In the figure below, ABCD is a rectangle. If the area of AEB : DS Archive
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# In the figure below, ABCD is a rectangle. If the area of AEB

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VP
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In the figure below, ABCD is a rectangle. If the area of AEB [#permalink]

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17 Oct 2005, 19:16
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In the figure below, ABCD is a rectangle. If the area of AEB is 8, what is the area of ACD?

(sorry if the picture is unclear. E is the intesection of two diagnals, ABCD goes from top left, top right, bottom right, bottom left)

A. 8
B. 12
C. 16
D. 24
E. 32
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17 Oct 2005, 19:20
c. 16
let a, b be the dimensions of ABCD

area of triangle = (1/2)*b*(a/2) = 8 (b -base & a/2 - height)
ab=32
so area of ACD = 0.5*ABCD = 16
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17 Oct 2005, 19:25
I get 16

Area (AEB) = 8
(1/2)AB*(h) = 8 => AB*h = 16 ... (1)
OR AB = 16/h
and h = 16/AB

DC = AB = 16/h

Area(ACD) = (1/2)*(32/AB)*(16/h) = (1/2)*(32)*(16)/16 (from 1 AB*h=16)
Area(ACD) = 16
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-Vikram

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17 Oct 2005, 19:28
16 is the OA.

I set up the equation but could not finish........Thanks
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17 Oct 2005, 19:28
gsr wrote:
c. 16
let a, b be the dimensions of ABCD

area of triangle = (1/2)*b*(a/2) = 8 (b -base & a/2 - height)
ab=32
so area of ACD = 0.5*ABCD = 16

Now why didn't I think of this easier method
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17 Oct 2005, 19:42
TeHCM wrote:
In the figure below, ABCD is a rectangle. If the area of AEB is 8, what is the area of ACD?

(sorry if the picture is unclear. E is the intesection of two diagnals, ABCD goes from top left, top right, bottom right, bottom left)

A. 8
B. 12
C. 16
D. 24
E. 32

Attachments

rec.JPG [ 13.65 KiB | Viewed 3830 times ]

Re: DS Geometry Rectangle   [#permalink] 17 Oct 2005, 19:42
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