From the diagram, XYZQ is a parallelogram. Thus, |YZ| = |XQ| = |PX|; \(\bigtriangleup\)PXY, Let the area of XYZQ = A1, the area of \(\bigtriangleup\)PXY = Area of \(\bigtriangleup\)YZR = A2 Area of \(\bigtriangleup\)PQR = A = A1 + 2A2 But from similarity of triangles \(\frac{\text{Area of PQR}}{\text{Area of PXY}} = (\frac{PQ}{PX})^2 = (\frac{QR}{XY})^2\) \(\frac{A}{A_2} = (\frac{2}{1})^2 = \frac{2}{1}\) A = 4A2 But, A = A1 + 2A A1 = 4A2 - 2A2 A1 = 2A2 \(\frac{A_1}{A_2}\) = 2 A1:A2 = 2:1 Area of XYZQ:Area of \(\bigtriangleup\)YZR = 2:1