Question

In a ∆ABC , D , E , F are the mid-points of sides BC, CA and AB respectively. If ar(∆ABC) = 16cm² ,then ar( trapezium FBCE) =

a) 4cm²
b)8cm²
c) 12cm²
d) 10cm²

Answer 1

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See the attached file
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According to the question D , E , F mid points on side BC , CA and AB respectively , By converse of mid-point theorem :-

»» BC  | |  FE

=> AF / FE = AB / BC

=> AF / FE = 2AF / BC , ( AB = 2AF )

=> BC = 2FE

Now , in. ∆ AFE and ∆ ABC
=> ∠ AFE = ∠ ABC
=> ∠ FAE = ∠ BAC

By ' AA Similarity ' - ∆ AFE ~ ∆ ABC  

Therefore we can say that

=> Area of ∆ AFE / Area of ∆ ABC = (FE)^2 / (BC)^2

here » ar.( ∆ ABC ) => 16 cm^2 and BC = 2FE

=> ar ( ∆ AFE ) / 16 = FE^2 / (2FE)^2

=> ar(∆AFE ) / 16 = 1/4

★=> ar( ∆ AFE ) = 4 cm^2


From the diagram ,, we know that :-

➡ Ar( Trapezium FBCE )  =  Ar( ∆ABC ) -  Ar( ∆AFE )

➡ ar( Trapezium FBCE ) = 16 - 4 = 12 cm^2


★ ar( Trapezium FBCE ) => 12 cm^2


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