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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.

Proof:

Since DR|| CB and DC || RB therefore DCBR is a parallelogram.

and DSQC is a parallelogram and SRBQ is a parallelogram.

So,

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;

$\angle \mathrm{SDO}=\angle \mathrm{QBO}[\mathrm{alternate}\mathrm{interior}\mathrm{angles}]\phantom{\rule{0ex}{0ex}}\angle \mathrm{DSO}=\angle \mathrm{BQO}[\mathrm{alternate}\mathrm{interior}\mathrm{angles}]\phantom{\rule{0ex}{0ex}}\mathrm{DS}=\mathrm{BQ}[\mathrm{proved}\mathrm{above}]\phantom{\rule{0ex}{0ex}}\u25b3\mathrm{DOS}\cong \u25b3\mathrm{BOQ}\left[\mathrm{ASA}\right]\phantom{\rule{0ex}{0ex}}\mathrm{DO}=\mathrm{OB}\left[\mathrm{CPCT}\right]$

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

PQ = PO+OQ

= 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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