Question

Find the area of an equilateral triangle with radius 2 square root 3

Answer 1

Draw the triangle and divide it into six congruent right triangles by drawing the three altitudes. The area of one of the triangles is \( [\text{opposite leg of 30°}][\text{adjacent leg of 30°}]/2 = (2\sqrt3 \cdot \sin(30\text{°}))(2\sqrt3 \cdot \cos(30\text{°}))/2\\ = (2\sqrt3\cdot\frac{\sqrt3}2)(2\sqrt3\cdot\frac{1}{2})/2 = 3 \sqrt3 / 2 \)

Answer 2

To get the area of the big triangle, multiply by 6. Final result: \(9\sqrt3\).