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Bunuel chetan2u
Please help to move this to the DS forum.


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You can pick a starting value and fill in the assumed vertices for the 4 points in terms of A and B

If (a , b) = (2 , 2)

The other 3 vertices would be:

(8,2) ——(0,7)——-(11,7)



No matter which (a , b) coordinates you choose, this 4 points will always be in the same relative position to each other and the shape will remain the same. The enclosed quadrilateral of these 4 points will therefore always take up the same area

So if you connect a horizontal line from:

(0 , 7) ———-> to ———> (11 , 7)

You will have a quadrilateral whose area is known. This quadrilateral will make up most of the unknown pentagon.

The only point/vertex of the Pentagon left unknown is (C , D)

And the only area of the pentagon left to find is the area above this quadrilateral which will be a Triangular Area with the base as (0 , 7) to (11 , 7) ———> 7 units

Concept: for any one straight line, there exists only 1 perpendicular line that can be drawn from a single Point to that straight line

In other words, the Area of the Triangle that finishes off the pentagon will only depend on the value of the Y coordinate of (C , D) , the unknown 5th vertex of the pentagon.

The height of the triangular region will be the difference from Coordinate D (Y coordinate) and our assumed Y coordinate of the Base = 7

If we know this difference, the area of the pentagon will be set and will not change. No matter what the value of C (X coordinate) is, if the value of D (Y coordinate) is known, the last Triangular Region will always have the same area.

S1: gives us the value of the X coordinate of the unknown 5th vertex of the pentagon — C — relative to all of the other coordinates.

At different values of the Y Coordinate, D, we will have different areas of this triangular region and thus different areas of the Pentagon.

S1 not sufficient

S2: the value of D, the Y coordinate of the 5th vertex, is fixed relative to the other 4 vertices. Therefore, no matter the value of C, the triangular area will remain the same.

The area of the pentagon is thus fixed.

S2 sufficient alone
B

Also - since we are given he values of both C and D relative to the other 4 vertices, the question stands out as a possible “C Trap”

Checking the other statements pays off in a such a scenario.

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