下图由一个扇形(四分之一圆)和一个圆相互重叠组成。该扇形在点 F 和 G 处与圆相切。
圆的圆心为 O,直径为 24 cm。求阴影部分的面积。
(取 π=3.14\pi = 3.14)

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完整解法

15

1. Find the area of the circle and the quadrant.

Radius of the circle, r=24÷2=12 cmr = 24 \div 2 = 12\text{ cm}.
Area of the circle =πr2=3.14×122=452.16 cm2= \pi r^2 = 3.14 \times 12^2 = 452.16\text{ cm}^2.
Radius of the quadrant, R=24 cmR = 24\text{ cm}.
Area of the quadrant =14πR2=14×3.14×242=452.16 cm2= \frac{1}{4} \pi R^2 = \frac{1}{4} \times 3.14 \times 24^2 = 452.16\text{ cm}^2.

2. Determine the area of the overlapping region.

The straight edges of the quadrant form a 90∘90^\circ angle at G. Since the angle in a semicircle is 90∘90^\circ, these edges pass exactly through the ends of the horizontal diameter of the circle.
The overlapping region consists of the upper semicircle and the triangle formed by the horizontal diameter and point G.
Area of upper semicircle =452.16÷2=226.08 cm2= 452.16 \div 2 = 226.08\text{ cm}^2.
Area of triangle =12×base×height=12×24×12=144 cm2= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 24 \times 12 = 144\text{ cm}^2.
Total overlap area =226.08+144=370.08 cm2= 226.08 + 144 = 370.08\text{ cm}^2.

3. Calculate the total area of the shaded parts.

The shaded parts are the regions in the quadrant outside the circle, and the regions in the circle outside the quadrant.
Shaded area =(Area of quadrant−Overlap)+(Area of circle−Overlap)= (\text{Area of quadrant} - \text{Overlap}) + (\text{Area of circle} - \text{Overlap})
Shaded area =(452.16−370.08)+(452.16−370.08)=82.08+82.08=164.16 cm2= (452.16 - 370.08) + (452.16 - 370.08) = 82.08 + 82.08 = 164.16\text{ cm}^2.

答案

164.16 cm2164.16\text{ cm}^2

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