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in a square of length 14 cm 4 quadrants are made inside the circle then find the area of the middle of all the segments |
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Answer» The green SHADED AREA = Area of sector =60360πx2=60360πx2 =16πx2=16πx2 The blue shaded area = Area of segment = Area of sector - Area of EQUILATERAL triangle =16πx2−3–√4x2=16πx2−34x2 A+E+D+H=blue+green=13πx2−3–√4x2A+E+D+H=blue+green=13πx2−34x2 A+E+D+H=13πx2−3–√4x2A+E+D+H=13πx2−34x2 A+E+D+H+C+G=A+E+D+H+C+G=Area of quarter CIRCLE =14πx2=14πx2 A+E+D+H+C+G=14πx2A+E+D+H+C+G=14πx2 Subtraction the 2 equations, C+G=3–√4x2−112πx2C+G=34x2−112πx2 Notice that, C+G=B+F=I+E=H+D=3–√4x2−112πx2C+G=B+F=I+E=H+D=34x2−112πx2 C+G+B+F+I+E+H+D=3–√x2−13πx2C+G+B+F+I+E+H+D=3x2−13πx2 Subtract this from area of square, A=x2−(3–√x2−13πx2)A=x2−(3x2−13πx2) A=x2−3–√x2+π3x2A=x2−3x2+π3x2 A=(1−3–√+π3)x2A=(1−3+π3)x2 A≈0.315147⋅x2A≈0.315147⋅x2 In your case, x=14x=14 A≈61.76876 cm2A≈61.76876 CM2 |
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