Question

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

• A. 3/2
• B. 4/5
• C. 9/8
• D. 10/7

Answer 1

The correct option is C

9/8

Drop a perpendicular PM on the side DC.
Let PN be 'x' and side of square be 2a
$\Rightarrow$$\frac{1}{2}\times 2a\times$ (2a + x) = 8a${}^2$
$\Rightarrow$x = 6a
$\frac{\Delta PQR}{\Delta PDC}={\left(\frac{6a}{8a}\right)}^2=\frac{9}{16}\Rightarrow \Delta PQR=\frac{9}{16}\Delta PDC$
$\Delta PQR=\frac{9}{16}(2\times ABCD)$
Required Ratio = $\frac{9}{8}$
Hence, option (c)