Let in ∆ABC, AB = BC = CA = a cm ON ⊥ AB OM ⊥ AC OP ⊥ BC Whose length respectively is ON = 9 cm, OM = 8 cm, OP = 7 cm. Area of ∆ABC = Area of ∆AOB + Area of ∆AOC + Area of ∆BOC √3/4 a² = 1/2 × ON × AB + 1/2 × OM × AC + 1/2 × OP × BC √3/4 a² = 1/2 × 9 × a + 1/2 × 8 × a + 1/2 × 7 × a √3/4 a² = 1/2a (9 + 8 + 7) √3/4 a² = 1/2 a × 24 √3/4 a² = 12 a a = (12 × 4)/√3 a = 4√3× 4 = 13√3 Side of ∆ABC, a = 16√3 ∴ Perimeter = 3 × 16√3 = 48√3
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