In the Cartesian plane four points P, Q, R, S have co-ordinates (1, 1), (4, 2), (4, 4) and (1, 4). The area of the quadrilateral PQRS is

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Q: 143 (IAS/1995)
In the Cartesian plane four points P, Q, R, S have co-ordinates (1, 1), (4, 2), (4, 4) and (1, 4). The area of the quadrilateral PQRS is

question_subject: 

Maths

question_exam: 

IAS

stats: 

0,3,5,1,3,2,2

keywords: 

{'quadrilateral pqrs': [0, 1, 0, 0], 'diagonals': [0, 1, 0, 0], 'cartesian plane': [0, 1, 0, 0], 'area': [0, 0, 0, 1], 'lengths': [0, 0, 1, 0]}

In the Cartesian plane, the points P, Q, R, S form a trapezoid whose parallel sides are PS and QR, and the perpendicular distance between the parallel sides is PR or QS. The coordinates of the points are given as P(1,1), Q(4,2), R(4,4) and S(1,4).

We can find the lengths of PS and QR by subtracting the x-coordinates, so PS = QR = 4 - 1 = 3. The height of the trapezoid, which is the distance between points Q and R or P and S, is found by subtracting the y-coordinates, so height = 4 - 1 = 3 again.

The area of a trapezoid is given by the formula 1/2*(sum of parallel sides)*height. Thus, area = 1/2*(3+3)*3 = 9.

Option 1 which suggests the area is 9 is incorrect. Option 3 suggesting 4.5 is also incorrect. Option 4 suggesting that we cannot find the area without knowing the lengths of the diagonals is irrelevant because the area of a trapezoid can be found with just the lengths of