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.
Answers
Answered by
13
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
Answered by
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)
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