In the figure PX XQ PQ YZ and XV QR What is the... - WAEC Mathematics 2008 Question

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