Answer:
Step-by-step explanation:
Given two parallelogram ABCD and EFCD that have the same base CD and lie between same parallel AF and CD.
We have prove that ar(ABCD)=ar(EFCD)
Since opposite sides of ∥gm are parallel AB∥CD and ED∥FC with transversal AB
⇒∠DAB=∠CBF [ Corresponding angles ]
with transversal EF
⇒∠DEA=∠CFE [ Corresponding angles ]
⇒AD=BC [ Opposite sides of parallelogram are equal ]
In △AED ξ △BFC
⇒∠DEA=∠CFE
∠DAB=∠CBF
∴AD=BC
⇒△AED≅△BFC [ AAS congruency ]
Hence, ar(△AED)=ar(△BFC)
( Areas of congruent figures are equal )
⇒ar(ABCD)=ar(△ADE)+ar(EBCD)
=ar(△BFC)+ar(EBCD)
=ar(EBCD)
∴ar(ABCD)=ar(EBCD)
Hence, the answer is proved.
GIVEN -- Two parallelogram ABCD and EFCD that have the same base CD and lie between same parallel AF and CD.
TO PROOF -- ar ( ABCD ) = ar ( EFCD )
SOLUTION -- Since opposite sides of ∥ gm are parallel AB ∥ CD and ED ∥ FC with transversal AB.
⇒ ∠DAB = ∠CBF [ Corresponding angles ]
With Transversal EF
⇒ ∠DEA = ∠CFE [ Corresponding angles ]
⇒ AD = BC [ Opposite sides of parallelogram are equal ]
⇒△AED ≅△BFC [ AAS congruency ]
Hence, ar (△AED ) = ar (△BFC )
⇒ar ( ABCD ) = ar (△ADE ) + ar ( EBCD )
= ar (△BFC ) + ar ( EBCD )
= ar (EBCD)
∴ ar (ABCD) = ar (EBCD)
hEnCe, tHe aNsWeR iS pROvEd.
Copyright © 2024 EHUB.TIPS team's - All rights reserved.
Answers & Comments
Answer:
Step-by-step explanation:
Given two parallelogram ABCD and EFCD that have the same base CD and lie between same parallel AF and CD.
We have prove that ar(ABCD)=ar(EFCD)
Since opposite sides of ∥gm are parallel AB∥CD and ED∥FC with transversal AB
⇒∠DAB=∠CBF [ Corresponding angles ]
with transversal EF
⇒∠DEA=∠CFE [ Corresponding angles ]
⇒AD=BC [ Opposite sides of parallelogram are equal ]
In △AED ξ △BFC
⇒∠DEA=∠CFE
∠DAB=∠CBF
∴AD=BC
⇒△AED≅△BFC [ AAS congruency ]
Hence, ar(△AED)=ar(△BFC)
( Areas of congruent figures are equal )
⇒ar(ABCD)=ar(△ADE)+ar(EBCD)
=ar(△BFC)+ar(EBCD)
=ar(EBCD)
∴ar(ABCD)=ar(EBCD)
Hence, the answer is proved.
Verified answer
Answer:
GIVEN -- Two parallelogram ABCD and EFCD that have the same base CD and lie between same parallel AF and CD.
TO PROOF -- ar ( ABCD ) = ar ( EFCD )
SOLUTION -- Since opposite sides of ∥ gm are parallel AB ∥ CD and ED ∥ FC with transversal AB.
⇒ ∠DAB = ∠CBF [ Corresponding angles ]
With Transversal EF
⇒ ∠DEA = ∠CFE [ Corresponding angles ]
⇒ AD = BC [ Opposite sides of parallelogram are equal ]
In △AED ξ △BFC
⇒∠DEA=∠CFE
∠DAB=∠CBF
∴AD=BC
⇒△AED ≅△BFC [ AAS congruency ]
Hence, ar (△AED ) = ar (△BFC )
( Areas of congruent figures are equal )
⇒ar ( ABCD ) = ar (△ADE ) + ar ( EBCD )
= ar (△BFC ) + ar ( EBCD )
= ar (EBCD)
∴ ar (ABCD) = ar (EBCD)
hEnCe, tHe aNsWeR iS pROvEd.