What fraction is shaded?

Question

The dots around the semicircle are equally spaced. What fraction is shaded?

This puzzle is by the amazing Catriona Agg.

Answer 1

This becomes very easy once we add an auxiliary

regular hexagon with its diagonals.

We will show that the largest of the three shaded regions (blue in figure a) is exactly

one third

of the total area

This is obvious because the triangles in b and c have the same area.

Because they have

• the same base and height

or some may prefer because

• they are made from the same two halves

or others still may skip c completely

• because the area of the triangle in b is clearly one sixth of the hexagon.

Easy.

Next we'll show that the blue shaded area and the non-shaded area are

the same

and we'll be done.

First, we move the tiny shaded bit (Figure d) and then we are left with a matching curved bit (Figure e) and again two triangles that share their base and have the same height, half the radius (Figure f).

To summarise

The blue, teal and non shaded regions split the total area evenly. In particular, 2/3 are shaded.

Answer 2

The shaded fraction is

2/3

Step by step explanation:

Answer 3

My answer is

Two-thirds of the semi-circle is shaded

The area of a sector is (area of the segment) - (area of triangle subtended)
Suppose the angle subtended is $\theta$, the segment area
$= (\pi r^2 \times \frac{\theta}{360}) - (r \sin \frac{\theta}{2} \times r \cos \frac{\theta}{2}) $
$= r^2 (\frac{\theta \pi}{360} - \frac{\sin \theta}{2})$

Each dot is spaced at 30°, so

Area of a 30° segment
A1 = $ r^2 ( \frac{30 \pi}{360} - \frac{\sin 30}{2})$

Area of a 60° segment
A2 = $ r^2 ( \frac{60 \pi}{360} - \frac{\sin 60}{2})$

Area of a 120° segment
A4 = $ r^2 ( \frac{120 \pi}{360} - \frac{\sin 120}{2})$

Area of a 150° segment
A5 = $ r^2 ( \frac{150 \pi}{360} - \frac{\sin 150}{2})$

Area of a 180° segment
A6 = $ r^2 ( \frac{180 \pi}{360} - \frac{\sin 180}{2})$

We want the area
A1 + (A4 - A2) + (A6 - A5)

Which is
$r^2( \frac{(30 + 120 - 60 + 180 - 150) \pi}{360} - \frac{(\sin 30 + \sin 120 - \sin 60 + \sin 180 - \sin 150)}{2})$

reducing to

$r^2( \frac{120 \pi}{360} - \frac{\sin 180}{2})$

and

$\frac{\pi r^2}{3}$

which is one third of a full circle.

Answer 4

Had to edit because I hadn't seen the tiny bit at the top at first:

2/3