Mathematics GCSE Question - The Dog Chain Problem?

Refer to the diagram below.
O is the point where the dog is chained. Then, arc PQR is the farthest point that the dog can reach.
The shaded area OAPQRD is the area that the dog about to access.

In right-angled ΔOAP:
cos∠AOP = 50/100
Then, ∠AOP = 60°

Also sin60° = AP/(100 ft)
Then, AP = 100 sin60° ft = 100 × (√3)/2 ft = 50√3 ft

Area of ΔAOP = (1/2) × OA × AP = (1/2) × 50 × 50√3 ft² = 1250√3 ft²

ΔOAP ≅ ΔODR (RHS)
Hence, ∠DOR = ∠AOP = 60°
also, Area of ΔDOP = Area of ΔAOP = 1250√3 ft²

∠POA = 180° - (∠DOR + ∠AOP) (adj. ∠s on a st. line)
∠POA = 180° - (60° + 60°) = 60°

Area of sector OPR = π × 100² × (60/360) ft² = 5000π/3 ft²

Total area that the dog can access
= (Area of sector OPR) + (Area of ΔAOP) + (Area of ΔDOR)
= (5000π/3) + (1250√3) + (1250√3) ft²
= (5000π/3) + (2500√3) ft²
= 2500(2π/3) + (2500√3) ft²
= 2500[(2π/3) + √3] ft²