The figure is made up of a big quadrant FHK a small quadrant FGL and a square GJLF. The radius of the big quadrant FHK is 13 m. The area of the big quadrant is twice the area of the small quadrant FGL. 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 13 x 13 
≈ 42.25π m²

Area of the big square
= ¹₂ x 13 x 13
= 84.5 m²

Area of the small quadrant
= 42.25π ÷ 2 
≈ 21.125π m²

Area of the triangle FGL
= 84.5 ÷ 2 
= 42.25 m²

Area of the shaded part
= (42.25π - 84.5) + (21.125π - 42.25)
= 42.25π + 21.125π  - 84.5 - 42.25
= 63.375π - 126.75
≈ 72.3 m² (Correct to 1 decimal place)

Answer(s): 72.3 m²