SSC CGL Tier-1 16-08-2015 Shift 1 Solved Paper

D and E are midpoints of BC and AC respectively.
∴ DE || BA ⇒ DE || BF, FE || BD
DF is the diagonal of parallelogram BDEF.
∴ Area of ΔBDF = Area of ΔDEF
Similarly
DE is the diagonal of parallelogram DCEF.
∴ Area of ΔDCE = Area of ΔDEF
∴ ΔBDF = ΔDCE = ΔAFE = ΔDEF
∴ On adding
4 × ΔDEF = ΔABC
Parallelogram BDEF
= 2 × ΔDEF =

1

2

× ΔABC
Quadrilateral CAFD = ΔABC – ΔBDF
= ΔABC –

1

4

ΔBDF
=

3

4

× ΔABC
∴ Required ratio
=

1

2

× ΔABC :

3

4

ΔABC = 2 : 3