The perimeter of a triangle is 10 text{ cm}. | vedclass

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(C) Let $a, b, c$ be the sides of the triangle. The perimeter of the triangle is $a+b+c = 10 	ext{ cm}$.Given $a = 4 	ext{ cm}$,so $b+c = 6 	ext{ c

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$3 	ext{ cm}, 3 	ext{ cm}$

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(C) Let $a, b, c$ be the sides of the triangle. The perimeter of the triangle is $a+b+c = 10 	ext{ cm}$.Given $a = 4 	ext{ cm}$,so $b+c = 6 	ext{ cm}$,which implies $c = 6-b$.The semi-perimeter $s = \frac{10}{2} = 5 	ext{ cm}$.Using Heron's formula,the area $\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{5(5-4)(5-b)(5-c)} = \sqrt{5(1)(5-b)(5-(6-b))} = \sqrt{5(5-b)(b-1)}$.To maximize the area,we maximize $f(b) = 5(5-b)(b-1) = 5(-b^2 + 6b - 5)$.Differentiating with respect to $b$,$f'(b) = 5(-2b + 6)$.Setting $f'(b) = 0$ gives $b = 3$.Since $f''(b) = -10 Then $c = 6 - 3 = 3 	ext{ cm}$.Thus,the remaining sides are $3 	ext{ cm}$ and $3 	ext{ cm}$.

The perimeter of a triangle is $10 \text{ cm}$. If one of its sides is $4 \text{ cm}$,then the remaining sides of the triangle,when the area of the triangle is maximum,are

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