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.
(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).