Question

prove that parallelogram on the same base and between the same pallells are equalin area

Answer 1

let ABCD and AEFB are two parallelograms on the same base and between same parallels
Now , As AE parallel FB and DF act as transversal . Therefore angle AED = angle CFB ( corresponding angles )
Now in ∆ ADE and ∆CFB
AD = BC ( opp. sides of parallelogram ABCD)
AE = FB ( opp. sides of parallelogram AEFB)
angle AED = angle CFB ( proved above )
therefore ∆ ADE congruent ∆ CFB
This implies ar. ( ∆ ADE )=ar.(CFB) - (1)
add ar . ( AECB ) on both sides in equation 1
then ar ( ABCD ) = ar. ( AEFB)
Hence Proved

Answer 2

Answer

Step-by-step

Given:-two parallelogram ANCD and EFCD of two parallelogram is CD and between same parallel AFandDc

To prove:-ar(ABCD)=(DCFE)

Proof:-AED=BFC (corresponding angle)

DAE=CEF (corresponding angle)

ADE=BCF (by angle sum property of tri.)

AD=BC (given)

Triangle ADE=BCF (by ASA rule)

Ar(triangle ADE)=Ar(triangleBCF)

Ar. Of triangle ADE + triangle BCF=ar. Of DCBF+Ar. Of DCBE