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# In the figure above, AD = DC, and BE = EC. The ratio of area of ∆ ABE

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Math Expert
Joined: 02 Sep 2009
Posts: 54376
In the figure above, AD = DC, and BE = EC. The ratio of area of ∆ ABE  [#permalink]

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08 Aug 2017, 10:32
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Difficulty:

45% (medium)

Question Stats:

61% (01:55) correct 39% (01:55) wrong based on 39 sessions

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In the figure above, AD = DC, and BE = EC. The ratio of area of ∆ ABE / area of ∆ BDC is

(A) greater than 0 and less than 1/2
(B) equal to 1/2
(C) greater than 1/2 and less than 1
(D) equal to 1
(E) greater than 1

Attachment:

2017-08-08_2131.png [ 9.96 KiB | Viewed 881 times ]

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Re: In the figure above, AD = DC, and BE = EC. The ratio of area of ∆ ABE  [#permalink]

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23 Aug 2017, 04:01
Bunuel wrote:

In the figure above, AD = DC, and BE = EC. The ratio of area of ∆ ABE / area of ∆ BDC is

(A) greater than 0 and less than 1/2
(B) equal to 1/2
(C) greater than 1/2 and less than 1
(D) equal to 1
(E) greater than 1

Attachment:
2017-08-08_2131.png

lets take area of ∆ ABC = 2 * ∆ ABE (BE=EC)

now if we compare ∆ ABC and ∆ BDC
DC = 1/2 * AC
BC = BC
hence area of ∆ ABC = 2* ∆ BDC

hence area of ∆ ABE = area of ∆ BDC
D
Please correct me if i am wrong
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In the figure above, AD = DC, and BE = EC. The ratio of area of ∆ ABE  [#permalink]

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23 Aug 2017, 04:06
2
1
The median bisects the triangle into two triangles of equal areas.The areas of all the four triangles are equal.So the ratio is equal to 1.(Option D). Please give me kudos . I need them badly to increase my score.
Math Expert
Joined: 02 Aug 2009
Posts: 7568
Re: In the figure above, AD = DC, and BE = EC. The ratio of area of ∆ ABE  [#permalink]

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23 Aug 2017, 04:09
1
Bunuel wrote:

In the figure above, AD = DC, and BE = EC. The ratio of area of ∆ ABE / area of ∆ BDC is

(A) greater than 0 and less than 1/2
(B) equal to 1/2
(C) greater than 1/2 and less than 1
(D) equal to 1
(E) greater than 1

Attachment:
2017-08-08_2131.png

Hi..
so area of ADB = area of DCB= $$\frac{1}{2}$$ area of ABC
BASE is equal and height is same

II. BE=EC
so area of AEB = area of AEC= $$\frac{1}{2}$$ area of ABC
BASE is equal and height is same..

from I and II area of AEB = Area of BDC
so ratio=area of AEB / Area of BDC = 1

D
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Re: In the figure above, AD = DC, and BE = EC. The ratio of area of ∆ ABE   [#permalink] 23 Aug 2017, 04:09
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