Find the area of an equilateral triangle that has perimeter of 24 m.

Mathematics

1 answer:

mafiozo [28]3 years ago

6 0

we know the perimeter is 24, and is an equilateral triangle, so it has three equal sides, so each side is 24 ÷ 3 = 8.

$\bf \textit{area of an equilateral triangle}\\\\ A=\cfrac{s^2\sqrt{3}}{4}~~ \begin{cases} s=side's\\ \qquad length\\ \cline{1-1} s = 8 \end{cases}\implies A=\cfrac{8^2\sqrt{3}}{4}\\\\\\ A=16\sqrt{3}\implies A\approx 24.71$

You might be interested in

(X^4)-(45x^2)+324 what are the zeroes? And how many turning points?

dmitriy555 [2]

Zeroes:

We must solve

$x^4-45x^2+324=0$

To do so, we define the auxiliary variable $t=x^2$ . The equation becomes

$t^2-45t+324=0$

The quadratic formula yields the solutions

$t=9,\quad t=36$

Substituting back $t=x^2$ gives

$x^2=9 \iff x=\pm 3,\quad x^2=36 \iff x=\pm 6$

So, the zeroes are -6, -3, 3, 6.

Turning points:

Turning points are points where a function stops being increasing to become decreasing, or vice versa. Since functions are increasing when their first derivative is positive and decreasing when it's negative, turning points are points where the first derivative is zero.

We have

$f(x)=x^4-45x^2+324 \implies f'(x)=4x^3-90x$