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What is the area of the quadrilateral bounded by the...

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by VJesus12 » Mon Jun 17, 2019 4:28 am
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What is the area of the quadrilateral bounded by the following lines?
\( l_1:\ y=\frac{3}{4}x+6\)
\(l_2:\ y=\frac{3}{4}x-6\)
\(l_3:\ y=-\frac{3}{4}x+6\)
\(l_4:\ y=-\frac{3}{4}x-6\)

(A) 48
(B) 64
(C) 96
(D) 100
(E) 140

[spoiler]OA=C[/spoiler]

Source: Manhattan GMAT
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by Jay@ManhattanReview » Tue Jun 18, 2019 12:38 am
VJesus12 wrote:What is the area of the quadrilateral bounded by the following lines?
\( l_1:\ y=\frac{3}{4}x+6\)
\(l_2:\ y=\frac{3}{4}x-6\)
\(l_3:\ y=-\frac{3}{4}x+6\)
\(l_4:\ y=-\frac{3}{4}x-6\)

(A) 48
(B) 64
(C) 96
(D) 100
(E) 140

[spoiler]OA=C[/spoiler]

Source: Manhattan GMAT
By plugging-in x = 0, in the first two equations, we get coordinates of two vertices: 1. (0, 6) and (0, –6). Similarly, plugging-in y = 0, in the first two equations, we get coordinates of the other two vertices: 1. (8, 0) and (–8, 0).

With these 4 vertices, we get a quadrilateral. The area of the quadrilateral = (6√2) *(8√2) = 96

The correct answer: C

Hope this helps!

-Jay
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