My Version Of The Deathly Hallows Symbol

$\large A = \pi r^2 = 9 \pi\\ \large \therefore r = 3$

All the angles are $60^{\circ}.$

Let, $BD = x, AB = 2x$ and given $AD= 6cm$

By Pythagorean theorem :

$\large AD^2 + BD^2 = AB^2$

$\large 36 + x^2 = (2x)^2\\ \large \implies 36+x^2 = 4x^2\\ \large 3x^2 = 36 \implies x^2 = \frac{36}{3} = 12$

$\large BD = x = \sqrt{12} = 2\sqrt{3}$

$\large \therefore AB = BC = 4\sqrt{3}$

$\large Area = \frac{1}{2} base \times height\\ \large \implies A = \frac{1}{2} \times 6 \times 4\sqrt{3}\\ \large \implies A = 12\sqrt{3} = 12 \times 1.73205 = \boxed{20.78460} \text{square units}.$

We can see also a special 30-60-90 triangle here.

So the height 9π=πr² Radius of a circle=3cm Doubling up will be the height of a triangle. Cutting the triangle into half will get a special right triangle which satisfies the formula

Longer leg= (1/2)hypotenuse √3 6=1/2 h*√3 6x2/√3=hypotenuse/side of the triangle

12/√3≈6.928cm= Sides of the triangle

So, Area of a triangle is 1/2(bh) A=1/2(6.928*6) A=1/2(41.568) A=20.784cm²

An altitude of an isosceles triangle bisects it into two congruent triangles. Therefore, the two triangles formed are 30-60-90 triangles. The radius is 3 (You can find it by doing $\sqrt { \frac { 9\pi }{ \pi } }$ ), so the height of the 30-60-90 triangles are 6. Tan(60) is $\frac { \sqrt { 3 } }{ 3 }$ , so the side length of the equilateral triangle is $4\sqrt { 3 }$ . Therefore, the area is $12\sqrt { 3 }$ $\approx$ 20.78460

Area of a circle is $\pi \cdot r^{2}$ . We can deduce simply that the radius is $3$ . Now, imagine drawing a line segment from O to $\overline{AB}$ and $\overline{AC}$ , at the point where each segment intersect's the circle. The result would form two small equalateral triangles, with a combined angle of $180$ . With circle theorems, we know that the angle above, $CAB$ is 60. That means each respective angle is 30 degrees. We can solve this by using two 30 60 90 triangles added together: $12 \cdot \sqrt{3} \approx \boxed{20.785}$ .