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M18-04

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M18-04 [#permalink]

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New post 16 Sep 2014, 01:02
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If in triangle \(ABC\) angle \(ABC\) is the largest and point \(D\) lies on segment \(AC\), is the area of triangle \(ABD\) larger than that of triangle \(DBC\)?


(1) \(AD \lt DC\)

(2) \(AB \lt BC\)

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Re M18-04 [#permalink]

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New post 16 Sep 2014, 01:03
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Official Solution:


Consider the diagram below:

Image

Notice that \(BE\) is the height of the triangle \(ABC\). Now, the area of triangle \(ABD\) is \(\frac{1}{2}*height*base=\frac{1}{2}*BE*AD\) and the area of triangle \(DBC\) is \(\frac{1}{2}*height*base=\frac{1}{2}*BE*DC\). So, we can see that the area of triangle \(ABD\) will be greater than the area of triangle \(DBC\) if \(AD\) is greater than \(DC\).

(1) \(AD \lt DC\). Sufficient.

(2) \(AB \lt BC\). Not sufficient.


Answer: A
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M18-04 [#permalink]

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New post 14 Jul 2015, 18:52
Hi,

Why can't i change my base as in either BD ,AB,BC.
In this case my height will vary.
Please explain how come you have height to be BE and not (BD,AB,BC).

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Re: M18-04 [#permalink]

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New post 14 Jul 2015, 20:28
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priyankasharma123 wrote:
Hi,

Why can't i change my base as in either BD ,AB,BC.
In this case my height will vary.
Please explain how come you have height to be BE and not (BD,AB,BC).

Thanks


hi,
irrespestive of what height depending on the base we take, the area will remain the same..
Since here we have a height and a base and the area of triangle depends on the base as the height in this case will be the same and thus only one variable involved, Why should we complicate the things by taking a bases , which are not the same resulting in different heights and thus two variables..
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Re M18-04 [#permalink]

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New post 09 Jul 2016, 04:14
I think this is a high-quality question and I agree with explanation.
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Re M18-04 [#permalink]

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New post 31 Jul 2016, 06:49
I think this is a high-quality question and I agree with explanation. I have another approach.

Triangle Median property
1. The median of the triangle divides the triangle into two smaller triangles which have the same area.
In this case point D is not a median. Applying above property on statement 1 we can say area of triangle ABD > area of triangle DBC.

Statement 2 clearly not sufficient.
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Re: M18-04 [#permalink]

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New post 06 Jun 2017, 09:48
The answer must be option A. My take :

The area of a triangle is 0.5 * base * height. Now we need to compare the areas of ABD and BDC.
The area of ABD is 0.5*AD*h.
The area of BDC is 0.5*CD*h.

To compare the area of ABD and BDC, we need to know the relation between AD and CD.

Option A - Sufficient
Option B - Not sufficient.

Hence option A is the answer.
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Re: M18-04 [#permalink]

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New post 10 Mar 2018, 09:43
Hi, What happens if the triangle is a right triangle with angle B = 90 and BC as the base?

Area of ABD is 1/2AB and DE where e is the height drawn from D (a point on the hypotenuse) to line AB
Area of DBC is 1/2 BC and DK where k is the height drawn from point D to BC

Help from here?
Re: M18-04   [#permalink] 10 Mar 2018, 09:43
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