The two moons

Note that when a triangle is inscribed in a circle or semi-circle such that one of its side is the diameter of the circle or semi-circle, it is a right triangle. Since the hypotenuse is $5$ and one of the legs is $4$ , the other leg must be $3$ (use pythagorean theorem). The area of the orange region is equal to the area of the inscribed triangle. We have, $\dfrac{1}{2}(3)(4)=6$

Here is the proof.

Consider my diagram. Let $x$ and $y$ be the legs and $z$ be the hypotenuse of the right triangle. The area of the orange region is equal to the area of the semi-circle with diameter $x$ plus the area of the semi-circle of diameter $y$ plus the area of the right triangle minus the area of the semi-circle with radius $z$ . We have

$A=\dfrac{\pi}{4} x^2+\dfrac{\pi}{4} y^2 +\dfrac{1}{2}xy - \dfrac{\pi}{4} z^2=\dfrac{\pi}{4}(x^2+y^2-z^2)+\dfrac{1}{2}xy$

But, by pythagorean theorem, we have, $z^2=x^2+y^2$

So, we have

$A=\dfrac{\pi}{4}(z^2-z^2)+\dfrac{1}{2}xy=\dfrac{1}{2}xy$

Hence, proved.

You won't believe I did not see the values of the segments AC, BC; I started algebraic:

• Assume x = AB, y = AC and z = BC.
• Orange will be half circle area of radious x/2 + half circle area of radious y/2 minus difference of half circle area of radious z/2 and triangle area.

Hence:

[ π * (x/2) ² ] / 2 + [ π * (y/2) ² ] / 2 - { [ π * (z/2) ² ] / 2 - xy/2 }

Factorizing:

π / 2 * [ (x/2) ² + (y/2) ² - (z/2) ² ] + xy/2 = π / 2 * [ x²/4 + y²/4 - z²/4 ] + xy/2 = π / 2 * [ ( x² + y² - z² ) / 4 ] + xy/2 =

replacing "z" by its value in terms of X and Y: z = sq[ x² + y² ] , hence z² = x² + y² ; we get:

π / 2 * { [ x² + y² - ( x² + y² ) ] / 4 } + xy/2 = π / 2 * ( 0 / 4 ) + xy/2 = π / 2 * 0 + xy/2 = xy/2

And because the blue triangle's area happens to be xy/2; hence, *no matter what the segments measure is, the Orange area is always the same one than the blue triangle.

Orange region is = Two small semicircles - (large semicircle - Triangle area) = Triangle area = 6.

By Pythagoras AB=√5²-4²=3 cm

Sum of the orange is (area) ABsemicrc+ACsemicrc-(BCsemicrc-ABCtri)

ABsemicrc=½π(AB/2)²=9/8π cm²

(half the area of a circle is ½(πr²))

ACsemicrc=½π(AC/2)²=2π cm²

BCsemicrc=½π(BC/2)²=25/8π cm²

Area of ABCtri=½(3)(4)=6 cm²

We notice that ABsemicrc+ACsemicrc=BCsemicrc=25/8π cm²

So area of orange equals area of triangle Or by substitution in the equaution of the sum

Orange area = ABsemicrc+ACsemicrc-(BCsemicrc-ABCtri) =9/8π+2π-(25/8π-6)=6 cm²

Therefore area of orange area equals 6cm².

The sum of the two orange regions = the sum of two orange semi-circles - [area of blue semi-circle - area of the triangle] =(1/2) pi (3/2)^2 + (1/2) pi (4/2)^2 - [(1/2) pi(5/2)^2 - (1/2) (3) (4)] =(pi/8) (9+16-25) +6 =6.Ed Gray

The Theorem of Hippocrates (ca. 430 B.C.) infers that the sum of the areas of such 'lunes' equals the area of the base triangle; along with Pythagoras' Theorem one arrives at 6 units squared.