1. ## Tough problem, geometery+algerbra

this is a tough problem that i have been stuck on for a while. Please look at the attached picture to see what it is. any help would greatly be appreciated.

2. Originally Posted by OnMyWayToBeAMathProffesor
this is a tough problem that i have been stuck on for a while. Please look at the attached picture to see what it is. any help would greatly be appreciated.
Hello,

let the side of the triangle b a

Then the height of the triangle is:
$h = \frac{1}{2}\sqrt{3} \cdot a$ (Use Pythagorean theorem)

The radius of the inner circle is 2/3 of the height: $r = \frac{2}{3} \cdot \frac{1}{2}\sqrt{3} \cdot a = \frac{1}{3}\sqrt{3} \cdot a$

Area of the triangle:
$A_{triangle} = \frac{1}{2} \cdot a \cdot \frac{1}{2}\sqrt{3} \cdot a = \frac{1}{4} \cdot \sqrt{3} \cdot a^2$

Area of the inner circle:
$A_{\text{inner circle}} = \pi \cdot \left( \frac{1}{3}\sqrt{3} \cdot a \right)^2 = \frac{1}{3}\pi \cdot a^2$

Area of outer half circle:
$A_{\text{outer half circle}} = \frac{1}{2} \cdot \pi \cdot \left( \frac{1}{2} a \right)^2 = \frac{1}{8} \pi a^2$

Area of the three moons:
$A_{\text{3 moons}} = A_{triangle} + 3 \cdot A_{\text{outer half circle}} - A_{\text{inner circle}}$

$A_{\text{3 moons}} = \frac{1}{4} \cdot \sqrt{3} \cdot a^2 + 3 \cdot \frac{1}{8} \pi a^2 - \frac{1}{3}\pi \cdot a^2 = \frac{1}{4} \cdot \sqrt{3} \cdot a^2 + \frac{1}{24}\pi \cdot a^2$

Obviously is:
$\frac{1}{8} \cdot A_{\text{inner circle}} = \frac{1}{8} \cdot \frac{1}{3}\pi \cdot a^2 = \frac{1}{24}\pi \cdot a^2$

and that's the area which is added to the area of the triangle.