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# Solution for In the Below Fig. Abcd and Aefd Are Two Parallelograms. Prove that (1) Pe = Fq (2) Ar (δ Ape) : Ar (δPfa) = Ar δ(Qfd) : Ar (δ Pfd) (3) Ar (δPea) = Ar (δQfd) - CBSE Class 9 - Mathematics

ConceptParallelograms on the Same Base and Between the Same Parallels

#### Question

In the below fig. ABCD and AEFD are two parallelograms. Prove that
(1) PE = FQ
(2) ar (Δ APE) : ar (ΔPFA) = ar Δ(QFD) : ar (Δ PFD)
(3) ar (ΔPEA) = ar (ΔQFD)

#### Solution

Given that, ABCD and AEFD are two parallelograms
To prove:   (1) PE = FQ

(2) "ar (ΔAPE)"/ "ar (ΔPFA)" = "ar (ΔQFD)"/"ar (ΔPED)"

(3) ar (ΔPEA) = ar (ΔQFD)

Proof:  (1) In  ΔEPA and ΔFQD

∠PEA  = ∠QFD                 [ ∴ Corresponding angles]
∠EPA  = ∠FQD                 [Corresponding angles]

PA = QD                    [opp .sides of 11gm]

Then,  ΔEPA  ≅  ΔFQD      [By. AAS condition]

∴ EP = FQ                       [c. p. c.t]

(2)  Since, ΔPEA and ΔQFD stand on the same base PEand FQlie between the same

∴  ar  (ΔPEA ) = ar (ΔQFD)  →  (1)

AD  ∴ ar (ΔPFA) = ar (PFD)        .....(2)

Divide the equation (1) by equation (2)

"area of (ΔPEA)"/"area of (ΔPFA)" = "ar Δ(QFD)"/"ar Δ(PFD)"

(3) From (1) part ΔEPA  ≅ FQD

Then, ar (ΔEDA) = ar (ΔFQD)

Is there an error in this question or solution?

#### APPEARS IN

Solution In the Below Fig. Abcd and Aefd Are Two Parallelograms. Prove that (1) Pe = Fq (2) Ar (δ Ape) : Ar (δPfa) = Ar δ(Qfd) : Ar (δ Pfd) (3) Ar (δPea) = Ar (δQfd) Concept: Parallelograms on the Same Base and Between the Same Parallels.
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