In Fig. 15.88, ABCD and AEFD are two parallelograms. Prove that: (i) PE=FQ (ii) ar(ΔAPE):ar(Δ PFA)= ar(Δ QFD)=ar(Δ PFD) (iii) ar(Δ PEA) = ar(Δ QFD)

Question

In Fig. 15.88, ABCD and AEFD are two parallelograms. Prove that: (i) PE=FQ (ii) ar(&#x394;APE):ar(&#x394; PFA)= ar(&#x394; QFD)=ar(&#x394; PFD) (iii) ar(&#x394; PEA) = ar(&#x394; QFD)

Philoid · Accepted Answer

(i) In and

∠PEA = ∠QFD (Corresponding angle)

∠EPA = ∠FQD (Corresponding angle)

PA = QD (Opposite sides of a parallelogram)

Then,

(By AAS congruence rule)

Therefore,

(c.p.c.t)

Since, and stand on equal bases PE and FQ lies between the same parallel EQ and FQ lies between the same parallel EQ and AD

Therefore,

() = Area (QFD) (1)

Since,

stand on the same base PF and lie between the same parallel PF and AD

Therefore,

Area ( = Area ( (2)

Divide the equation (1) by (2), we get

=

(iii) From (i) part,

Then,

In Fig. 15.88, ABCD and AEFD are two parallelograms. Prove that:(i) PE=FQ(ii) ar(ΔAPE):ar(Δ PFA)= ar(Δ QFD)=ar(Δ PFD)(iii) ar(Δ PEA) = ar(Δ QFD)

In Fig. 15.88, ABCD and AEFD are two parallelograms. Prove that:

(i) PE=FQ

(ii) ar(ΔAPE):ar(Δ PFA)= ar(Δ QFD)=ar(Δ PFD)

(iii) ar(Δ PEA) = ar(Δ QFD)