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

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

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15 Aug 2011, 02:57
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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.
[Reveal] Spoiler: OA

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Geomtery_Median_DS.JPG [ 10.3 KiB | Viewed 2911 times ]

Geometry.docx [14.79 KiB]

Last edited by DeeptiM on 15 Aug 2011, 07:06, edited 1 time in total.
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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, 06:58
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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.

(18*3)*2 or (24*2)*2
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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, 07:10
well u wld hve to b a lil patient wid me...

somehow um still unclear with the explanation
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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, 08:08
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, 08:40
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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.

Area(OAF)=Area(OAB)=24
Area(OBF)=2*Area(OAB)=2*24=48

A diagonal of a rectangle divides the rectangle in two equal halves such that the area remains same for both halves.

Thus, Area(ABEO)=2*Area(BOF)=2*48=96
Sufficient.

2.
Same concept as statement 1.
OD is the median of OCE because CD=DE
AND
OC is the median of OBD because CD=BC

Area(ODE)=Area(OCD)=Area(OCB)=16
Area(OBE)=3*16=48
Area(FBEO)=2*Area(OBE)=2*48=96
Sufficient.

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

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31 Jul 2014, 03:16
Need a explanation of this.
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Re: In the rectangle above, A is the midpoint of the side, and [#permalink]

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31 Jul 2014, 07:33
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Vijayeta wrote:
Need a explanation of this.

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.

Hope it's clear.
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Re: In the rectangle above, A is the midpoint of the side, and [#permalink]

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14 Nov 2015, 16:31
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In the rectangle above, A is the midpoint of the side, and [#permalink]

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06 May 2016, 11:33
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.

Am I correct?
In the rectangle above, A is the midpoint of the side, and   [#permalink] 06 May 2016, 11:33
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