Question

In the given figure find area of triangle ECF.

$△EBC=480sq.cm$

Answer 1

In figure parallelogram ABCD And BCEF line bc is a common base and BC II ADII EF

Then EF = BC and height both parallelogram is same

Then area of$△ECF$=$\frac{1}{2}EF\times hieght$

Or area of $△EBC$=$\frac{1}{2}BC\times hieght$

But height of $△ECFand△EBC$ is equal and base BC=EF

Then area of both triangle is equal

Then area of $△EBC=areaof△EBF$=480 sq cm