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Mastering Important Concept on Triangles – III - Question #1

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Mastering Important Concept on Triangles – III - Question #1  [#permalink]

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Updated on: 15 Dec 2016, 23:50
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65% (hard)

Question Stats:

68% (02:40) correct 32% (02:53) wrong based on 290 sessions

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Mastering Important Concept on Triangles – III - Question #1

In the given figure, BD is the median of triangle ABC and the lines BD and CE intersect at point F. If the area of triangle BFC is $$6cm^2$$ and area of the quadrilateral FEAD is $$18cm^2$$. Find the area of triangle FDC, if the ratio of the area of triangle FDC and triangle FEB is $$3:2$$.

a. $$12 cm^2$$
b. $$18 cm^2$$
c. $$24 cm^2$$
d. $$36 cm^2$$
e. $$42 cm^2$$

Key concepts on Triangles are explained in detail in the following posts:

1. Mastering Important Concepts Tested By GMAT in Triangle - I

2. Mastering Important Concepts Tested By GMAT in Triangle - II

3.Mastering Important Concepts Tested By GMAT in Triangle - III

Detailed solution will be posted soon.

Thanks,
Saquib
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e-GMAT

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Originally posted by EgmatQuantExpert on 11 Dec 2016, 08:29.
Last edited by EgmatQuantExpert on 15 Dec 2016, 23:50, edited 1 time in total.
e-GMAT Representative
Joined: 04 Jan 2015
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Re: Mastering Important Concept on Triangles – III - Question #1  [#permalink]

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Updated on: 15 Dec 2016, 23:49
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Hey,

PFB the official solution

It is given that BD is the median of the triangle ABC.

From the given information we can conclude that:
Area of Triangle ABD = Area of Triangle BDC
Which is equal to,
Area of Triangle FEB + Area of Quadrilateral FEAD = Area of Triangle BFC + Area of Triangle FDC…. (i)

We are also given the following information –
• Area of BFC = $$6 cm^2$$
• Area of Quadrilateral FEAD = $$18cm^2$$
• And area of FDC: Area of FEB = 3: 2
o Let us consider the area of FDC and FEB to be $$3x$$ and $$2x$$ respectively.

Let us substitute the above values in equation (i)
Area of Triangle FEB + Area of Quadrilateral FEAD = Area of Triangle BFC + Area of Triangle FDC
o $$2x + 18 = 6 + 3x$$
o $$x = 12 cm^2$$
• Thus the Area of Triangle FDC = $$3x = 36 cm^2$$

Hence the correct answer option is D

Thanks,
Saquib
Quant Expert
e-GMAT
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Originally posted by EgmatQuantExpert on 11 Dec 2016, 08:31.
Last edited by EgmatQuantExpert on 15 Dec 2016, 23:49, edited 1 time in total.
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Re: Mastering Important Concept on Triangles – III - Question #1  [#permalink]

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14 Dec 2016, 22:14
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Since BD is median , The area of triangle BDA= Area of triangle BDC

The area of triangle BDA= Area of triangle BDC

Area of BFE+ Area EFAD = Area BFC+ Area CFD

2x + 18 = 3x+ 6

X= 12.

we need 3x = 36 cm^2
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Mastering Important Concept on Triangles – III - Question #1  [#permalink]

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15 Dec 2016, 23:54
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Joined: 07 Dec 2016
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Mastering Important Concept on Triangles – III - Question #1  [#permalink]

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31 Aug 2017, 14:31
As BD is the median , it divides triangle ABC into two triangles of equal area.

Area of triangle ABD = Area of triangle BDC
Area of triangle BFE + Area of quadilateral EAFD = Area of triangle BFC + Area of triangle FDC ........(1)

Also, Area of FDC/Area of FEB = 3/2
which gives us , Area of FEB = 2*Area of FDC / 3

Substituting above relation in our equation (1)

2*Area of FDC / 3 +Area of EFAD = Area of BFC + Area of FDC
Area of FDC/3 = Area of EFAD - Area of BFC
Area of FDC = 3( 18-6)
Area of FDC = 36 cm^2

Option D
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Re: Mastering Important Concept on Triangles – III - Question #1  [#permalink]

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15 Sep 2018, 11:34
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Re: Mastering Important Concept on Triangles – III - Question #1   [#permalink] 15 Sep 2018, 11:34
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