KJW said:

ChrispenEvan said:

here’s me thinking maths was unambiguous and then we get this dog’s breakfast. disappointed doesn’t cover it.

The maths was not ambiguous. I just made a silly mistake that in hindsight isn’t worth explaining. What’s particularly bothersome about the mistake was that it refused to reveal itself by my checks. However, my mathematical intuition should’ve told me that the formula was wrong. It just didn’t press hard enough.

dv said:

To generalise

The base b of the rectangle would be 6 cosθ
The height of the triangle would be 6 sinθ
Pythag:
r^2 = h^2 + (b-r)^2
r^2 = 36 (sinθ)^2 + (6cosθ – r)^2
r^2 = 36 (sinθ)^2 + 36 (cosθ)^2 + – 12cosθr + r^2
36 (sinθ^2 + cosθ^2) = 12cosθr
36 = 12cosθr
r = 3/cosθ

A = r b = 3/cosθ * 6 cosθ = 18

Actually, there is a much simpler approach:

Let D be the diameter of the semicircle, and B be the length of the base of the rectangle. Then:

6/D = B/6 = cosθ (Note that the two triangles (one shown, the other implied) are similar)

Multiply by 6D:

DB = 36

Area = ½ DB = 18

No Pythagoras’s theorem or trigonometry required, just the recognition and properties of similar triangles, noting that the angle subtended by a diameter at any point on the circle is a right angle (Thales’s theorem).

Nice one :)

Shows it works for θ > 45 degrees as well.