SSC CGL 26 Oct 2014 Shift 1 Paper

Question : 135 of 200

Marks: +1, -0

Given an equilateral D ABC, D, E and F are the mid-points of AB, BC and AC respectively. Then the quadrilateral BEFD is exactly a :

Solution:

Points E and F are mid-points of sides BC and AC respectively.
∴ EF || AB || BD and EF =

1/2

AB
Again, D and F are mid-points of sides AB and AC repectively.
∴ DF || BE || BC and DF =

1/2

BC
Clearly, BE = EF = DF = BD
Hence, BEFD is a rhombus.

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