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Re: What is the area of the quadrilateral bounded by the lines and y = 3/4 [#permalink]
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Bunuel wrote:
Bunuel wrote:
What is the area of the quadrilateral bounded by the lines and y = 3/4*x + 6, y= 3/4*x - 6, y= -3/4*x + 6, y = -3/4*x - 6 ?

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


MANHATTAN GMAT OFFICIAL SOLUTION:

All of these line equations are of the form y = mx + b, where m is the slope and b is the y-intercept. Two of these lines have a slope of 3/4 and are thus parallel to each other. The other two lines are parallel to one another with a slope of –3/4. Two of the lines have a y-intercept of 6 while the other two lines have a y-intercept of –6.
Sketch the lines:

In each quadrant, we have a triangle with the dimensions 6–8–10, a multiple of the common 3–4–5 right triangle:
Area = 4(1/2*bh) = 2bh = 2*6*8 = 96.

Alternatively, recognize that the quadrilateral is a rhombus (four equal sides of length 10), and use the formula for the area of a rhombus: \(\frac{D_1*D_2}{2}\), where D indicates the length of the diagonals:

The correct answer is C.

Attachment:
2015-06-08_1520.png


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What is the area of the quadrilateral bounded by the lines and y = 3/4 [#permalink]
VeritasKarishma Is there any simpler way to solve this question? I'm really finding it difficult to understand the solutions given above.
Thanks a lot in advance :)
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Re: What is the area of the quadrilateral bounded by the lines and y = 3/4 [#permalink]
it forms a quad with diagonals 16 and 12.
Hence, not a rectangle or a square. Hence rhombus.
Ar: (d1*d2)/2 = 16*12/2 = 96
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Re: What is the area of the quadrilateral bounded by the lines and y = 3/4 [#permalink]
How do you conclude the other side is 10 do they form right angle triangle?

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Re: What is the area of the quadrilateral bounded by the lines and y = 3/4 [#permalink]
Tomnovhere wrote:
How do you conclude the other side is 10 do they form right angle triangle?

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yes ..every quadrant is rt. triangle
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Re: What is the area of the quadrilateral bounded by the lines and y = 3/4 [#permalink]
Area of a quadrilateral = ½ | x1 – x3 y1-y3 |
x2-x4 y2- y4
= ½ | 0-0 6+6|
-8-8 0-0
=1/2 { 0*0 – (12* -16)
=1/2 * 12 * 16 = 96 (C)
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Re: What is the area of the quadrilateral bounded by the lines and y = 3/4 [#permalink]
Plugging All the Intercepts for Each Line:

y = 3/4x + 6

(0 , 6) and (-8 , 0)


y = 3/4x - 6

(0 , -6) and (8 , 0)


y = (-)3/4x + 6

(0 , 6) and (8 , 0)


y = (-)3/4x - 6

(0 , -6) and (-8 , 0)



4 Vertices bound the Quadrilateral:

(-8 , 0)
(0 , 6)
(8 , 0)
(0 , -6)


Method 1: You can Add up the AREAS of the 4 Right Triangles that are formed by Origin (0 ,0) as the Right Angle

4 * [ (1/2) * (8) * (6) ] = 96


Method 2:

Rule: In a Square, the 2 Diagonal are CONGRUENT and are Perpendicular, Bisectors of each Other

While the 2 Diagonals are Perpendicular to Each Other and DO Bisect Each Other:

Diagonal 1 runs from (-8 , 0) -------> to (8 , 0) = 16 Units ----- 8 on Each Side of the Origin running on the X-Axis

Diagonal 2 runs from (0 , -6) -------> to (0 , 6) = 12 Units ---- 6 on Each Side of the Origin running on the Y-Axis


Rule: in a Rhombus, the 2 Diagonals are NON-Congruent ---- but the 2 Diagonals are Perpendicular, Bisectors of Each Other

Based on these Fact, the Quadrilateral is a Rhombus

Area of a Rhombus = (Diagonal 1) * (Diagonal 2) * (1/2)

Area of this Rhombus = (16) * (12) * (1/2) = 16 * 6 =

96

-C-
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Re: What is the area of the quadrilateral bounded by the lines and y = 3/4 [#permalink]
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Re: What is the area of the quadrilateral bounded by the lines and y = 3/4 [#permalink]
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