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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: 56307
In the figure above, what is the area of the quadrilateral OACB?  [#permalink]

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13 Dec 2017, 21:02
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15% (low)

Question Stats:

90% (01:31) correct 10% (01:30) wrong based on 41 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

Attachment:

2017-12-12_2125_001.png [ 11.66 KiB | Viewed 777 times ]

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

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14 Dec 2017, 09:38
Bunuel wrote:

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

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

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

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18 Jul 2018, 23:16
OA:B

$$O(x_{1},y_{1}) =(0,0)$$
$$B(x_{2},y_{2}) =(4,0)$$
$$C(x_{3},y_{3}) =(8,2)$$
$$A(x_{4},y_{4}) =(0,6)$$

Area of quad OBCA $$=$$ $$\frac{1}{2}|x_{1}y_{2}+x_{2}y_{3}+x_{3}y_{4}+x_{4}y_{1}-x_{2}y_{1}-x_{3}y_{2}-x_{4}y_{3}-x_{1}y_{4}|$$
$$=\frac{1}{2}|0*0+4*2+8*6+0*0-4*0-8*0-0*2-0*6|$$
$$=\frac{1}{2}|0+8+48+0-0-0-0-0|$$$$=\frac{1}{2}|8+48|= \frac{56}{2}=28$$
In the figure above, what is the area of the quadrilateral OACB?   [#permalink] 18 Jul 2018, 23:16
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