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    why are the semicircles are multiplied by 1/sqrt 2

    Module 2 Day 3 Your Turn Part 3
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    • M
      mirthfulostrich M0 M1 M2
      last edited by debbie

      Module 2 Week 1 Day 3 Your Turn Explanation Part 3

      Maybe I am missing something but I seriously don't understand

      debbieD 1 Reply Last reply Reply Quote 3
      • debbieD
        debbie ADMIN M0★ M1 M5 @mirthfulostrich
        last edited by debbie

        @mirthfulostrich Thanks for asking, and don't worry, the video explanation goes very quickly through this rather complex logic sequence. 👨‍🏫

        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.

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

        🙂

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