Half Area
The horizontal red line divides this equilateral triangle into two shapes of equal area. How long is the red line?
Problem
This equilateral triangle has sides of length 2 cm.
Image
The red line, which is parallel to the base, divides this equilateral triangle into two shapes of equal area.
How long is the red line?
Student Solutions
Using the relationship between area and length scale factors
The area scale factor between the smaller triangle and the larger triangle is $2$.
The area scale factor is always the square of the length scale factor, so the length scale factor must be $\sqrt{2}$.
So the red length must be $2\div \sqrt{2}=\sqrt{2}$ cm.
Finding the areas of the pieces
Let the red length be $x$ cm.
The perpendicular heights of the equilateral triangles can be found using Pythagoras' theorem.
Image
For the larger triangle, half of the base is 1 cm, and the hypotenuse is 2 cm, so, calling the height $h$ cm, $1^2+h^2=2^2$, so $h^2=3$, so $h=\sqrt{3}.$
Using the scale factor $\dfrac{x}{2}$ between the two triangles, or applying Pythagoras' theorem again to the smaller triangle, the height of the smaller triangle must be $\dfrac{x\sqrt{3}}{2}.$
So the area of the larger triangle is $\dfrac{1}{2}\times 2\times\sqrt{3}$ cm$^2$ and the area of the smaller triangle is $\dfrac{1}{2}\times x\times \dfrac{x\sqrt{3}}{2}$ cm$^2.$
So, since the area of the smaller triangle is half of the area of the larger triangle, $$\begin{align}\frac{1}{2}\times 2\times\sqrt{3}&=x\times \dfrac{x\sqrt{3}}{2}\\
\Rightarrow 1&=\dfrac{x^2}{2}\\
\Rightarrow x^2&=2\\
\Rightarrow x&=\sqrt{2}\end{align}$$
So the red length must be $2\div \sqrt{2}=\sqrt{2}$ cm.