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For a surveying instrument, see Graphometer.

In mathematics (more specifically geometry), a semicircle is a two-dimensional geometric shape that forms half of a circle. Being half of a circle's 360°, the arc of a semicircle always measures 180°. A triangle inscribed in a semicircle is always a right triangle.

## Uses

A semicircle can be used to construct arithmetic and geometric means of two lengths using straight-edge and compass. If we make a semicircle with a diameter of a+b, then the length its radius is the arithmetic mean (since it's half of the diameter). The geometric mean can be found by dividing the diameter into two segments of lengths a and b, and then connecting their common end and the semicircle with a segment perpendicular to the diameter. The length of the resulting segment is the geometric mean,[1] which can be proved using Pythagorean theorem.

This method can be used to accomplish quadrature of a rectangle (since a square whose sides are equal to geometric mean of sides of a rectangle has the same area as the rectangle), and thus any figure for which we can construct a rectangle with equal area, such as any polygon (but not a circle).

### Area and Perimeter

The area is:

$A=\frac{\pi}{2}r^2$

The perimeter is:

$P=\pi r$

## Circumscribed Semicircle

The area is:

$A=\frac{\pi}{32}\left(\frac{abc}{A}\right)^2$
$A=\frac{\pi}{2}\left(\frac{a^2b^2c^2}{(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)}\right)$
$A=\frac{\pi}{8}s^2$
$A=\frac{\pi}{2}\left(1-\frac{2}{n}\right)^2\tan^2\left(\frac{180}{n}\right)s^6$
$A=\frac{\pi}{8}\frac{s^2}{\sin^2\left(\frac{180}{n}\right)}$

The perimeter is:

$P=\frac{\pi}{4}\frac{abc}{A}$
$P=\pi\frac{abc}{\sqrt{(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)}}$
$P=\pi\left(1-\frac{2}{n}\right)\tan\left(\frac{180}{n}\right)s^3$
$P=\frac{\pi}{2}s$
$P=\frac{\pi}{2\sin\left(\frac{180}{n}\right)}s$

## Inscribed Semicircle

The area is:

$a=2\pi\left(\frac{A}{P}\right)^2$
$a=\frac{\pi}{8}\left(\frac{(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)}{(a+b+c)^2}\right)$
$a=\frac{\pi}{8}\left(\frac{\sin\left(\frac{360}{n}\right)}{1+\cos\left(\frac{180}{n}\right)+\sin\left(\frac{180}{n}\right)}\right)^2s^2$
$a=\frac{\pi}{72}\left(\frac{n}{(n-2)\tan\left(\frac{180}{n}\right)}\right)^2s^2$
$a=\frac{\pi}{8\tan^2\left(\frac{180}{n}\right)}s^2$

The perimeter is:

$p=2\pi\frac{A}{P}$
$p=\frac{\pi}{4}\left(\frac{\sqrt{(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)}}{a+b+c}\right)$
$p=\pi\frac{\sin\left(\frac{360}{n}\right)}{2\left(1+\cos\left(\frac{180}{n}\right)+\sin\left(\frac{180}{n}\right)\right)}s$
$p=\pi\frac{n}{(6n-12)\tan\left(\frac{180}{n}\right)}s$
$p=\frac{\pi}{2\tan\left(\frac{180}{n}\right)}s$