In a Δ*ABC*, *D*, *E*, *F* are the mid-points of sides *BC*, *CA* and *AB* respectively. If ar (Δ*ABC*) = 16cm^{2}, then ar (trapezium *FBCE*) =

#### Options

4 cm

^{2}8 cm

^{2}12 cm

^{2}10 cm

^{2}

#### Solution

**Given:** In ΔABC

(1) D is the midpoint of BC

(2) E is the midpoint of CA

(3) F is the midpoint of AB

(4) Area of ΔABC = 16 cm^{2}

**To find:** The area of Trapezium FBCE

**Calculation: **Here we can see that in the given figure,

Area of trapezium FBCE = Area of ||^{gm} FBDE + Area of ΔCDE

Since D and E are the midpoints of BC and AC respectively.

∴ DE || BA ⇒ DE || BF

Similarly, FE || BD. So BDEF is a parallelogram.

Now, DF is a diagonal of ||^{gm} BDEF.

∴ Area of ΔBDF = Area of ΔDEF ……(1)

Similarly,

DE is a diagonal of ||^{gm} DCEF

∴ Area of ΔDCE = Area of ΔDEF ……(2)

FE is the diagonal of ||^{gm} AFDE

∴ Area of ΔAFE = Area of ΔDEF ……(3)

From (1), (2), (3) we have

Area of ΔBDF = Area of ΔDCF = Area of ΔAFE = Area of ΔDEF

But

Area of ΔBDF + Area of ΔDCE + Area of ΔAFE + Area of ΔDEF = Area of ΔABC

∴ 4 Area of ΔBDF = Area of ΔABC

Area of ΔBDF = `1/4` Area of ΔABC

= `1/4 (16)`

= 4 cm^{2 }

Area of ΔBDF = Area of ΔDCE = Area of ΔAFE = Area of ΔDEF = 4 cm^{2} …….(4)

Now

Area of trapezium FBCE = Area of || FBDE + Area of ΔCDE

=(Area of ΔBDF + Area of ΔDEF ) + Area of ΔCDE

= 4 + 4+ 4 (from 4)

= 12 cm^{2}

Hence we get

Area of trapezium FBCE = **12 cm**^{2 }