Answer :

Given: ABCD is a parallelogram in which

AG = 2 GB

CE = 2 DE

BF = 2 FC

(i) Since ABCD is a parallelogram, we have AB ‖ CD and AB = CD

Therefore,

BG = AB

And,

DE = CD = AB

Therefore,

BG = DE

ADEH is a parallelogram (Since, AH is parallel to DE and AD is parallel to HE)

Area of parallelogram ADEH = Area of parallelogram BCIG (i)

(Since, DE = BG and AD = BC parallelogram with corresponding sides equal)

Area (ΔHEG) = Area (ΔEGI) (ii)

(Diagonals of a parallelogram divide it into two equal areas)

From (i) and (ii), we get,

Area of parallelogram ADEH + Area (ΔHEG) = Area of parallelogram BCIG + Area (ΔEGI)

Therefore,

Area of parallelogram ADEG = Area of parallelogram GBCE

(ii) Height, h of parallelogram ABCD and ΔEGB is the same

Base of ΔEGB = AB

Area of parallelogram ABCD = h * AB

Area (EGB) = * AB * h

= (h) * AB

= * Area of parallelogram ABCD

(iii) Let the distance between EH and CB = x

Area (EBF) = * BF * x

= * BC * x

= * BC * x

Area (EFC) = * CF * x

= * * BC * x

= * Area (EBF)

Area (EFC) = * Area (EBF)

(iv) As, it has been proved that

Area (EGB) = = * Area of parallelogram ABCD (iii)

Area ( = Area ()

Area ( = * * CE * EP

= * * * CD * EP

= * * Area of parallelogram ABCD

Area ( = * Area ( [By using (iii)]

Area ( = Area (

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