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 9 cm. 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 9 x 9 
≈ 20.25π cm²

Area of the big square
= ¹₂ x 9 x 9
= 40.5 cm²

Area of the small quadrant
= 20.25π ÷ 2 
≈ 10.125π cm²

Area of the triangle FGL
= 40.5 ÷ 2 
= 20.25 cm²

Area of the shaded part
= (20.25π - 40.5) + (10.125π - 20.25)
= 20.25π + 10.125π  - 40.5 - 20.25
= 30.375π - 60.75
≈ 34.7 cm² (Correct to 1 decimal place)

Answer(s): 34.7 cm²