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In a semicircle of diameter $CD$ there's a chord $AB$ of length 7, and it's parallel to the diameter. There's also a small semicircle that is tangent to $AB$ and its diameter is a segment in $CD$ . Find the area of the semicircle without the small semicircle.

I'm pretty curious about this problem, i've tried many things, like drawing triangles that are similar and also rectangles, so i tries with pithagorean theorem but i got nothing. I'd really like to know how to solve it, thanks.

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  • $\begingroup$ "There's also another semicircle that is tangent to AB" does not give enough information... Why dont you show a figure ? $\endgroup$ – Jean Marie Apr 29 '17 at 10:34
  • $\begingroup$ I edited the question a little, is it more clear now? $\endgroup$ – SonodaUmi Apr 29 '17 at 10:43
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Place the small semicircle center on the center of the big semicircle, for simplicity. This won't change the area you are looking for.
Let us call $r$, $R$ and $d$ the radius of the small semicircle, the radius of the big semicircle and half the chord, respectively.
Using your notation, join $A$ with the center $O$ of the semicircles: that is of course a radius of the big semicircle. If you erect a perpendicular from the center to the chord, intersecting at $H$, you draw a right triangle $AOH$, whose hypotenuse is the radius $R$ and whose cathetuses are $r$ and $d$. The area you are looking for is: $$\frac{\pi}{2}R^2 - \frac{\pi}{2} r^2$$ Though, by Pythagorean Theorem, you have that $R^2 = r^2 + d^2$. By plugging this into the preivous expression, you find: $$A= \frac{\pi}{2}(r^2 + d^2) - \frac{\pi}{2}r^2 = \frac{\pi}{2}d^2$$ Now you can calculate that area, since you have the length of the chord and, thus, $d=7/2$. As a result, $$A = \frac{49 \pi}{8}$$

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Since we're told nothing about the radius of the small semicircle, let's assume its radius is zero, and hence its area is also zero. Then $AB$ is the diameter of the large semicircle and the radius is $7/2$.

The area of a semicircle is given by $\frac{\pi}{2}r^2$, so the area is $\frac{49\pi}{8}$.

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  • $\begingroup$ +1 for the “aha!” method. This problem is a relative of the old “a hole 1 meter deep is drilled through a sphere...” aha problem. $\endgroup$ – amd Apr 29 '17 at 17:43

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