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    Incorrect Answer For Mini-Question?

    Module 2 Day 16 Challenge Part 3
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    • A
      Aaron Wang M0★ M1★ M2★
      last edited by

      For the mini question, shouldn't the answer be

      $$r={\sqrt{(s-a)(s-b)(s-c)}\over s} $$

      instead of

      $$r={\sqrt{(s-a)(s-b)(s-c)\over s}} $$

      The denominator shouldn't be inside the square root, right?

      debbieD 1 Reply Last reply Reply Quote 5
      • debbieD
        debbie ADMIN M0★ M1 M5 @Aaron Wang
        last edited by

        @aaron-wang Thanks for asking! Actually, the area formula equals \( \sqrt{ s(s-a)(s-b)(s-c)},\) which I can write like this: \(\sqrt{s}\sqrt{(s-a)(s-b)(s-c)}.\) So since the area of the triangle is equal to \(rs,\) dividing this by \(s\) makes the \(\sqrt{s}\) cancel out but leaves a \(\sqrt{s}\) on the bottom.

        $$\begin{aligned} r &= \frac{\textcolor{red}{\sqrt{s}} \sqrt{(s-a)(s-b)(s-c)}}{\textcolor{red}{s}} \\\\ &= \frac{\textcolor{red}{\sqrt{s}} \sqrt{(s-a)(s-b)(s-c)}}{\textcolor{red}{\sqrt{s}\sqrt{s}}} \\\\ &= \frac{\sqrt{(s-a)(s-b)(s-c)} }{\sqrt{s}}\\ \end{aligned} $$
        A 1 Reply Last reply Reply Quote 3
        • A
          Aaron Wang M0★ M1★ M2★ @debbie
          last edited by

          @debbie Ah, ok! So since the denominator is

          $$\sqrt{s} $$

          it gets combined into that bigger square root.

          A debbieD 2 Replies Last reply Reply Quote 4
          • A
            Aaron Wang M0★ M1★ M2★ @Aaron Wang
            last edited by

            This post is deleted!
            1 Reply Last reply Reply Quote 0
            • debbieD
              debbie ADMIN M0★ M1 M5 @Aaron Wang
              last edited by debbie

              @aaron-wang

              temp-20210104.png

              You got it!

              1 Reply Last reply Reply Quote 3

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