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sudhir

· started a discussion

· 1 Months ago

solution (Question: 187399.0 )

i m nt getting it...provide solution

Question:
The parallel side of a trapezium are p and q respectively. The length of line joining the mid points of its non-parallel side will be –
Options:
A) \(\sqrt{pq}\)
B) \(\cfrac{2 \ pq}{p+q}\)
C) \(\cfrac{p+q}{2}\)
D) \(\cfrac{1}{2}(p-q)\)
Solution:

Ans: (C) ABCD is a trapezium in which AB || DC and E, F are mid points of AD, BC respectively.


Join CE and produce it to meet BA produced at G.

In ΔEDC and ΔEAG,

ED = EA ( E is mid point of AD)

∠CED = ∠GEC ( Vertically opposite angles)

∠ECD = ∠EGA ( alternate angles) ( DC||AB, DC||GB and CG transversal)

∴ ΔEDC ≅ ΔEAG

CD = GA and EC = EG 

In ΔCGB,

E is mid point of CG ( EC = EG proved)

F is a mid point of BC (given)

∴ By mid point theorem EF ||AB and EF = (1/2)GB.

But GB = GA + AB = CD + AB

Hence EF||AB and EF = (1/2)( AB + CD).

Knowledge Expert

· commented

· 1 Months ago

ABCD is a trapezium in which AB || DC and E, F are mid points of AD, BC respectively.

Join CE and produce it to meet BA produced at G.

In ΔEDC and ΔEAG,

ED = EA ( E is mid point of AD)

∠CED = ∠GEC ( Vertically opposite angles)

∠ECD = ∠EGA ( alternate angles) ( DC||AB, DC||GB and CG transversal)

∴ ΔEDC ≅ ΔEAG

CD = GA and EC = EG

In ΔCGB,

E is mid point of CG ( EC = EG proved)

F is a mid point of BC (given)

∴ By mid point theorem EF ||AB and EF = (1/2)GB.

But GB = GA + AB = CD + AB

Hence EF||AB and EF = (1/2)( AB + CD).

Knowledge Expert

· commented

· 1 Months ago

Hi its just a theorem, like: E and F are respectively mid points of non parallel sides AD and BC of a trapezium ABCD. Prove that EF||AB and EF = 1/2(AB + CD).
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