Triangular Function

Geometry Level pending

Let $g(x)$ be a function defined on the interval $[-1,1]$ so that the area of the equilateral triangle with two of its vertices at $(0,0)$ and $(x,g(x))$ is $\dfrac{\sqrt{3}}{4}$ , then what is the function $g(x)$ equal to?

\sqrt{1-x^{3}}

\sqrt{1+x^{2}}

\sqrt{1-x^{2}}

\sqrt{1+x^{3}}

1 solution

Marta Reece
Apr 5, 2017

Area of an equilateral triangle with side $a$ is $A=\frac{\sqrt{3}}{4}a^2=\frac{\sqrt{3}}{4}$ . So we need $a^2=1$ .

From Pythagorean theorem, the square of the distance between the two points mentioned is $a^2=x^2+(g(x))^2$ .

$x^2+(g(x))^2=1$

$g(x)=\sqrt{1-x^2}$

Triangular Function

1 solution

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