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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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New post Updated on: 15 Dec 2016, 23:50
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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 \(3:2\).


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Answer Choices

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

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New post 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


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

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New post 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 :) :)

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Number Properties – Even Odd | LCM GCD
Word Problems – Percentage 1 | Percentage 2 | Time and Work 1 | Time and Work 2 | Time, Speed and Distance 1 | Time, Speed and Distance 2
Advanced Topics- Permutation and Combination 1 | Permutation and Combination 2 | Permutation and Combination 3 | Probability
Geometry- Triangles 1 | Triangles 2 | Triangles 3 | Common Mistakes in Geometry
Algebra- Wavy line

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Number Properties 1 | Number Properties 2 | Algebra 1 | Geometry | Prime Numbers | Absolute value equations | Sets



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

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