Question

.... . ............q1

Answer 1

Hey dear!!

Consult The Picture For Diagram!!

Given: CG||BF||AE
CD=DE

Solution: CG||BF||AE
CA & CE are transversals such that,
CE=DE

By intercept theorem:
AB=BC
BC=AB
BC=7.2 cm [ Ans (1) ]

CG||BF||AE
CE & GE are transversals such that:
CD=DE

By intercept theorem:
GF=FE
GF=EF
GF = 4 cm

So,
GE=GF+FE
GE = 4 + 4
GE = 8 cm [ Ans (2) ]

In ∆ CAE
B is mid point of CA
D is mid point of CE

By mid point theorem:
$bd \: = \frac{1}{2} ae$
$4.1 = \frac{1}{2} ae$
AE = 8.2 cm [ Ans (3) ]

In ∆ CEG,
D is the mid point of CE
F is the mid point of GE

By mid point theorem:
$df \: = \frac{1}{2} cg$
$df = \frac{1}{2} \times 11$

df = 5.5 cm [ Ans (4) ]

Hope it helps...