Question

ABCD is a parallelogram in which AE is perpendicular to CD (see figure). Also AC = 5 cm, DE = 4 cm and area of ΔAED = 6 cm². Find the perimeter and area of ABCD.

Answer 1

Given, area of ΔAED = 6 cm² and AC = 5 cm and DE = 4 cm

∴ Area of ΔAED = `1/2` × DE × AE  ...[∵ Area of triangle = Base × Height]

⇒ `1/2` × 4 × AE = 6

⇒ AE = `(6 xx 2)/4`

⇒ AE = 3 cm

Now, In right-angled ΔAEC,

AE = 3 cm and AC = 5 cm

So, (EC)² = (AC)² – (AE)²  ...[By Pythagoras theorem]

⇒ (EC)² = 5² – 3² = 25 – 9

⇒ EC = `sqrt(16)`

⇒ EC = 4 cm

∵ DE + EC = DC

⇒ DC = 4 + 4 = 8 cm

∵ ABCD is a parallelogram.

So, AB = DC = 8 cm

Now, In right-angled ΔAED,

AD² = AE² + ED²  ...[By Pythagoras theorem]

⇒ AD² = 3² + 4² = 9 + 16

⇒ AD = `sqrt(25)`

⇒ AD = 5 cm

So, AD = BC = 5 cm  ...[∵ ABCD is a parallelogram]

∴  Perimeter of parallelogram ABCD = 2(l + b)

= 2(DC + AD)

= 2(8 + 5)

= 2 × 13 

= 26 cm 

Area of parallelogram ABCD = Base × Height

= DC × AE

= 8 × 3

= 24 cm²