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

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

Post Mon Nov 27, 2017 6:50 am
What is the area of the quadrilateral bounded by the lines $$y=\frac{3}{4}x+6,\ y=\frac{3}{4}x-6,\ y=-\frac{3}{4}x+6,\ y=-\frac{3}{4}x-6$$

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

The OA is C.

Please, can any expert explain this PS question for me? I have many difficulties to understand why that is the correct answer. Thanks.

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Post Mon Nov 27, 2017 9:33 pm
swerve wrote:
What is the area of the quadrilateral bounded by the lines $$y=\frac{3}{4}x+6,\ y=\frac{3}{4}x-6,\ y=-\frac{3}{4}x+6,\ y=-\frac{3}{4}x-6$$

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

The OA is C.

Please, can any expert explain this PS question for me? I have many difficulties to understand why that is the correct answer. Thanks.
I can do half work for you; the rest you could follow.

1. Put x = 0 in the given equations, you would get two points (0, 6) and (0, -6).
2. Put y = 0 in the given equations, you would get two points (8, 0) and (-8, 0).
3. Plot them in the coordinate system.
4. You would get a quadrilateral bounded by the given lines.
5. You would find that the quadrilateral can be divided into four equal rectangles whose base and heights are 6 and 8.
6. Calculate the area of one rectangle = 1/2*6*8 = 24
7. Area of the quadrilateral = 4*24 = 96

The correct answer: C

Hope this helps!

-Jay
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