Question

In the figure AB || DC and AB = 2DC. If AD = 3 cm,
BC = 4cm and AD, BC produced meet at E, find
(1) ED (ii) BE (iii) area of AEDC : area of trapezium ABCD.​

Answer 1

In the figure AB || DC and AB = 2DC. If AD = 3 cm,

BC = 4cm and AD, BC produced meet at E, find

(1) ED (ii) BE (iii) area of AEDC : area of trapezium ABCD.​

Step-by-step explanation:

We are given AB || DC AB = 2DC    AD = 3 cm,

BC = 4cm

Since ABE is a triangle AB || DC D and C are points

AD=DE=3

(1) ED

ED=3cm

C is mid-point of BE

(iii) area of AEDC : area of trapezium ABCD.​

ECD/ABCD     (DC/AB)²   = (1\2)²   =1/4

Answer 2

(i) ED = 3 cm

(ii) BE = 8 cm

(iii) Area of (AEDC : trapezium ABCD) = 3 : 1

Given:

AB || DC

AB = 2DC

AD = 3 cm,

BC = 4cm

Step-by-step explanation:

(i)

In ΔEAB,

EA/DA = EB/CB = AB/DC = 2DC/DC = 2/1

DA = 2 × 3 = 6

ED = EA - DA = 6 - 3 = 3 cm

(ii)

EB/CB = 2/1

EB = 2 CB = 2 × 4 = 8 cm = BE

(iii)

ΔEDC ≈ ΔEAB

(Area of ΔEDC)/(Area of ΔABE) = (DC)²/(AB)²

(Area of ΔEDC)/(Area of ΔABE) = (DC)²/(2DC)²

(Area of ΔEDC)/(Area of ΔABE) = (DC)²/4(DC)²

(Area of ΔEDC)/(Area of ΔABE) = 1/4

∴ Area of ΔABE = 4 Area of ΔEDC

⇒ Area of ΔEDC + Area of ΔABCD = 4 Area of ΔEDC

Area of ΔABCD = 4 Area of ΔEDC - Area of ΔEDC

Area of ΔABCD = 3 Area of ΔEDC

(Area of ΔABCD)/(Area of ΔEDC) = 3/1

∴ Area of ΔABCD : Area of ΔEDC = 3 : 1