept, there is a way to get to the answer rather quickly (though it is beyond the GMAT).

See  BrentGMATPrepNow  for the best solution.

Ignoring the shaded part, the concept is the following.

Property: when all 3 medians are drawn from each vertex of the triangle, they will intersect at the centroid. The centroid and the 3 medians will create 6 triangles of equal area.

So draw in the 3rd median from Vertex A to side BC through point G.

If we let the area of the triangle be 6, then 4 of the 6 triangles will be below the shaded portion.

If you’ve left out line DE for the time being, you can see that triangle BGC represents 2 of the 6 triangles that are created. Call this Triangle BGC Area = 2 for the 2 out of 6 equal triangles it represents.

Mid segment Theorem: when a line segment is drawn from the midpoint of one side of a triangle to the midpoint of the other side of a triangle, it will be parallel to the opposite side. Furthermore, the line segment will be 1/2 length of that opposite side

Hence, DE is parallel to BG. Using the parallel line properties when intersected by a transversal, we can find that triangle DGE (the shaded triangle) is similar to triangle BGC.

Since DE is 1/2 length of BC and fhey are corresponding sides of the similar triangles, we can say the areas are in the Ratio of: (1)^2 ; (2)^2 = 1:4

This means triangle DGE is 1/4th the area of triangle BGC.

We called triangle BGC Area = 2

1/4 * 2 = Area of 1/2 for the shaded portion

And because of the “median-centroid rule” above, we called the entire triangle Area = 6

(Shaded portion) / (entire triangle area) = (1/2) / (6) =

1/12

Posted from my mobile device

bumpbot · Answer

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Given a triangle ABC, where G is the intersection of point of medians

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Given a triangle ABC, where G is the intersection of point of medians

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