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In the figure given below ABCD is a square and ΔPDC is isosceles with PD = PC. If the area of ΔPDC is twice the area of the square, then find the ratio of the area of ΔPDC that lies outside the square to the area of the square.


In The Figure Given Below ABCD Is A Square And ΔPDC Is Isos...


Options:
A .   32
B .   45
C .   98
D .   107
Answer: Option C
:
C

In The Figure Given Below ABCD Is A Square And ΔPDC Is Isos...
Drop a perpendicular PM on the side DC.
Let PN be 'x' and side of square be 2a
12×2a× (2a + x) = 8a2
x = 6a
ΔPQRΔPDC=(6a8a)2=916ΔPQR=916ΔPDC
ΔPQR=916(2×ABCD)
Required Ratio = 98
Hence, option (c)



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