Yes, that's a perfectly good method. If we let s be the length of AB then, since this is an equilateral triangle, BC also has length s and the length of the segment from B to the foot of the orthogonal is s/2. By the Pythagorean theorem, s^2= (s/2)^2+ 9. Solve that for s.
For (b) use the fact that all angles in an equilateral triangle are \(\displaystyle \pi/3\) radians. 1/6 of the \(\displaystyle 2\pi\) of an entire circle so the area of the badge is 1/6 of the area of the area of a circle with radius 3 cm.
(c) is simply wrong! The outer, circular, edge is 1/6 of the circumference of the circle of radius 3 cm. That is \(\displaystyle (1/6)(2\pi (3))= (1/6)(6\pi)= \pi\) cm. The two straight sides, AB and AC, each have
For length 3 cm so the perimeter of the badge is \(\displaystyle \pi+ 6\) cm.
I have no idea where that "\(\displaystyle \frac{2\sqrt{3}}{3}\) is from!
