ABCD is a rectangle. P, Q and R are the midpoint of BC, CD and DA. The point S lies on the line QR in such a way that SR: QS = 1:3. The ratio of the area of triangle APS to area of rectangle ABCD is
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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