如图所示,一个四分之一圆 ABC 和一个长方形 CDEF。AC 的长度为 10 cm。EF 和 ED 的长度之和为 14 cm。 (取 π=3.14\pi = 3.14)

  1. (a)

    求四分之一圆 ABC 的面积。

  2. (b)

    求阴影部分 DABF 的周长。

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完整解法

9(a)

1. Identify the radius of the quarter circle

The length of ACAC is given as 10 cm10\text{ cm}. Since ABCABC is a quarter circle with centre AA, the radius r=AC=10 cmr = AC = 10\text{ cm}.

2. Calculate the area of the quarter circle

Area=14×π×r2\text{Area} = \frac{1}{4} \times \pi \times r^2
Area=14×3.14×102\text{Area} = \frac{1}{4} \times 3.14 \times 10^2
Area=14×3.14×100\text{Area} = \frac{1}{4} \times 3.14 \times 100
Area=14×314=78.5 cm2\text{Area} = \frac{1}{4} \times 314 = 78.5\text{ cm}^2

答案

78.5 cm278.5\text{ cm}^2

9(b)

1. Find the length of the diagonal DF and the arc AB.

In rectangle CDEFCDEF, the diagonals are equal in length, so DF=CEDF = CE. Since EE lies on the arc of the quarter circle with center CC, CECE is a radius, meaning DF=10DF = 10 cm.
The length of arc AB=14×2×3.14×10=15.7AB = \frac{1}{4} \times 2 \times 3.14 \times 10 = 15.7 cm.

2. Find the sum of lengths DA and BF, and calculate the total perimeter.

Since CA=CB=10CA = CB = 10 cm, we have DA=10−CDDA = 10 - CD and BF=10−CFBF = 10 - CF. Thus, the sum of their lengths is DA+BF=20−(CD+CF)DA + BF = 20 - (CD + CF).
In rectangle CDEFCDEF, opposite sides are equal, so CD=EFCD = EF and CF=EDCF = ED. Given EF+ED=14EF + ED = 14 cm, we have CD+CF=14CD + CF = 14 cm.
Therefore, DA+BF=20−14=6DA + BF = 20 - 14 = 6 cm.
The perimeter of the shaded part DABF=arc AB+DF+(DA+BF)=15.7+10+6=31.7DABF = \text{arc } AB + DF + (DA + BF) = 15.7 + 10 + 6 = 31.7 cm.

答案

31.731.7 cm

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