Question

In a trapezium ABCD, AB || DC and AB=7cm and DC=5cm.The mid points of AD and

BC are E and F respectively. Find the length of EF.​

Answer 1

Answer:

6cm

Step-by-step explanation:

join A and C such that EF intersects AC at G

in ∆ADC, EG = ½DC (by mid-point theorem)

in ∆ABC, GF = ½AB (by mid-point theorem)

=> EG + GF = ½DC + ½AB

=> EF = (5+7)/2 = 6cm

Answer 2

ABCD is a trapezium and E,F are mid-points of diagonal AC and BD

AB∥CD [ one par of opposite side is parallel in trapezium ]

In △CDF and △GBF

⇒ DF=BF [ Since, F is mid-point of diagonal BD ]

⇒ ∠DCF=∠BGF [ DC∥GB and CG is a transversal ]

⇒ ∠CDF=∠GBF [ DC∥GB and BD is a transversal ]

∴ △CDF≅△GBF [ By ASA congruence rule ]

⇒ CD=GB [ C.P.C.T ] ---- ( 1 )

In △CAG, the points E and F are the mid-points of AC and CG respectively.

∴ EF=

2

1

(AG)

⇒ EF=

2

1

(AB−GB)

From ( 1 )

⇒ EF=

2

1

(AB−CD)