abcd is a trapezium AB=70 , DC=50.p&q are mid points of ad and bc.prove
1 pq=60
2 dcqp&abqp are trapezium
2 ar(dcqp) = 9/11 ar(abqp)

Given:ABCD is a trapezium , where AB = 70 and DC = 50
P and Q are the mid-points of sides AB and BC respectively.
Construction :
Draw a line DR || CB and join DB .
let DB intersect PQ at O.
Since DR|| CB and DC || RB therefore DCBR is a parallelogram.
and DSQC is a parallelogram and SRBQ is a parallelogram.
DS= CQ [opposite sides of a parallelogram are equal]

CQ = BQ [as Q is the mid-point of BC]

thus DS=BQ

now in triangle DOS and triangle BOQ;
SDO=QBO [alternate interior angles]DSO=BQO [alternate interior angles]      DS=BQ       [proved above] DOSBOQ ASA      DO = OB   CPCT

now in the triangle DAB;
P and O are the mid-points of sides DA and DB
therefore by mid-point theorem,
PO || AB and PO = 1/2 AB 
similary OQ =1/2 DC

= 1/2 AB+1/2 DC

= 1/2(70)+1/2(50)

= 35+25 = 60

and DCQP and ABQP are trapezium

hope this helps you


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