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<title>Pythagoras' Theorem and Areas</title>
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<h1>Pythagoras' Theorem and Areas</h1>
<h2>Pythagoras' Theorem</h2>
<p>Let's start with a quick refresher of the famous Pythagoras' Theorem.</p>
<p style="float:left; margin: 0 10px 5px 0;"><img src="images/triangle-abc.svg" alt="triangle abc" style="width: 201px; height: 115px;"></p>
<p style="text-align: center;">Pythagoras' Theorem says that, in a
right angled triangle:<br>
the square of the hypotenuse (<strong>c</strong>) is equal to the sum of
the squares of the other two sides (<strong>a</strong> and <strong>b</strong>).</p>
<p class="center large">a<sup>2</sup> + b<sup>2</sup> = c<sup>2</sup></p>
<p>That means we can draw squares on each side:</p>
<p class="center"><img src="images/pythagoras-squares-area.svg" alt="Pythagoras Areas" class="center"></p>
<p>And this will be true:</p>
<p class="largest center">A + B = C</p>
<p>You can learn more about the <a href="../pythagoras.html">Pythagorean
Theorem</a> and review its <a href="pythagorean-theorem-proof.html">algebraic
proof</a>.</p>
<h2>A More Powerful Pythagorean Theorem&nbsp;</h2>
<p>Say we want to draw semicircles on each side of a right triangle:</p>
<div style="text-align: center;"><img src="images/pythagoras-circle.svg" alt="Pythagoras semicircle"><br>
<strong>A</strong>, <strong>B</strong> and<strong> C</strong>
are the areas of each<br>
semicircle with diameters <strong>a</strong>,
<strong>b</strong> and <strong>c</strong>.</div>
<p>Maybe A + B = C ?</p>
<p>But they aren't squares! Yet let's go ahead anyway to see where it leads us.</p>
<p>OK, the area of a <a href="circle.html">circle</a> with diameter "D" is:</p>
<p class="center larger"><strong>Area of Circle</strong> =&nbsp;<span class="intbl"><em>1</em> <strong>4</strong>&nbsp;</span> <span class="times">π</span> D<sup>2</sup></p>
<p>So the area of a semicircle is <b>half</b> of that:</p>
<p class="center larger"><strong>Area of Semicircle</strong> =&nbsp;<span class="intbl"><em>1</em> <strong>8</strong>&nbsp;</span> <span class="times">π</span> D<sup>2</sup></p>
<p>And so the area of each semicircle is:</p>
<p class="center larger"><strong>A</strong> =&nbsp;<span class="intbl"><em>1</em>
<strong>8</strong>&nbsp;</span> <span class="times">π</span>a<sup>2</sup></p>
<p class="center larger"><strong>B</strong> =&nbsp;<span class="intbl"><em>1</em>
<strong>8</strong>&nbsp;</span> <span class="times">π</span>b<sup>2</sup></p>
<p class="center larger"><strong>C</strong> =&nbsp;<span class="intbl"><em>1</em>
<strong>8</strong>&nbsp;</span> <span class="times">π</span>c<sup>2</sup></p>
<p>Now our question:</p>
<div class="fun">
<p class="center larger">Does A + B = C ?</p>
</div>
<p>Let's substitute the values:</p>
<p class="center larger">Does <span class="intbl"><em>1</em> <strong>8</strong>
</span><span class="times">π</span>a<sup>2 </sup>+ <span class="intbl"><em>1</em>
<strong>8</strong> </span><span class="times">π</span>b<sup>2</sup>
=&nbsp;<span class="intbl"><em>1</em> <strong>8</strong> </span><span class="times">π</span>c<sup>2</sup>&nbsp; ?</p>
<p>We can <a href="../algebra/factoring.html">factor out</a>&nbsp;<span class="intbl"><em>1</em>
<strong>8</strong> </span><span class="times">π</span> and we get:</p>
<p class="center larger">a<sup>2 </sup>+ b<sup>2</sup> =&nbsp;c<sup>2</sup></p>
<p>Yes! It is simply Pythagoras' Theorem.</p>
<p>Therefore, we have shown that Pythagoras' Theorem is true for
semicircles.</p>
<p>Will it work for any other shape?</p>
<div style="text-align: center;"><img src="images/pythagoras-star.svg" alt="Pythagoras' Star"></div>
<p>Yes! The Pythagorean Theorem can be taken further into a
shape-generalized form as long as the shapes are <a href="similar.html">similar</a> (has a special meaning in Geometry).</p>
<p><br></p>
<div class="def">
<p class="larger"><b> Shape-Generalization Form of the Pythagorean
Theorem:<br>
</b><br>
Given a right triangle, we can draw <b>similar</b> shapes on
each side so that the area of the shape constructed on the
hypotenuse is the sum of the areas of similar shapes constructed on
the legs of the triangle.</p>
<p class="largest center">A + B = C</p>
<p class="larger">Where:</p>
<ul style="text-align: left;">
<li><strong>A</strong> is the area of the shape on the
hypotenuse.<strong></strong></li>
<li><strong>B</strong> and <strong>C</strong> are the areas of the
shapes on the legs.</li>
</ul>
</div>
<p><br></p>
<p>The Theorem still holds for cool shapes that are not polygons, such as
this amazing dragon!</p>
<div style="text-align: center;"><img src="images/pythagoras-dragon.svg" alt="Pythagoras' Dragon"></div><br>
<p><br></p>
<p><br></p>
<div class="related">
<a href="../pythagoras.html">Pythagorean Theorem</a>
<a href="pythagorean-theorem-proof.html">Pythagorean Theorem Algebraic Proof</a>
<a href="pythagoras-3d.html">Pythagorean Theorem in 3D</a>
<a href="index.html">Geometry Index</a>
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