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Re: In the rectangle above, A is the midpoint of the side, and [#permalink]

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15 Aug 2011, 05:58

1

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Consider this, area for triangle is BH/2.

the two bigger triangles have the same B & H, the three small triangles have the same base and height. thus we can say the three smaller triangles or the 2 larger are equal. with (1) or (2) we can see the area of the full rectangle is 96.

Re: In the rectangle above, A is the midpoint of the side, and [#permalink]

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15 Aug 2011, 07:08

DeeptiM wrote:

well u wld hve to b a lil patient wid me...

somehow um still unclear with the explanation

D from my side.

1. Area of rectangle = 2*2*Area(AOB) - Suff. 2. Area of rectangle = 2*3*Area COD (because Area BOC= Area COD= Area DOC) - Suff.

Hope it helps.

Cheers, Aj.
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Re: In the rectangle above, A is the midpoint of the side, and [#permalink]

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15 Aug 2011, 07:40

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DeeptiM wrote:

In the rectangle above, A is the midpoint of the side, and BC=CD=DE. What is the area of the rectangle? (1) The area of the shaded region is 24. (2) The area of triangle CDO is 16.

I got the answer but need a more concise approach..

Attachment:

Geomtery_Median_DS.JPG

Let's call the vertex diagonally opposite of E as F.

1. OA is the median of \(\triangle OBF\), because A is the mid-point of FB. A median divides a triangle in two-halves such that the areas of two newly formed triangles are equal.

In the rectangle above, A is the midpoint of the side, and BC = CD = DE. What is the area of the rectangle?

The area of a rectangle is (width)*(length).

Notice that since A is the midpoint of the width of the rectangle, then AB = (width)/2 Also, since BC = CD = DE, then CD = (length)/3.

(1) The area of the shaded region is 24. The area of the shaded triangle = 1/2*BE*AB = 1/2*(length)*(width)/2 = 24 --> (length)*(width) = 96. Sufficient.

(2) The area of triangle CDO is 16. The area of triangle CDO = 1/2*OE*CD = 1/2*(width)*(length)/3 = 16 --> (width)*(length) = 96. Sufficient.

Re: In the rectangle above, A is the midpoint of the side, and [#permalink]

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14 Nov 2015, 15:31

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In the rectangle above, A is the midpoint of the side, and [#permalink]

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06 May 2016, 10:33

A question,

assume that the upper left vertice is called X. So according to 1) since B is the mid point, would it be correct to assume that the shaded region has the same area as AXO? So the shaded region is 1/4th of the whole rectangle.

Same goes to 2) since the base is 1/3rd of the length, so the region CDO is 1/3rd of the "lower" right triangle BOE and 1/6th of the rectangle.

Re: In the rectangle above, A is the midpoint of the side, and [#permalink]

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23 Sep 2017, 09:54

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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