Geometry

(E)
Trianle QPS could be whateve it wanted to be. Point P is not fixed

jainrahul1985 wrote:In the figure above, is quadrilateral PQRS a parallelogram?
(1) The area of ΔPQS is equal to the area of ΔQRS.
(2) QR = RS
A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is
sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.

A: An quadrilateral is a parallelogram when a diagonal bisects the quadrilateral,

So A is correct.

If any of the opposite side is equal then also it is a parallelogram, but here the two given side are not opposite so nothing can be said.

A

The answer is E

IMO - E

The statements are individually not sufficient and even when taken together , the conditions are satisfied by an isosceles trapezium with QR = RS

@maihuna

I am hearing this rule for the first time...wrt parallelograms

Anyway, this is valid for rectangles also...

So now how can you say PQRS is a parallelogram?

It is saying that the two triangle bisected by the diagonal are equal it means they are congruent.
Using the congruency and the common diagonal it can be shown that the two lines touched by the diagonal which are opoosite sides as well are parallel, and they are equal as triangles are congruent so the opposite sites are parallel and equal means the whole fig is a paralllogram.

Sorry I just realized it is saying the area are equal, its not talking abt the congruency, I mistook it earlier, A is not correct, E is ok

as in a parallelogram opposite sides are equal...if QR = RS..RS= QP (OPPO. SIDE)
Therefore RS= PS..and so all sides are equal..hence it should be a square...therefore B shud be the right answer.Please help....am i missing something!!!!!?????

Why not A?

Without attaching any pictures (because I'm not so great in MS paint) just visualize in your head with strings, tacks, rubber bands and rails.

(A) If you know that the areas are equal you know the height*base of each triangle is equal. Take the segment QS to be the base of each triangle. Now you know that the two heights must be equal. Imagine a "rail" that intersects P and runs parallel to QS and imagine QPS is a string that goes around a tack at P. You can move P anywhere along that rail. Sometimes PS is parallel to QR and sometimes not. So this is INSUFFICIENT.

(B) You know that QR=RS and thus QRS is an isosceles. Imagine QRS is a rubber band going around a tack at R and that there is a "rail" running through R and perpendicular to QS. You may move R anywhere along this rail. Sometimes QR is parallel to PS and sometimes not. So this is INSUFFICIENT.

(C) Same setup. You may still move R anywhere along its rail and P anywhere along its rail (compensating for the fact that rail P must move away from QS whenever point R moves away from QS. Sometimes the answer is yes and sometimes no. So this is INSUFFICIENT.

thnk u Luke....