In the figure, how do I find the value of the variable?

Question

According to the problem, I have to find the value of the variable in simplest radical form. I began with the equation 18√9 = 6 (1/3)x, so x = 9√3. Now, I know we have to find the area of the base in terms of x. But how do I find the height; do I use Pythagorean Theorem?

Answer 1

You can use Heron's formula for example.

Or since it is an equilateral triangle by the sine formula:

$$A=\frac12x^2\sin60°=\frac{x^2\sqrt 3}{4}$$

You can find many other formulas here Equilateral triangle.

Answer 2

In an equilateral triangle, the heights are exactly the medians. Whence, the heights have length: $$\frac{\sqrt{3}}{2}x.$$

Answer 3

You are asking how to find the height of an equilateral triangle.

Pythagorean theorem is a good idea. Drop the height to the base; it will hit the base in the midpoint (by symmetry of the equilateral triangle), so it divides the equilateral triangle into two right triangles, each of which has a leg length $\frac{x}{2}$ and hypotenuse length $x$. The last leg is the height.

Answer 4

Your initial equation is a little strange, it uses $x$ to designate the area of the base of the pyramid, while in fact on the picture $x$ has another meaning, which is the length of the edges of the base. Also some $\sqrt{3}$ became $\sqrt{9}$, but the result still good, I suspect a typo here.

I would write for a pyramid $V=\frac 13 BH\implies B=\dfrac{3V}H=\dfrac{3\times 18\sqrt{3}}6=9\sqrt{3}$

Note: on this picture I found on the web, they used $a$ in place of $x$.

Now the base is an equilateral triangle.

Area is $B=h\times \frac x2$

And by Pythagoras $(\frac x2)^2+h^2=x^2$ so $h=\frac{\sqrt{3}}2x$

Thus you get $B=\frac{\sqrt{3}}4x^2=9\sqrt{3}\implies x^2=36\implies x=6$

Answer 5

\begin{gathered} V\ =\ \frac{Bh}{3} \ \ \ where\ B\ is\ the\ base\ area\ of\ the\ equilateral\ triangle\ of\ length\ x\\ \\ B\ =\ \frac{1}{2} \ ( base\ triangle\ length)( base\ triangle\ height) \ =\ \ \frac{1}{2} \ ( x)\left(\frac{\sqrt{3} x}{2} \ \right) \ =\ \frac{\sqrt{3} x^{2}}{4}\\ \\ V\ =\ \frac{1}{3} \ *\ \frac{\sqrt{3} x^{2}}{4} \ *\ 6\ \ =\ \frac{6\sqrt{3} x^{2}}{12} \ =\ \frac{\sqrt{3} x^{2}}{2}\\ \ \ \\ \ 18\sqrt{3} \ =\ \frac{\sqrt{3} x^{2}}{2}\\ \\ 36\ =\ x^{2} \ \ \ \ \ \ \ \ \ so\ \ \ \ \ \ \ \ x\ =\ 6\\ \end{gathered}