The story starts with this right isosceles triangle here:

M2W1D3-y-part-3-why-divide-by-sqrt-2-squared1.png

We saw this triangle in the Day 2 lesson; its side lengths can be derived using the Pythagorean Theorem.

Now let's introduce our two semicircles:

M2W1D3-y-part-3-why-divide-by-sqrt-2-squared2.png

They are similar shapes with diameters in the ratio \( 4\sqrt{2} : 8,\) or \(\sqrt{2} : 2,\) which we can see more clearly if we take away the triangle.

M2W1D3-y-part-3-why-divide-by-sqrt-2-squared3.png

As we saw in the video, if we know the ratio of the semicircles' diameters, we can find the ratio of their areas by simply squaring the length ratio.

temp-20200806-2.png

Why does this work? One way to observe this is to imagine a simpler shape, such as a square. Draw two squares, one of which has \(5\) times the side of the other; you would need \(25\) times as many tiles to cover the larger square as the smaller square.

M0W2-exam-cubes-why-30-cubed2-50-percent.png

This forum post by @thomas discusses this idea of area scaling. ⬅ ⬅ I recommend taking a look.

Thus Prof. Loh got his factor of \(\frac{1}{\sqrt{2}}\) based on the ratio of the diameters of the two semicircles.

$$\begin{aligned} \text{small diameter} &= \frac{1}{\sqrt{2}} \times \text{ large diameter} \\ \implies \text{ small semicircle area } &= \left( \frac{1}{\sqrt{2}}\right)^2 \times \text{ large semicircle area } \\ \end{aligned} $$🙂

]]>I think earlier in this module Prof Loh proved that a hexagon can be made of 6 equilateral congruent triangles.

So the \(r\) is half the side length of the triangles, and \(2r\) is the length of the triangle.

🙂 ]]>