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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  [#permalink]

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New post 15 Jan 2019, 04:57
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E

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Question Stats:

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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. \(\frac{1}{9}\)

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Re: Given a triangle ABC, where G is the intersection of point of medians  [#permalink]

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New post 15 Jan 2019, 16:05
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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.
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Re: Given a triangle ABC, where G is the intersection of point of medians  [#permalink]

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New post 16 Jan 2019, 09:41
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2asc wrote:
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.


2asc I dont understand how you get the height from G to base.
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Re: Given a triangle ABC, where G is the intersection of point of medians  [#permalink]

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New post 16 Jan 2019, 11:29
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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
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Re: Given a triangle ABC, where G is the intersection of point of medians  [#permalink]

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New post 16 Jan 2019, 12:37
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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!
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Re: Given a triangle ABC, where G is the intersection of point of medians  [#permalink]

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New post 15 Jan 2019, 09:07
Could you please explain this
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Re: Given a triangle ABC, where G is the intersection of point of medians  [#permalink]

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New post 16 Jan 2019, 15:46
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Yuridias wrote:
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. \(\frac{1}{9}\)


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
Image

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

Now let's draw our medians
Image

We can also add some angles
Image

We need to find the area of the RED triangle in the middle
Image

So, let's flip that red triangle like so...
Image

If we draw an altitude from the top vertex, the line divides the triangle into two 30-60-90 special right triangles.
Image
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
Image
Area of triangle = (base)(height)/2
= (1)[1/(2√3)]/2
= 1/(4√3)

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

Answer: B

Cheers,
Brent
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Re: Given a triangle ABC, where G is the intersection of point of medians  [#permalink]

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New post 16 Jan 2019, 15:49
Top Contributor
Yuridias wrote:
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. \(\frac{1}{9}\)


Image

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.


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Given a triangle ABC, where G is the intersection of point of medians  [#permalink]

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New post 22 Jan 2019, 12:08
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.
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Given a triangle ABC, where G is the intersection of point of medians   [#permalink] 22 Jan 2019, 12:08
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