The perimeter of a triangle is 10 ~cm . If one of its side is 4 ~cm , then remaining sides of the triangle, when area of triangle is maximum are

Step-by-step solution

Let a , b , c be the sides of the triangle. The perimeter of triangle (2 ~s )= a + b + c Let a =4 and 2 ~s =10 i.e. s =5 10=4+b+c b=6-c Now, area of triangle = = s(s-a)(s-b)(s-c) = 5(5-4)(5-6+c)(5-c) = 5(1)( c -1)(5- c ) ²=5( c -1)(5- c )=5 (5 c-c²-5+c )=5 (-c²+6 c-5 ) Let f(c)=5 (-c²+6 c-5 ) f^ (c)=5(-2 c+6) f^ (c)=5(-2)=-10 For extreme value, f^ (c)=0 i.e. 5(-2 c+6)=0 c=3 Thus f^ (3)=-10 < 0 f(c) has maximum value at c=3 b =6-3=3 i.e. the lengths of the remaining sides are 3 ~cm and 3 ~cm .