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# 'Impossible' GSCE Maths question?

See e.g. The Daily Express.

1. Here’s the basic question. Let’s assume side length is 10, for sake of argument. 2. Draw the overlapping bits (dotted outlines) and an isosceles triangle (red outlined), 3. Work out area of the sector (pink). If you can’t guess that this is one-third of the area of a circle (which seems a bit obvious, with the benefit of hindsight), you need to work out the angle of the right-hand corner of the triangle (which touches the circle in the centre). From 3. below, it is 120 deg, which is a third of a full circle 360 deg, so the pink sector has area = pi x r^2 ÷ 3 = 100/3 pi (about 105, in old money). 4. Focus on the triangle (from 2. above).
Sides are all the same length r = 10, so it is an isosceles triangle and angles = 60 deg.
Fold it in half (middle triangle) and the bottom side is half r = 5. The vertical side is the square root of r^2 – 1/2 r^2 = square root of 3/4 r^2 = square root of 75.
The area of the triangle on the right (or on the left, for that matter) = base x height x 1/2
= 5 x square root of 75 = about 43 in old money.
Angle of the triangle on the right where it touches the centre of the circle = 120 deg. (needed for 3. above)
5. Subtract this from the area of the sector. The pink segment left over thus has area = (100/3 pi) – (5 x square root of 75)
= about 62 in old money. 6. Then subtract four of these segments from the area of a full circle = 100 pi – 4 x ((100/3 pi) – (5 x square root of 75))
= 100 pi – 400/3 pi + 20 x square root of 75
= – 100/3 pi + 20 x square root of 75
= which is about 68.5 in old money.
Looks about right, that’s a bit more than one-fifth of the area of the circle (which is about 314).

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