The area of pentagon ABCDE is equal to the sum of the area of square ABCE and the area of triangle CDE. The area of ABCE is is 4 square units and the area of CDE is 0.5 x 2 x 2 x sin 60^osquare units, that is √3 square units.

Therefore area of ABCDE = 4 + √3 square units.

Clearly O, the centre of circle G, lies on the axis of symmetry of pentagon ABCDE.

Let the perpendicular distance of O from AB be h.

Then as all of OA, OB and OD are radii of circle G, we can equate expressions for OA and OD in terms of h.

OA = √(1 + h²) and OD = √3 + 2 – h, so

√(1 + h²) = √3 + 2 – h

Squaring both sides,

1 + h² = 3 + 4 + h² + 4√3 – 2h√3 –4h

Simplifying,

2h√3 + 4h = 6 + 4√3

h√3 + 2h = 3 + 2√3

h(2 + √3) = 3 + 2√3

h = (3 + 2√3)/(2 + √3) = [(3 + 2√3) (2 - √3)]/[(2 + √3) (2 - √3)] = (6 + 4√3 - 6 - 3√3)/1 = √3

Therefore the radius of circle G is √3 + 2 – h =√3 + 2 – √3 = 2 units.

The area of circle G outside pentagon ABCDE is therefore

π x 2² – (4 + √3) = 4( π – 1) – √3 square units.

Challenge 307: Outside the Pentagon