# The given figure shows a parallelogram PQRS in which A and B are any two points in the interior region. AS and BP intersect each other at C, while AR and QB intersect each other at D. If area (ASBR) = 35 cm2 and area (ACBD) = 20 cm2, then what is the area of the shaded portion? 35 cm2 50 cm2 55 cm2 70 cm2 Ans : Draw a parallel through point A to PQ, which intersects PS and QR at X and Y respectively. Consider quadrilateral PXYQ.PQ||XY [by construction]PX||QY [PS||QR]Hence, PXYQ is a parallelogram.Parallelograms PXYQ and ΔPAQ are lying on the same base PQ and between the same parallels PQ and XY.∴ Area (ΔPAQ) = Similarly, it can be proved thatArea (ΔSAR) = On adding equations (i) and (ii), we obtainArea (ΔPAQ) + Area (ΔSAR) =  ∴Area (ΔPAQ) + Area (ΔSAR) = Similarly, by drawing a line passing through B parallel to RS, it can be proved thatArea (ΔPBQ) + Area (ΔSBR) = On comparing equations (iii) and (iv), we obtainArea (ΔPAQ) + Area (ΔSAR) = Area (ΔPBQ) + Area (SBR)⇒ Area (ΔSAR) Area (ΔSBR) = Area (ΔPBQ) Area (ΔPAQ)⇒ Area (ASBR) = Area (BPAQ)⇒ Area (BPAQ) = 35 cm2 [Area (ASBR) = 35 cm2]Area (ΔPAC) + Area (ΔAQD) = Area (BPAQ) Area (ABCD)It is given that area (ABCD) = 20 cm2 ∴Area (ΔPAC) + Area (ΔAQD) = 35 cm2 20 cm2 = 15 cm2 Thus, area of the shaded portion= Area (ASBR) + [Area (ΔPAC) + Area (AQD)]= 35 cm2 + 15 cm2 = 50 cm2 The correct answer is B.hey...plz explain me..on adding 1 and 2 how do we get 12ar(abcd)plz reply fast...........

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