This question can be solved in 30 seconds just keeping in mind simple rules of geometry such as the triplets.

If the triangle has a 90° angle and one side is 6 and the other is 8 then PR must be 10. The triangle is a 90°,60°,30°

triplet: 3,4,5; 6,8,10...so forth

We do also know that x is exactly half away in PQ which means that PX is 3. Similarly, PY is 5. Now, it does not matter that point T is everywhere on the side QR. We do know the length of the two sides of the polygon inside the triangle.

The area is \(3 \times 5 = 15\)

the answer is A

can you please elaborate how you arrived at "The area is 3×5=15" ? how do we know it is a rectangle?

PQ=6 and QR = 8
AND

PR = 10

PX is = 3 and PY=5

You know these two sides. \(3 \times 5 = 15\) regardless you do not know where is T. Actually is it a rectangle.

Regards

AREA=bh
15

A is greater than B

How can angle QPR be 90 degree as PQR=90. Are you sure its A? I think the answer is D

I saw it as 3 cases :
Case 1 : Assuming T is the midpoint,
Area of PXTY = Area of PQR - [Area of XQT + Area of YTR ] = (1/2 * 6 * 8) - (1/2 * 3 * 4 * 2) = 12 units

Case 2 : T is at point Q, so cannot figure out area of triangle PQY

Case 3 : T is at point R, in which case
Area of PXTY = Area of triangle PQR - Area of triangle QXR = (1/2 * 6 * 8) - (1/2 * 3 * 8) = 12 units

Because of case 2, I thought answer is D

THANKS

I struggle to understand how the quadrilateral is 3*5 = 15.

I got D as well.

Since \(T\) can be any point on \(QR\), we can push \(T\) all the way to R or all the way to \(Q\).

The area of \(PQR\) = 24.

If we push T all the way to \(Q\), the line \(PT\) collapses onto \(PR\). So we can bring the quadrilateral infinitely close to the line \(PR\), minimizing it's area to 0.

In that case, the quadrilateral's area is less than 12.

With the logic from above, we can maximize the area of the quadrilateral by having \(T\) be the midpoint of \(QR\). This

So \(QT\) = 4 and \(XQ\) = 3, which implies that \(XT\) = 5.

This leads to triangle the area of triangle \(QXT\) = 6.

Since \(PQR\) is a 3-4-5 triangle, we can deduce that \(PR\) = 10, or that \(PY\) = 5.

Notice how \(XT\) = 5, \(PY\) = 5, and \(PX\) = 3. This must mean that \(YT\) = 3.

\(TR\) = 4 , and so we have another 3-4-5 triangle in \(TYR\), whose area is 6.

The area of the quadrilateral is:

the area of triangle \(QPR\) - the area of triangle \(QXT\) - the area of triangle \(TYR\).
\(24 - 6 - 6 = 12\)

So if \(T\) is the midpoint of \(QR\), the area's are equal.


Since we have two different answers, the answer is D

No matter how i look at it the answer is C.

For better understanding use Midpoint theorem where the line passing through the mid point of 2 sides is parallel to the third. XY || QR

Also XY = 0.5 * QR

Same here! Used midpoint theorem to assume C as the final answer.

Is midpoint theorem required knowledge for GRE math? I won't forget it now but just curious

Great explanation :-D . Moreover, fdundo can you please mention the source as well so that the questions can be segregated

Further, added few links in regards to quadrilateral.



Although I had used the same approach, I was not so sure whether I should have used area formula 3*5 because I was not so sure about what kind of quadrilateral it is. Can someone help?

That is absolutely correct, The answer must be C

Updated the OA

Thank you guys

Among all the solution, Midsegment Theorem is good way to tackle this and related to this type of the question. If in any triangle, two mid point is joined, then it is parallel to the third side and it is exactly half. It seems this is very underrated concept, but can be used in various problem like this .

Umm, are questions based on Midsegment theorem asked ? Because this is an extremely easy problem if you know the theorem but I don't think if it is in the syllabus. Or is it absolutely rare like quintiles in stats ?

You have to admire how they drew the triangle. Y is in the wrong place, and T looks like exactly the midpoint. Perfect for throwing off your answer. Well done, GRE.