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# Mastering Important Concept on Triangles – III - Question #1

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e-GMAT Representative
Joined: 04 Jan 2015
Posts: 746

Kudos [?]: 2080 [0], given: 123

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

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11 Dec 2016, 08:29
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Question Stats:

63% (01:49) correct 37% (01:56) wrong based on 97 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
Quant Expert
e-GMAT

[Reveal] Spoiler: OA

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Last edited by EgmatQuantExpert on 15 Dec 2016, 23:50, edited 1 time in total.

Kudos [?]: 2080 [0], given: 123

e-GMAT Representative
Joined: 04 Jan 2015
Posts: 746

Kudos [?]: 2080 [2], given: 123

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

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11 Dec 2016, 08:31
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Expert's post
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
_________________

| '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com

Last edited by EgmatQuantExpert on 15 Dec 2016, 23:49, edited 1 time in total.

Kudos [?]: 2080 [2], given: 123

Manager
Joined: 07 Mar 2015
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Location: India
Concentration: General Management, Operations
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Re: Mastering Important Concept on Triangles – III - Question #1 [#permalink]

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14 Dec 2016, 22:14
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

Kudos [?]: 27 [0], given: 48

e-GMAT Representative
Joined: 04 Jan 2015
Posts: 746

Kudos [?]: 2080 [0], given: 123

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

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15 Dec 2016, 23:54
Hey Everyone,

The official solution has been posted. Please go through the solution once. In case of any doubts please feel to ask

Thanks,
Saquib
Quant Expert
e-GMAT
_________________

| '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com

Kudos [?]: 2080 [0], given: 123

Intern
Joined: 07 Dec 2016
Posts: 41

Kudos [?]: 9 [0], given: 74

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

Kudos [?]: 9 [0], given: 74

Mastering Important Concept on Triangles – III - Question #1   [#permalink] 31 Aug 2017, 14:31
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