GIVEN: a + b + c = 11 ab + bc + ca = 36 CONCEPT: Algebra FORMULA USED: (a+b+c)2=a2+b2+c2+2ab+2bc+2ca CALCULATION: a + b + c = 11 ab + bc + ca = 36 We know that: (a+b+c)2=a2+b2+c2+2ab+2bc+2ca ⇒a2+b2+c2=121−72 ⇒a2+b2+c2=49 ⇒a2+b2=49−c2 We know that sum of two squares are always greater than or equal to zero ‘0’. a2+b2≥0 ⇒49−c2≥0 ⇒49≥c2 ⇒c2≤49 ⇒c≤7 ∴ Maximum value which ‘c’ can take is 7.