The given figure shows a parallelogram ABCD. E is a point in AD and CE produced meets BA produced at point F. If AE = 4 cm, AF = 8 cm and AB = 12 cm, find the perimeter of the parallelogram ABCD.

From figure,

FB = AF + AB = 8 + 12 = 20 cm.

In △DEC and △EAF

⇒ ∠DEC = ∠FEA [Vertically opposite angles are equal]

⇒ ∠EDC = ∠EAF [Alternate angles are equal]

∴ △DEC ~ △EAF [By AA]

Since, corresponding sides of similar triangles are proportional we have :

$\Rightarrow \dfrac{DE}{AE} = \dfrac{DC}{AF} \\[1em] \Rightarrow \dfrac{DE}{AE} = \dfrac{AB}{AF} \space \Big[\because\text{ AB = CD}\Big] \\[1em] \Rightarrow \dfrac{DE}{4} = \dfrac{12}{8} \\[1em] \Rightarrow DE = \dfrac{48}{8} = 6 \text{ cm}.$

Since, ABCD is a ||gm.

AB = CD and BC = AD.

From figure,

AD = AE + ED = 4 + 6 = 10 cm.

Perimeter of ||gm ABCD = AB + BC + CD + AD

= 12 + 10 + 12 + 10

= 44 cm.

Hence, perimeter of ||gm ABCD = 44 cm.