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Let's have some fun with a geometry puzzle. Given: it's a partial solar eclipse, and the Moon is covering half of the Sun's diameter. For the sake of simplicity, we'll assume that the angular diameters of the Sun and Moon are equal. Problem to be solved: what percentage of the solar disk is covered by the Moon?

Incidentally, if you're up for learning some astronomical jargon, I should familiarize you with a few terms. Themagnitudeof a solar eclipse refers to the portion of the sun's diameter that's covered over by the sun.Obscurationis defined as the portion of the solar disk that's covered over.

So given the magnitude of the solar eclipse, we are now seeking to solve for obscuration.

Looking at the illustration below, the solution comes in finding the area of the circular segment in blue above the chord AB and the circular segment in orange below the chord AB. Because the circular segment above equals the area of the circular segment below, we can solve for one to solve for the other.

That being the case, let's focus on the circular segment in blue. We draw in two more line segments (radii) from the center of the circle to the points A and B. So we now find out the total area of the circular sector (the circular segment in blueplusthe triangle in brown).

Subtract the area of the triangle in brown from the area of the circular sector (in brownplusblue) to find out the circular segment in blue. Multiply this answer by two then divide by the area of circle to find out the total area of the Sun's disk covered over by the Moon.

**Solving for the Triangle in Brown: Think Two 30 ^{o}-90^{o}-60^{o} Triangles**

To solve for the isosceles triangle in brown, think two equal-sized 30^{o}-90^{o}-60^{o}triangles. Let's have the radius of the circle = 1 = hypotenuse of each right triangle. Thereby, the lesser side = 0.5 and the greater side = 0.866 (square root of 3/2). The longest side of the entire 30^{o}-120^{o}-30^{o}isosceles triangle = 1.732 (2 x 0.886) = square root of 3.

Area of triangle in brown = 1/2 x base x height Area of triangle in brown = 1/2 x 1.732 x 1/2 = 0.433

**Solving for the Circular Sector in Brown plus Blue: Think One-Third Circle**

The two 60^{o}angles added together = 120^{o}= 1/3 the central angle of a 360^{o}- circle. Let the radius = 1 and pi = 3.1416.

Area of circular sector = 1/3 x pi ^{.}r^{2}Area of circular sector = 1/3 x 3.1416 x 1 ^{2}Area of circular sector = 1/3 x 3.1416 x 1 Area of circular sector = 1.0472

Area of circular segment = area of circular sector in blue and brown - area of triangle in brown Area of circular segment = 1.0472 - 0.433 = 0.6142

Area covered over = 2 x circular segment/area of sun's disk Area covered over = 2 x 0.6142/pi ^{.}r^{2}Area covered over = 1.2284/3.1416 Area covered over = 0.391 = 39.1%

copyright 2012 by Bruce McClure