Question

In the given figure 1.66, seg PQ || seg DE, A(∆PQF) = 20 units, PF = 2 DP, then Find A(◻DPQE) by completing the following activity.

Answer 1

Given:
seg PQ || seg DE
A(∆PQF) = 20 units
PF = 2 DP
Let us assume DP = x
∴ PF = 2x
$\mathrm{DF}=\mathrm{DP}+\boxed{\mathrm{PF}}=\boxed{x}+\boxed{2x}=3x$
In △FDE and △FPQ
∠FDE = ∠FPQ (Corresponding angles)
∠FED = ∠FQP (Corresponding angles)
By AA test of similarity
△FDE ∼ △FPQ
$$\begin{gathered} \therefore \frac{\mathrm{A}\left(△\mathrm{FDE}\right)}{\mathrm{A}\left(△\mathrm{FPQ}\right)}=\frac{{\displaystyle \boxed{{\mathrm{FD}}^2}}}{{\displaystyle \boxed{{\mathrm{FP}}^2}}}=\frac{{\left(3x\right)}^2}{{\left(2x\right)}^2}=\frac{9}{4} \\ \mathrm{A}\left(△\mathrm{FDE}\right)=\frac{9}{4}\mathrm{A}\left(△\mathrm{FPQ}\right)=\frac{9}{4}\times \boxed{20}=\boxed{45} \end{gathered}$$
$$\begin{gathered} \therefore \mathrm{A}\left(\square \mathrm{DPQE}\right)=\mathrm{A}\left(△\mathrm{FDE}\right)-\mathrm{A}\left(△\mathrm{FPQ}\right) \\ =\boxed{45}-\boxed{20} \\ =\boxed{25} \end{gathered}$$