D, E, F are midpoints of the sides BC, CA, AB of a ∆ABC. If area of ∆ABC is 20 cm²​ , then find area of the trapezium FBCE.

D,E,F are mid points of sides BC ,AC,AB

Area of ∆ABC =20 cm² 

Concept Used: 

The line joining the midpoints of two sides of a triangle is parallel to the third side and is half its length (Midpoint Theorem).

The median of a triangle divides it into two triangles of equal area.

Using these properties:

D, E, F divide △ABC into four smaller triangles of equal area. 

Area of triangle AEF = DEF = BDF = CED = $$\frac{1}{4}$$​of Area of triangle ABC

 $$ar. \triangle AEF = ar. \triangle DEF = ar. \triangle CED = \frac{1}{4} \times ar. \triangle ABC = \frac{1}{4} \times 20 = 5\, cm^2$$ 

Now,  

Since,  trapezium FBCE is made of triangle FBD + DEF + DEC
Area of Trapezium FBCE = 5+ 5 + 5  = 15 
So, Area of Trapezium is 15 cm²​ ​​​