In figure 1.66, seg PQ ∥ seg DE, A(ΔPQF) = 20 units, PF = 2 DP, then find A(□DPQE) by completing the following activity.

In figure 1.66, seg PQ ∥ seg DE, A(\(\triangle\)PQF) = 20 units, PF = 2 DP, then find A(□DPQE) by completing the following activity.

Figure: \(\triangle\)DEF with point P on side DF and point Q on side EF, seg PQ ∥ seg DE.

A(\(\triangle\)PQF) = 20 units, PF = 2 DP. Let us assume DP = x. \(\therefore\) PF = 2x

DF = DP + □ = □ + □ = 3x

In \(\triangle\)FDE and \(\triangle\)FPQ, \(\angle\)FDE ≅ \(\angle\)…… corresponding angles; \(\angle\)FED ≅ \(\angle\)…… corresponding angles

\(\therefore \triangle FDE \sim \triangle FPQ\) …… AA test

\(\therefore \dfrac{A(\triangle FDE)}{A(\triangle FPQ)} = \dfrac{\square}{\square} = \dfrac{(3x)^2}{(2x)^2} = \dfrac{9}{4}\)

A(\(\triangle\)FDE) = \(\frac{9}{4}\) A(\(\triangle\)FPQ) = \(\frac{9}{4}\) × □ = □

A(□DPQE) = A(\(\triangle\)FDE) − A(\(\triangle\)FPQ) = □ − □ = □