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There is a trap ABCD and E and F are mid points.AB//BC.AB=50,CD=30.prove ar(DEFC) = 7/9 ar (AEFB)

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:)

How is AB//BC ???

she meant DC

yeah,i meant AB//DC

okay .....

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:)

How is AB//BC ???

she meant DC

yeah,i meant AB//DC

okay .....

Log in to add a comment

The Brainliest Answer!

Const : Join EF, AC and construct a perpendicular from A on CD and mark the point on CD as G and the point on AG intersecting EF as H.

Now, In triangle ADC, let O be the mid point of AC

Thus, EO // CD and EO = 1/2 CD = 25

Similarly, OF = 1/2 AB =15

So, EF = OE + OF = 25 +15 = 40

In triangle ADG,

EH // CD and E is the mid point.

Therefore, H is also the mid point of AG (converse MPT)

AH = GH ................ (i)

Now we compare the area of the trapeziums

ar(ABFE) = (1/2(40+30) x AH : ar(DEFC) = 1/2(40+50) x GH

as AH = GH,

= 1/2 x 70 = 1/2 x 90

= 70 : 90

= 7 : 9

Hope it helps !!

Hope it helps . Make sure to do the construction and prove AB half of the side