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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.
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.
Half Area
Age 14 to 16
ShortChallenge Level 
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.
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.
You can find more short problems, arranged by curriculum topic, in our short problems collection.
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