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# In the figure above, what is the area of the quadrilateral OACB?

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Math Expert
Joined: 02 Sep 2009
Posts: 43348

Kudos [?]: 139709 [0], given: 12794

In the figure above, what is the area of the quadrilateral OACB? [#permalink]

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13 Dec 2017, 20:02
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95% (00:43) correct 5% (00:48) wrong based on 22 sessions

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In the figure above, what is the area of the quadrilateral OACB?

(A) 12
(B) 28
(C) 56
(D) 72
(E) 80

[Reveal] Spoiler:
Attachment:

2017-12-12_2125_001.png [ 11.66 KiB | Viewed 314 times ]
[Reveal] Spoiler: OA

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Kudos [?]: 139709 [0], given: 12794

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Joined: 22 May 2016
Posts: 1258

Kudos [?]: 467 [0], given: 683

In the figure above, what is the area of the quadrilateral OACB? [#permalink]

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14 Dec 2017, 08:38
Bunuel wrote:

In the figure above, what is the area of the quadrilateral OACB?

(A) 12
(B) 28
(C) 56
(D) 72
(E) 80

[Reveal] Spoiler:
Attachment:
2017-12-12_2125_001.png

Let the unlabeled point at (12,0) be called X

The area of the quadrilateral OACB equals the (area of large ∆ AOX) - (area of small ∆ BCO).

Area of large ∆ AOX:

Base OX = 12*
Height AO = 6
Area = $$\frac{b*h}{2}=\frac{(12*6)}{2}= 36$$

Area of small ∆ BCO:

Base BX = 8
Height = 2 (from y-coordinate of point C)
Area: $$\frac{(8*2)}{2}= 8$$

(Area of large ∆) = (area of small ∆)
(36 - 8) = 28

*subtract x-coordinates of points O and X to get distance: (12 - 0) = 12. Same method for all other distances.
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Formerly genxer123

Kudos [?]: 467 [0], given: 683

In the figure above, what is the area of the quadrilateral OACB?   [#permalink] 14 Dec 2017, 08:38
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