In the figure, CD = 2AB and AB is parallel to CD. If the area of ABE i

Look at the figure above \(\Delta\) ABE and \(\Delta\) CED are similar by property AAA.

\(\angle E = \angle E\) common in both the \(\Delta's\)
\(\angle A = \angle D\) since AB // DC and \(\angle A\) and \(\angle D\) are corresponding \(\angle's\)
\(\angle B = \angle C\)since AB // DC and \(\angle B\) and \(\angle C\) are corresponding \(\angle's\)

Now in two similar \(\Delta's\):

Ratio of square of the sides = ratio of area of the triangles
Since we know \(\frac{CD}{AB} =\frac{2}{1}\) we can write \(\frac{CD^2}{AB^2}\) =\(\frac{4}{1}\)=\(\frac{area \hspace{1mm} of \hspace{1mm}\Delta CED }{area \hspace{1mm} of \hspace{1mm} \Delta ABE}\) ,given area of \(\Delta\) ABE =3
\(\frac{4}{1}\) = \(\frac{area \hspace{1mm} of \hspace{1mm}\Delta CED }{3}\)

area of \(\Delta\)CED =12
Ans-A