Source: Official Guide Revised GRE 1st Ed. Part 6; Set 2; #3

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Lelia Gessner

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Jun 7, 2018 • Comment

Sam Kinsman, Magoosh Tutor

Hi Leila,

Unfortunately, that approach works only by coincidence.

If you go back to the video, and go to the 3:00 mark at the video, on the left hand side you will see a diagram of the square. Let's consider the uppermost triangle.

This is a 45 - 45 - 90 triangle. To use the formula b*h/2, we will have to choose a base and a height. Note that if both the base and the height are sides, we have to choose the two equal sides (because they form a right angle). We cannot choose the hypotenuse of the triangle.

The two non-hypotenuse sides of the triangle are of equal length. So the base is (equal side #1), and the height is (equal side #2).

Now, the length of these sides is neither r (the radius) nor d (the diameter). Instead, to find the length of the sides, we'd have to use the pythagorean theorem:

h^2 = s^2 + s^2

d^2 = 2 * s^2

10^2 = 2 * s^2

50 = s^2

sqrt 50 = s

So as you can see, the sides are both sqrt (50), not d or r.

I hope that helps! :)

Jun 12, 2018 • Reply

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Roshan Shrestha

I am not sure if I am visualizing it properly, but suggest me if I am wrong in here. Since we can make the quadrilateral as skinny as possible, its area tend to be 0 and for the largest area of quadrilateral, we can make it as a square. From chris's explanation, the largest area it can have is 50. So, the answer is [D]

Jan 20, 2017 • Comment

Sam Kinsman, Magoosh Tutor

Yes, that's right! As we make the quadrilateral skinny, its area gets close to zero - but it never quite reaches zero. :-)

Jan 20, 2017 • Reply

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Vamshi Bhagirath Katta

Can we assume AC to be a diameter??

Nov 6, 2016 • Comment

Sam Kinsman, Magoosh Tutor

AC could be the diameter of the circle - but we cannot be sure about this, since we don't know if A is directly above C. The diagram makes it look like AC is the diameter, but it's possible that AC doesn't go exactly through the middle of the circle - in which case it is not the diameter.

As Cydney explained below, we can maximize the area of the quadrilateral by assuming that it is a square. If that's the case, AC must go through the middle of the circle, and it is the diameter. We can use this information to find that the area of the square would be 50. Keep in mind that, as Cydney explained below, this is the maximum area that the quadrilateral could possibly have.

Nov 11, 2016 • Reply

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Arpita Nehra

Can't we simply say 'D' because we have no clue about how the quadilateral lies in the circle (in comparison to the centre) and hence there's no way even to approximate the area?

Sep 8, 2015 • Comment

Cydney Seigerman, Magoosh Tutor

In this problem, we can see that the quadrilateral is inscribed by the circle, or in other words that the four vertices of the quadrilateral touch the circumference of the circle. That gives us the limit to the area of the quadrilateral. With that information, we can determine that the largest quadrilateral possible in the diagram is a square and we must figure out if the area of the square is less than 40. Using the diameter of the circle as the measure of the diagonal of the square, we can solve for the area of the square, 50, which is larger than the Quantity B, 40. At the same time, by visualizing a very thin quadrilateral, we can also see that the area of the quadrilateral could be less than 40. Therefore, we can say that the relationship between A and B cannot be determined.

Nov 3, 2015 • Reply

Jacob Xavier

"Can't we simply say 'D' because we have no clue about how the quadrilateral lies in the circle."

If the diameter had been 5 (instead of 10), then the area of even the largest possible quadrilateral would have been less than 40. Then the answer would have been B.

By doing the math above, we are able to see that the area of the quad ranges from near-0, to a maximum of 50. Thus the answer must be D.

Apr 7, 2017 • Reply

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John Bohne

If you make it a "skinny little kite thing", aren't the sides still the same? How does that affect the answer? Yes the angles might be different, but shouldn't the ratio be the same? I put A when I did this question.

May 30, 2014 • Comment

Lucas Fink, Magoosh Tutor

Because this is only a "quadrilateral," not a square, rectangle, rhombus, or even a parallelogram, we don't know much about it at all. The sides could be very differentâ€”there's no ratio we can use. Notice how small the sides BC and DC could become in Chris's illustration. They could actually be infinitely small, as B and D get closer to C.

But even that doesn't really matter, because we're looking at the **area**. The sides only matter when they're used to find the area. The area of the kite is not found by multiplying sides.

In this case, to see the possible areas, it's better to just visualize. If we can make that kite infinitely thin, we can make the area infinitely small. The closer B and D get to C, the smaller the area.

Jun 6, 2014 • Reply

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Official Guide Revised GRE 1st Ed.

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