Question

please solve this question using area therom

Answer 1

THANKS FOR THE QUESTION !

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GIVEN :

=> QUADRILATERAL ABCD IS A PARALLELOGRAM

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=> AB IS PARALLEL TO CD

=> AD IS PARALLEL TO BC

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=> AB = CD

=> AD = BC

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TAKE POINT R ON SIDE AP

=> TAKE DR AS THE HEIGHT OF ∆ ADP

=> A ( ANY ∆ ) = 1/2 * BASE * HEIGHT

=> A ( ∆ ADP ) = 1/2 * BASE * HEIGHT

=> A ( ∆ ADP ) = 1/2 * DP * DR

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TAKE POINT S ON SIDE PQ

=> TAKE CS AS THE HEIGHT OF ∆ QCP

=> A ( ∆ QCP ) :

=> 1/2 * BASE * HEIGHT

=> 1/2 * CP * CS

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SIMILARLY CS IS THE HEIGHT OF ∆ BCP ,

=> AS IT IS AN OBTUSE ANGLED TRIANGLE

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SIMILARLY DR IS THE HEIGHT OF ∆ DQP

=> AS IT IA AN OBTUSE ANGLED TRIANGLE
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IN ∆ BCP ,

=> A ( ∆ BCP ) :

=> 1/2 * BASE * HEIGHT

=> 1/2 * CP * CS

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IN ∆ DQP

=> A ( ∆ DQP )

=> 1/2 * BASE * HEIGHT

=> 1/2 * DP * DR

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SO,

=> A ( ∆ ADP ) = A ( ∆ DPQ )

=> A ( ∆ QCP ) = A ( ∆ BCP )

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DRAW SEG AC PARALLEL TO SEG DQ

=> NOW,

=> QUADRILATERAL ACQD IS A PARALLELOGRAM .....

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=> AD IS PARALLEL TO CQ

=> AC IS PARALLEL TO DQ

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IN THE PARALLELOGRAM ,

=> AD = CQ

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DUE TO AD IS PARALLEL TO CQ ,

=> HEIGHTS OF ∆ ADP AND ∆ QCP IS CONGRUENT. .......

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A ( ∆ ADP ) :

=> 1/2 * BASE * HEIGHT

=> 1/2 * AD * PO

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A ( ∆ QCP ) :

=> 1/2 * BASE * HEIGHT

=> 1/2 * CQ * PG

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BUT,

=> AD = CQ

&

=> HEIGHTS ( PO = PG )

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SO,

=> AREAS OF ∆ ADP = ∆ QCP

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ACCORDING TO THE ABOVE INFO :

=> A ( ∆ ADP ) = A ( ∆ DPQ )

&

=> A ( ∆ QCP ) = A ( ∆ BCP )

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THAT'S WHY :

=> IT IS PROVED THAT,

=> A ( ∆ DPQ ) = A ( ∆ BCP )

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HOPE IT WILL HELP U.......

THANKS AGAIN ........

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