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?
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...........