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]


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 that

Area (ΔSAR) =

On adding equations (i) and (ii), we obtain

Area (ΔPAQ) + Area (ΔSAR) =

∴Area (ΔPAQ) + Area (ΔSAR) =

Similarly, by drawing a line passing through B parallel to RS, it can be proved that

Area (ΔPBQ) + Area (ΔSBR) =

On comparing equations (iii) and (iv), we obtain

Area (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...........

ar. PAQ = 12ar. PXYQ-----(1)ar. SAR = 12ar.SXYR-------(2)Adding (1) and (2)ar. PAQ + ar. SAR = 12(ar. PXYQ+ar.SXYR)ar. PAQ +ar. SAR = 12ar. ABCD

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