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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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

Similar Questions

The condition that $f(x) = ax^3 + bx^2 + cx + d$ has no extreme value is

Prove that the volume of the largest cone that can be inscribed in a sphere of radius $R$ is $\frac{8}{27}$ of the volume of the sphere.

Find the minimum value of $64 \sec x + 27 \csc x$ for $0 < x < \frac{\pi}{2}$.

The minimum value of the function $f(x) = x \log x$ is

The function $f(x) = x^{25}(1 - x)^{75}$ for $x \in [0, 1]$ attains its maximum at $x = \dots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module