Discussion

Explanation:

Let the length of the rectangle be ‘l’ and breadth be ‘b’.

The given figure is.

Let us draw a line MS parallel to DC

​​​​​​​

∆RMS ~ ∆RDQ

∴ RM : MD = RS : SQ = 1 : 3

Let the length of the rectangle (i.e., AB = CD) ABCD be 8l and width be 8b (i.e., AD = BC).

∴ AR = 4b and RM = b

Also, MS = l and SL = 7l.

Also, PL : LC = 1 : 3 [∵ RS : SQ = 1 : 3]

Area (∆ARS) = ½ × AR × SM = 2lb

Area (∆RDQ) = ½ × RD × DQ = 8lb

Area (∆ABP) = ½ × AB × BP = 16lb

Area (PSQC) = Area (∆PSL) + Area(∆SLQC) 
                         = ½ × SL × PL + ½ × (SL + QC) × LC 
                         = 3.5lb + 16.5lb
                         = 20lb

Area of ∆ASP = Area of rectangle ABCD – [Area (∆ARS) + Area (∆RDQ) + Area (∆ABP) + Area (PSQC)]
                          = 64lb – [2lb + 8lb + 16lb + 20lb]
                          = 64lb - 46lb = 18lb

Area(ASP)Area(ABCD) = 18lb64lb = 36128

Hence, option (a).

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