Q. 1025.0( 2 Votes )

# ABCD is a parallelogram in which AE is perpendicular to CD (Fig. 9.54). Also AC = 5 cm, DE = 4 cm, and the area of ∆AED = 6 cm2. Find the perimeter and area of ABCD.

Answer :

Given, area of ∆AED = 6 cm^{2}

× 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 cm^{2}

Since the diagonal divides the parallelogram into two congruent triangles,

Area of parallelogram ABCD = 2 × Area of ∆ADC = 2 × 12 = 24 cm^{2}

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 cm^{2}

Perimeter of parallelogram ABCD = 26 cm

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