The figure is made up of a big quadrant BDF a small quadrant BCG and a square CEGB. The radius of the big quadrant BDF is 23 m. The area of the big quadrant is twice the area of the small quadrant BCG. Using the calculator value of π, find the area of the shaded parts, correct to 1 decimal place.
Area of the big quadrant
= ¹₄ x π x 23 x 23 
≈ 132.25π m²

Area of the big square
= ¹₂ x 23 x 23
= 264.5 m²

Area of the small quadrant
= 132.25π ÷ 2 
≈ 66.125π m²

Area of the triangle BCG
= 264.5 ÷ 2 
= 132.25 m²

Area of the shaded part
= (132.25π - 264.5) + (66.125π - 132.25)
= 132.25π + 66.125π  - 264.5 - 132.25
= 198.375π - 396.75
≈ 226.5 m² (Correct to 1 decimal place)

Answer(s): 226.5 m²