In the figure given below ABCD is a square and ΔPDC is isosceles with PD = PC.

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In the figure given below ABCD is a square and ΔPDC is isosceles with PD = PC. If the area of

Δ

PDC is twice the area of the square, then find the ratio of the area of

Δ

PDC 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

\Rightarrow

\frac{1}{2} \times 2 a \times

(2a + x) = 8a

^{2}

\Rightarrow

x = 6a

\frac{Δ P Q R}{Δ P D C} = {(\frac{6 a}{8 a})}^{2} = \frac{9}{16} \Rightarrow Δ P Q R = \frac{9}{16} Δ P D C

Δ P Q R = \frac{9}{16} (2 \times A B C D)

Required Ratio =

\frac{9}{8}

Hence, option (c)

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