Answers

2016-03-12T23:29:53+05:30
Given QE=2PE
SF=2RF

QE+PE=PQ
so we can say that,
2PE+PE=PQ
3PE=PQ---eq 1

similarly
3RF=RS---eq 2

from 1 and 2
3RF=3PE [PQ=RS]
therefore RF=PE and RF is parallel to PE
so PERF is a parallelogram

area of parallelogram=b*h
b=3PE,h=sd
area of PQRS=(3PE)(SD)
PQRS=3PERF
therefore,
PERF=1/3PQRS
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