Revision 601217 of Proving the Pythagorean theorem

Revision slug: Web/MathML/Examples/MathML_Pythagorean_Theorem
Revision title: MathML Pythagorean Theorem
Revision id: 601217
Created: May 28, 2014, 8:26:21 AM
Creator: nielsdg
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We will now prove the Pythogorian theorem: $a^{2} + b^{2} = c^{2}$

We can prove the theorem algebraically by showing that the area of the big square equals the area of the inner square (hypotenuse squared) plus the area of the four triangles: $\begin{array}{rcl} {(a + b)}^{2} & = & c^{2} + 4 \cdot (\frac{1}{2} a b) \\ a^{2} + 2 a b + b^{2} & = & c^{2} + 2 a b \\ a^{2} + b^{2} & = & c^{2} \end{array}$

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<p>
  We will now prove the Pythogorian theorem:
  <math>
    <mrow>
      <msup><mi> a </mi><mn>2</mn></msup>
      <mo> + </mo>
      <msup><mi> b </mi><mn>2</mn></msup>
      <mo> = </mo>
      <msup><mi> c </mi><mn>2</mn></msup>
    </mrow>
  </math>
</p>

<p>
  We can prove the theorem algebraically by showing that the area of the big square equals the area of the inner square (hypotenuse squared) plus the area of the four triangles:

  <math style="display:block">
    <mtable columnalign="right center left">
      <mtr>
        <mtd>
          <!-- We want to square the whole expression between the braces so we need to wrap it -->
          <msup>
            <mrow>
              <mo> ( </mo>
              <mi> a </mi> <mo> + </mo> <mi> b </mi>
              <mo> ) </mo>
            </mrow>
            <mn> 2 </mn>
          </msup>
        </mtd>

        <mtd>
          <mo> = </mo>
        </mtd>

        <mtd>
          <msup><mi> c </mi><mn>2</mn></msup>
          <mo> + </mo>
          <mn> 4 </mn>
          <mo> &sdot; </mo>
          <!-- the area of the small triangles-->
          <mo>(</mo>
          <mfrac>
            <mn> 1 </mn>
            <mn> 2 </mn>
          </mfrac>
          <mi> a </mi><mi> b </mi>
          <mo>)</mo>
        </mtd>

      </mtr>

      <mtr>
        <mtd>
          <msup><mi> a </mi><mn>2</mn></msup>
          <mo> + </mo>
          <mn> 2 </mn><mi> a </mi><mi> b </mi>
          <mo> + </mo>
          <msup><mi> b </mi><mn>2</mn></msup>
        </mtd>

        <mtd>
          <mo> = </mo>
        </mtd>

        <mtd>
          <msup><mi> c </mi><mn>2</mn></msup>
          <mo> + </mo>
          <mn> 2 </mn><mi> a </mi><mi> b </mi>
        </mtd>
      </mtr>

      <mtr>
        <mtd>
          <msup><mi> a </mi><mn>2</mn></msup>
          <mo> + </mo>
          <msup><mi> b </mi><mn>2</mn></msup>
        </mtd>

        <mtd>
          <mo> = </mo>
        </mtd>

        <mtd>
          <msup><mi> c </mi><mn>2</mn></msup>
        </mtd>
      </mtr> 

    </mtable>
  </math>
</p>

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