Another one to calculate area of the shadow

Saw on TikTok that this is a high-school problem in New Zealand — calculate the area of the shaded region. It looked like it should yield to a bit of trigonometry. The blue lines below are helper lines I added.

Shadow area problem setup

To get the shaded region, area b, we can take the area of the half-circle minus areas a and c.

Decomposing the shadow

Area a is easy. It equals the circle area minus triangle ADC, divided by 2:

S_a = \frac{(\pi - 2) r^2}{2}

Area a reasoning

To get area c, we use area of DFC minus area e plus area f. Because triangle DAF is similar to EFC and AF = 2·EF, we know area e = 4·f.

Area c reasoning

To find area e, take the circular sector ADF minus triangle ADF. The angle ∠DAE equals arctan(0.5), so the sector area ADF is (2r)² · π · 2·arctan(0.5) / (2π), which simplifies to 4·arctan(0.5)·r².

\angle DAE = \arctan(0.5)

S_{\frown DAF} = 4 \arctan(0.5)\, r^2

For triangle DFA, the area is DG · AG, where DG = 2r·sin(arctan(0.5)) and AG = 2r·cos(arctan(0.5)):

S_{\triangle DAF} = 4 r^2 \sin\bigl(\arctan(0.5)\bigr) \cos\bigl(\arctan(0.5)\bigr)

So area e:

S_e = 4 r^2\Bigl(\arctan(0.5) - \sin\bigl(\arctan(0.5)\bigr) \cos\bigl(\arctan(0.5)\bigr)\Bigr)

And since e = 4·f:

S_f = r^2\Bigl(\arctan(0.5) - \sin\bigl(\arctan(0.5)\bigr) \cos\bigl(\arctan(0.5)\bigr)\Bigr)

The area of triangle DFC equals DF · CF / 2. By similar triangles, CF = DG = GF, so:

S_{\triangle DFC} = 4 r^2 \sin^2\bigl(\arctan(0.5)\bigr)

Now area c = DFC − e + f:

S_c = 4 r^2 \sin^2\bigl(\arctan(0.5)\bigr) - 3 r^2 \Bigl(\arctan(0.5) - \sin\bigl(\arctan(0.5)\bigr) \cos\bigl(\arctan(0.5)\bigr)\Bigr)

Finally, b = half-circle − a − c:

S_b = \frac{\pi r^2}{2} - \frac{(\pi - 2) r^2}{2} - \Biggl(4 r^2 \sin^2\bigl(\arctan(0.5)\bigr) - 3 r^2 \Bigl(\arctan(0.5) - \sin\bigl(\arctan(0.5)\bigr) \cos\bigl(\arctan(0.5)\bigr)\Bigr)\Biggr)

Simplified:

S_b = r^2 \Biggl(1 - 4 \sin^2\bigl(\arctan(0.5)\bigr) + 3 \Bigl(\arctan(0.5) - \sin\bigl(\arctan(0.5)\bigr) \cos\bigl(\arctan(0.5)\bigr)\Bigr)\Biggr)

Which comes out to:

S_b \approx 0.394\, r^2