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17 May 2011, 04:07
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
52% (02:17) correct
48%
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wrong
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On the xy-coordinate plane, a quadrilateral is bounded by the x-axis, y-axis, the line \(y = 6\), and the line \(y = ax + 8\). What is the area of this region?
(1) The point (1,6) is on the boundary of the quadrilateral.
(2) The point (6,5) is on the boundary of the quadrilateral.
©Grockit
Attachment:
screen_shot_2011_03_11_at_3.28.58_pm.png [ 18.66 KiB | Viewed 13396 times ]
There are a few inferences we can start with before we turn to the statements. Since the lines y = 6 and the x-axis are parallel to one another (and the y-axis and the line y = ax + 8 are not parallel), this area is a trapezoid. Furthermore, since the x-axis and y-axis are parallel, this area is a right trapezoid (a trapezoid with two right angles). It is relatively easy to find the area of a right trapezoid; they can be divided into a rectangle and a triangle. All we need is the value of a. Then we will have the equations of all of the lines that form this trapezoid. It is neither advisable nor necessary to actually calculate the area of this trapezoid. It is better just to determine what information we will need to do so. In this case, we just need to find the equation of that fourth line. Now, turning to the statements:
(1) Remember that the line y = 6 is one of the borders of the trapezoid. This line is the horizontal line that passes through (1,6). Therefore, we don't know if the point given is on y = 6 or the vertex intersection with the line y = ax + 8. INSUFFICIENT
(2) The point (6,5) does not lie on the x-axis, the y-axis, or the line y = 6. Thus, it must lie on the line y = ax + 8. Substituting x and y into the equation we have:
5 = a(6) + 8
-3 = 6a
-0.5 = a
The equation of the line is y = -0.5x + 8
Now that we have this equation, we can solve the questions. SUFFICIENT.
Although we don't want to waste time finding the area, here's how we would do it for review purposes: the line we found crosses the x-axis and the line y = -0.5x + 8. Let's first find the intersection points.
6 = -0.5x + 8
-2 = -0.5x
x = 4
This is where the line we found meets the line y = 6. Draw an imaginary line from this point to the x-axis. this will split the trapezoid into a rectangle and a triangle. The area of the rectangle is 24 (4 wide x 6 long). Now let's find the other intersection point, the x-intercept of y = -0.5x + 8:
0 = -0.5x + 8
-8 = -0.5x
16 = x
So we have a right triangle with a base of 12 {from our imaginary line at (4,0) to the x-intercept at (16,0)} and a height of 6. Plugging these values into the formula for the area of a triangle, we get:
0.5(12 x 6) = 0.5(72) = 36
Adding up the two areas gives us a final value of 60.
The credited response is (B): Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
17 May 2011, 05:09
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See the attached images. All co-ordinates mentioned can be found out by substituting correct co-ordinates.
Stmt1: The point (1,6) is on the boundary of the quadrilateral.
That means point is on line y=6. But we don't know the value of a. Insufficient.
Stmt2:The point (6,5) is on the boundary of the quadrilateral.
That means point is on y=ax+8. Substituting (6,5)
5=6a+8
a=-1/2
We know the value of a. We can find all the co-ordinates with a in attached image.
(-8/a,0)=> 16,0
(-2/a,6)=> 4,0
Area of shaded region will be bigger triangle minus smaller triangle.
Area of bigger triangle= 1/2 * 16 * 8 =64
Area of smaller triangle=1/2 * 4 * 2 =4
Area of shaded region= 64-4=60.
OA B.
Stmt1: The point (1,6) is on the boundary of the quadrilateral.
That means point is on line y=6. But we don't know the value of a. Insufficient.
Stmt2:The point (6,5) is on the boundary of the quadrilateral.
That means point is on y=ax+8. Substituting (6,5)
5=6a+8
a=-1/2
We know the value of a. We can find all the co-ordinates with a in attached image.
(-8/a,0)=> 16,0
(-2/a,6)=> 4,0
Area of shaded region will be bigger triangle minus smaller triangle.
Area of bigger triangle= 1/2 * 16 * 8 =64
Area of smaller triangle=1/2 * 4 * 2 =4
Area of shaded region= 64-4=60.
OA B.
Attachments
soln5.JPG [ 10.3 KiB | Viewed 13225 times ]
17 May 2011, 09:00
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a (1,6) lies on y= ax + 8 or y=6 as 6= ax + 8 means ax = -2.
Hence not sufficient.
b (6,5) on y = ax+8 gives 5 = 6a + 8 giving a = -1/2
also ax= -2 means x = 4
means y= ax + 8 when y = 0 = -1/2 x + 8 thus x = 16
thus area of trapezium = 1/2 * altitude * (sum of parallel sides )
= 1/2 * 6 * (16 + 4) = 60 Sufficient.
Hence B.
Hence not sufficient.
b (6,5) on y = ax+8 gives 5 = 6a + 8 giving a = -1/2
also ax= -2 means x = 4
means y= ax + 8 when y = 0 = -1/2 x + 8 thus x = 16
thus area of trapezium = 1/2 * altitude * (sum of parallel sides )
= 1/2 * 6 * (16 + 4) = 60 Sufficient.
Hence B.
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