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Question Number 193256 by cherokeesay last updated on 08/Jun/23

Answered by som(math1967) last updated on 09/Jun/23

Commented by som(math1967) last updated on 09/Jun/23

$$DE=\sqrt{5^2-3^2}=4cm$$$$AD=5cm,AB=AF+FB=5cm+5cm=10cm$$$$let\mathrm{\angle }ABC=\theta \therefore \mathrm{\angle }DAB=180-\theta$$$$purpleregionarea$$$$=Arofsemicircle+Ar.ofABCD$$$$-Ar.ofsectorADF-ArofsectorBCF$$$$=\frac{1}{2}\pi \times 5^2+10\times 4-\frac{180-\theta }{360}\times \pi \times 5^2$$$$-\frac{\theta }{360}\times \pi \times 5^2$$$$=\frac{25\pi }{2}+40-25\pi \left(\frac{180-\theta +\theta }{360}\right)$$$$=40{cm}^2$$

Commented by cherokeesay last updated on 09/Jun/23

$$thankyou!$$

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