Sir Mike Ambrose is the radius of the question.
Area R is;
Area triangle with height 4 cm and base (4tan52.5) cm - Area sector with angle 52.5° and radius 4 cm.
= ½*4(4tan52.5) - (52.5*4*4π)/360
= 8tan52.5 - ⅓(7π)
= ⅓(24tan52.5-7π)
= 3.09542012435 cm²
≈ 3.10 cm² to 2 decimal places.
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