In the figure, seg `DE||` side `AB`. `DC=2BD`, `A(DeltaCDE)=20cm^(2)`. Find `A(squareABDE)`.

Explanation

`DC=2BD`……(given)
Let `BD` be `x`. Then `DC=2x`
`BC=BD+DC=x+2x=3x`
Seg `DE||` side `AB`
`CA` is the transversal
`:. /_CED~=/_CAB` …..(Corresponding angles)
In `DeltaCED` and `DeltaCAB`,
`/_ECD~=/_ACB`……(Common angle)
`:.DeltaCED~DeltaCAB`......("AA" test for similarity)
By the theorem on areas of similar triangles,
`(A(DeltaCED))/(A(DeltaCAB))=(CD^(2))/(CB^(2))=((CD)/(CB))^(2)=((2x)/(3x))^(2)=((2)/(3))^(2)=(4)/(9)`
`:.(20)/(A(DeltaCAB))=(4)/(9)` `:'A(DeltaCAB)=(20xx9)/(4)=45cm^(2)`
`A(DeltaCAB)=A(DeltaCED)+A(squareABDE)`........(Area addition postulate)
`:.45=20+A(squareABDE)`
`:.A(squareABDE)=45-20`
`:.A(squareABDE)=25`