RRB Junior Engineer Stage 1 Model Papers 4

Question : 19 of 99

Marks: +1, -0

If D, E and F are the mid points of BC, CA and AB respectively of the ∆ABC then the ratio of area of the parallelogram DEFB and area of the trapezium CAFD is :

1 : 3
1 : 2
3 : 4
2 : 3

Validate

Solution:

Area of

\DEFB=BD×ED

Area of trapezium CAFD

=1/2(AC+FD)×DE

Here AC = 2 AE = 2FD

\CAFD=1/2 3FD×DE

Now

{\text "(Area of DEFB)"}/{( \text "Area of trapezium CAFD")}={BD×EF}/{3/2 FD×DE}={BD×ED}/{3/2×BD×DE}=2/3

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