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On a graph the four corners of a certain quadrilateral [#permalink]

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04 Jan 2013, 03:06

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Question Stats:

64% (02:29) correct
36% (01:53) wrong based on 75 sessions

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On a graph the four corners of a certain quadrilateral are (a,5), (b,5), (a,0) and (b,0). If a + c = 12, a < b and both a and b are positive values then what is the area of the quadrilateral?

(1) b + c = 6 (2) The quadrilateral is a rectangle.

Re: On a graph the four corners of a certain quadrilateral [#permalink]

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04 Jan 2013, 04:41

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This post received KUDOS

Expert's post

trex16864 wrote:

On a graph the four corners of a certain quadrilateral are (a,5), (b,5), (a,0) and (b,0). If a + c = 12, a < b and both a and b are positive values then what is the area of the quadrilateral?

(1) b + c = 6 (2) The quadrilateral is a rectangle.

There is a problem with this question. We are given that a + c = 12 and (1) says that b + c = 6, thus a = b + 6, which implies that a > b. But the stem says that a < b. I guess it should be a > b instead of a < b.

In this case the question would be: On a graph the four corners of a certain quadrilateral are (a,5), (b,5), (a,0) and (b,0). If a + c = 12, a > b and both a and b are positive values then what is the area of the quadrilateral?

Look at the diagram below:

Attachment:

Quadrilateral.PNG [ 9.43 KiB | Viewed 2455 times ]

As we can see given quadrilateral is a rectangle and its area = 5*(a-b).

(1) b + c = 6. Since a + c = 12 and b + c = 6, then a - b = 6. Area = 5*6=30. Sufficient.

(2) The quadrilateral is a rectangle. We already know that. Not sufficient.

Re: On a graph the four corners of a certain quadrilateral [#permalink]

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04 Jan 2013, 03:54

My solution:

Given the information in the question we can already conclude, that the quadriliteral is a rectangle with a width of 5. We need to figure out the length of the rectangle in order to find the area.

Re: On a graph the four corners of a certain quadrilateral [#permalink]

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01 Jun 2013, 01:16

Bunuel wrote:

trex16864 wrote:

On a graph the four corners of a certain quadrilateral are (a,5), (b,5), (a,0) and (b,0). If a + c = 12, a < b and both a and b are positive values then what is the area of the quadrilateral?

(1) b + c = 6 (2) The quadrilateral is a rectangle.

There is a problem with this question. We are given that a + c = 12 and (1) says that b + c = 6, thus a = b + 6, which implies that a > b. But the stem says that a < b. I guess it should be a > b instead of a < b.

In this case the question would be: On a graph the four corners of a certain quadrilateral are (a,5), (b,5), (a,0) and (b,0). If a + c = 12, a > b and both a and b are positive values then what is the area of the quadrilateral?

Look at the diagram below:

Attachment:

Quadrilateral.PNG

As we can see given quadrilateral is a rectangle and its area = 5*(a-b).

(1) b + c = 6. Since a + c = 12 and b + c = 6, then a - b = 6. Area = 5*6=30. Sufficient.

(2) The quadrilateral is a rectangle. We already know that. Not sufficient.

Answer: A.

Hope it's clear.

Hi,

Sorry to re-open the thread. But Can you please explain how does the question stem tells us that it is a rectangle?

Re: On a graph the four corners of a certain quadrilateral [#permalink]

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01 Jun 2013, 04:17

Expert's post

Genfi wrote:

Bunuel wrote:

trex16864 wrote:

On a graph the four corners of a certain quadrilateral are (a,5), (b,5), (a,0) and (b,0). If a + c = 12, a < b and both a and b are positive values then what is the area of the quadrilateral?

(1) b + c = 6 (2) The quadrilateral is a rectangle.

There is a problem with this question. We are given that a + c = 12 and (1) says that b + c = 6, thus a = b + 6, which implies that a > b. But the stem says that a < b. I guess it should be a > b instead of a < b.

In this case the question would be: On a graph the four corners of a certain quadrilateral are (a,5), (b,5), (a,0) and (b,0). If a + c = 12, a > b and both a and b are positive values then what is the area of the quadrilateral?

Look at the diagram below:

Attachment:

Quadrilateral.PNG

As we can see given quadrilateral is a rectangle and its area = 5*(a-b).

(1) b + c = 6. Since a + c = 12 and b + c = 6, then a - b = 6. Area = 5*6=30. Sufficient.

(2) The quadrilateral is a rectangle. We already know that. Not sufficient.

Answer: A.

Hope it's clear.

Hi,

Sorry to re-open the thread. But Can you please explain how does the question stem tells us that it is a rectangle?

Thank you

Mark (a,5), (b,5), (a,0) and (b,0) on the plane and you'll see that you'll get a rectangle for any values of a and b (a>b). _________________

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