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Answer :

Given, area of ∆AED = 6 cm2


× DE × AE = 6


× 4 × AE = 6


AE = 3 cm


In ∆AEC, AE = 3 cm, AC = 5 cm


Applying Pythagoras theorem in ∆AEC,


(AE)2 + (EC)2 = (AC)2


(3)2 + (EC)2 = (5)2


(EC)2 = 25 - 9 = 16


EC =√16 = 4 cm


DC = DE + EC = 4 + 4 = 8 cm


Area of ∆ADC = × DC × AE = × 8 × 3 = 12 cm2


Since the diagonal divides the parallelogram into two congruent triangles,


Area of parallelogram ABCD = 2 × Area of ∆ADC = 2 × 12 = 24 cm2


Applying Pythagoras theorem in ∆AED,


(AE)2 + (DE)2 = (AD)2


(3)2 + (4)2 = (AD)2


(AD)2 = 9 + 16 = 25


AD =√25 = 5 cm


Perimeter of parallelogram ABCD = 2 (l + b) = 2(DC + AD) = 2(8 + 5) = 26 cm


Area of parallelogram ABCD = 24 cm2


Perimeter of parallelogram ABCD = 26 cm


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