Name: Sum of Infinite Terms of G.P.
Uploaded: 2019-07-19
Duration: 3 min 36 s
Description: Do you know that we can find the sum of an infinite number of terms of G.P.? Of course, there will be some conditions. Check this video to learn more.

Question

Prove that among all triangles with a given perimeter, the equilateral triangle has the maximum area.

Accepted Answer

Let, the sides of the triangle are $a, b, c$ and also let the perimeter of the triangle is $2 s .$

Then, we have $a + b + c = 2 s .$

The area of a triangle with sides $a, b, c$ is $△ = \sqrt{s (s - a) (s - b) (s - c)},$ where $s = \frac{a + b + c}{2}$ is the semiperimeter.

Hence, $△^{2} = s (s - a) (s - b) (s - c)$

$\Rightarrow \frac{△^{2}}{s} = (s - a) (s - b) (s - c) . . . (i)$

Since, $s - a, s - b$ and $s - c$ are positive numbers, hence using $A . M . \geq G . M .,$ where $A . M .$ is the arithmetic mean and $G . M .$ is the geometric mean of the numbers, also we know that, the equality holds if all the numbers are same, thus we have

$\frac{(s - a) + (s - b) + (s - c)}{3} \geq \sqrt[3]{(s - a) (s - b) (s - c)}$

$\Rightarrow \frac{3 s - (a + b + c)}{3} \geq \sqrt[3]{(s - a) (s - b) (s - c)}$

$\Rightarrow \frac{3 s - (2 s)}{3} \geq \sqrt[3]{(s - a) (s - b) (s - c)}$

$\Rightarrow \frac{s}{3} \geq \sqrt[3]{(s - a) (s - b) (s - c)}$

$\Rightarrow \frac{s}{3} \geq \sqrt[3]{\frac{△^{2}}{s}}$

$\Rightarrow \sqrt[3]{\frac{△^{2}}{s}} \leq \frac{s}{3}$

$\Rightarrow \frac{△^{2}}{s} \leq {(\frac{s}{3})}^{3}$

$\Rightarrow △^{2} \leq \frac{s^{4}}{27}$

$\Rightarrow △ \leq \frac{s^{2}}{3 \sqrt{3}}$

Thus, the maximum value of the area of the triangle is $\frac{s^{2}}{3 \sqrt{3}}$ and we know that, the equality holds, when all the numbers are same, i.e.

$s - a = s - b = s - c$

$\Rightarrow a = b = c .$

Hence, the triangle is an equilateral triangle and the maximum area is $△_{\max} = \frac{{(\frac{a + a + a}{2})}^{3}}{3 \sqrt{3}}$

$\Rightarrow △_{\max} = \frac{{(\frac{3 a}{2})}^{2}}{3 \sqrt{3}}$

$\Rightarrow △_{\max} = \frac{9 a^{2}}{4 \times 3 \sqrt{3}} = \frac{\sqrt{3} a^{2}}{4} .$

Prove that among all triangles with a given perimeter, the equilateral triangle has the maximum area.

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