Let ∆ABC be an isosceles triangle, where AB = 8 cm
Case I:
If AB = BC
AB = BC = 8 cm
Perimeter of the triangle ABC = 28 cm
⇒ AB + BC + CA = 28 cm
⇒ 8 cm + 8 cm + CA = 28 cm
⇒ 16 cm + CA = 28 cm
⇒ CA = 28 cm – 16 cm = 12 cm
∴ Sides are 8 cm, 8 cm and 12 cm
Case II:
If BC = CA
Perimeter of the triangle ABC = 28 cm
⇒ AB + BC + CA = 28 cm
8 cm + 2BC = 28 cm
⇒ 2BC = 28 cm – 8cm = 20 cm
⇒ BC = 10 cm
∴ Sides are 10 cm, 10 cm and 8 cm