<p>This is from the Big Blue book 2e, test 3 section 2 question 17 on page 518.</p>
<p>Semicircular arcs AB, AC, BD, and CD divide the circle above into regions. The points shown along the diameter AD divide it into 6 equal parts. If AD = 6, what is the total area of the shaded regions?</p>
<p>Since the two halves of the circle are identical, consider only the area inside the upper semicircle.</p>
<p>If AD = 6, then the radius of the circle is 3, so the area of the upper semicircle is (1/2)(3^2)(pi), or (9/2)(pi).</p>
<p>But taken out of that area, there’s a smaller semicircle with a radius of 2. (That is to say, the unshaded region from A to C. Yes, I know there’s some blue stuff in there; we’re going to add that back in later). That white region (ignoring the blue stuff) is the region inside a semicircle with a radius of 4. So its area is (1/2)(2^2)(pi), or 2(pi). Subtract that area from the area inside the larger semicircle, and the remaining area (which is the blue region OUTSIDE the white semicircle) is [(9/2)(pi) - 2(pi)], or (5/2)(pi).</p>
<p>But now we have to add back in that small blue region that’s inside the white region. It is bounded by a semicircle with radius 1, so its area is (1/2)(1^2)(pi), or simply (1/2)(pi). Add that to the other blue area: (1/2)(pi) + (5/2)(pi) = (6/2)(pi), or 3(pi).</p>
<p>But that’s only the blue area inside the upper semicircle. There’s another blue area the same size inside the lower semicircle. So the sum of the areas of all the blue regions inside the whole circle is 3(pi) + 3(pi), or 6(pi).</p>
