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  • Find the difference of the areas of two segments of a circle formed by a chord of length 5 cm subtending an angle of 90° at the centre.

Find the difference of the areas of two segments of a circle formed by a chord of length 5 cm subtending an angle of 90° at the centre.

Answers (1)

Answer [32.16cm2]

Solution

            

By using Pythagoras in \triangleABC
   \\(AB)^{2}+(BC)^{2}=(AC)^{2}\\ r^{2}+r^{2}=25\\ 2r^{2}=25\\ r=\frac{5}{\sqrt{2}}cm

 Area of circle =\pi r^{2}=\frac{22}{7} \times \frac{5}{\sqrt{2}}\times\frac{5}{\sqrt{2}}=39.28cm^{2}

Area of sector =\frac{\pi r^{2} \theta}{360}
        \\=3.14 \times \frac{25}{2}\times\frac{90}{360}\\ =9.81cm^{2}

Area of \triangle ABC=\frac{1}{2}\times base \times height

                         \\=\frac{1}{2}\times \frac{5}{\sqrt{2}} \times \frac{5}{\sqrt{2}}\\ =\frac{25}{4} =6.25cm^{2}

Area of minor segment = Area of sector – Area of DABC

                        =9.81-6.25

                        =3.56cm2

Area of major segment = Area of circle – Area of minor segment

                        =39.28-3.56=35.72cm2

Required difference = Area of major segment – Area of minor segment

                             =35.72-3.56=32.16cm2

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