In the figure seg `PQ||` seg DE, `A(DeltaPQF)=20` units `PF=2DP`,then find `A(square DPQE)` by completing the following activity:

Activity: `A(DeltaPQF)=20` sq units, `PF=2DP`.
Let us assume `DP=x`
`:.PF=2x`
`DF=DE+square=square+square=3x`
In `DeltaFDE` and `DeltaFPQ`.
`/_FDE~=/_square` ..........(Corresponding angles)
`/_FED~=/_square` .....(Corresponding angles)
`:.DeltaFDE~DeltaFPQ` .....(AA test)
`:.(A(DeltaFDE))/(A(DeltaFPQ))=(square)/(square)=((3x)^(2))/((2x)^(2))=9/4`
`A(DeltaFDE)=9/4A(DeltaFPQ)=9/4xxsquare=square`
`A(squareDPQE)=A(DeltaFDE)-A(DeltaFPQ)`
`=square-square`
`=square`

Explanation

`A(DeltaPQF)=20` sq units, `PF=2DP`.
Let us assume `DP=x`
`:.PF=2x`
`DF=DP+PF=x+2x=3x`
In `DeltaFDE` and `DeltaFPQ`
`/_Fde~=/_FPQ` ……(Corresponding angles)
`/_FED~=/_FQP` …..(Corresponding angles)
`:.DeltaFDE~DeltaFPQ` ...(AA test)
`:.(A(DeltaFDE))/(A(DeltaFPQ))=(DF^(2))/(PF^(2))=((3x^(2)))/((2x)^(2))=9/4`
`A(DeltaFDE)=9/4A(DeltaFPQ)=9/4xx20=45` sq units
`A(square DPQE)=A(DeltaFDE)-A(DeltaFPQ)`
`=45-20`
`=25` sq units.