# If you split an equilateral triangle as it sits on its base horizontally, at what height would the top and bottom areas be equal? Using variables?

Dec 25, 2015

Let altitude be $h$. Note that a trapezoid and a smaller triangle will occur after you split the triangle. The splitting height (trapezoid's height) is
$\textcolor{w h i t e}{\times} h \left(2 - h\right)$

#### Explanation:

Let altitude be $h$, one side length be $a$, and area be $A$.
$\textcolor{w h i t e}{\times} \frac{h}{a} = \sin 60$
$\textcolor{w h i t e}{\times x} = \frac{\sqrt{3}}{2}$

Then one side length of the triangle will be
$\textcolor{w h i t e}{\times} a = \frac{2 \sqrt{3} h}{3}$

$\textcolor{w h i t e}{\times} A = h \times 2 \frac{\sqrt{3}}{3} h \times \frac{1}{2}$
$\textcolor{w h i t e}{\times x} = \frac{\sqrt{3}}{3} {h}^{2}$

$\implies \frac{A}{2} = \frac{\sqrt{3}}{6} {h}^{2}$

Note that a trapezoid and a smaller triangle will occur after you split the triangle. Suppose that splitting height (trapezoid's height) is $x$ and splitting base (trapezoid's upper base) is $b$:

If height of smalll triangle is $h - x$, then:
$\textcolor{w h i t e}{\times} h - x = \frac{\sqrt{b}}{2}$

$\implies b = 2 \frac{\sqrt{3}}{3} \left(h - x\right)$
$\implies {A}_{\text{Lower}} = 2 \frac{\sqrt{3}}{3} \left(h - x\right)$

$\textcolor{w h i t e}{\times} {A}_{\text{Lower}} = \frac{A}{2}$
$\implies \frac{\sqrt{3}}{6} {h}^{2} = \frac{\sqrt{3}}{3} \left(h - x\right)$
$\implies {h}^{2} = 2 h - 2 x$

$\implies x = h \left(2 - h\right)$