里昂(Leon)设计了一个如图所示的标志。该标志由两个相同的半圆、一个圆和一个大扇形(四分之一圆)组成。YZ 是圆的直径,O 是圆心。X 是大扇形的圆心,且 XY = XZ。小半圆的半径为 7 cm。 (取 π=227\pi = \tfrac{22}{7})

  1. (a)

    求 YZ 的长度。

  2. (b)

    求三角形 XYZ 的面积。

  3. (c)

    求里昂设计的标志中阴影部分的总面积。

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完整解法

16(a)

1. Determine the diameter of the small semi-circle

Diameter of small semi-circle = 2×7=142 \times 7 = 14 cm

2. Determine the length of YZ

The two small semi-circles span the diameter of the large circle. Thus, YZ=2×14=28YZ = 2 \times 14 = 28 cm.

答案

2828 cm

16(b)

1. Determine the length of YZ from part (a)

From part (a), the diameter of the circle YZYZ is equal to 4×7=28 cm4 \times 7 = 28\text{ cm}.

2. Identify the properties of triangle XYZ

Since XX is the centre of the big quadrant and XY,XZXY, XZ are its radii, ∠YXZ=90∘\angle YXZ = 90^\circ. Thus, triangle XYZXYZ is a right-angled isosceles triangle with hypotenuse YZ=28 cmYZ = 28\text{ cm}.

3. Calculate the area of triangle XYZ

Let the height of triangle XYZXYZ from XX to YZYZ be hh. Since it is a right-angled isosceles triangle, the height to the hypotenuse is half the hypotenuse. So h=282=14 cmh = \frac{28}{2} = 14\text{ cm}.
Area of △XYZ=12×base×height=12×28×14=196 cm2\triangle XYZ = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 28 \times 14 = 196\text{ cm}^2.
Alternatively, XY2+XZ2=YZ2  ⟹  2r2=282=784  ⟹  r2=392XY^2 + XZ^2 = YZ^2 \implies 2r^2 = 28^2 = 784 \implies r^2 = 392.
Area of △XYZ=12×XY×XZ=12r2=12×392=196 cm2\triangle XYZ = \frac{1}{2} \times XY \times XZ = \frac{1}{2} r^2 = \frac{1}{2} \times 392 = 196\text{ cm}^2.

答案

392 cm2392\text{ cm}^2

16(c)

1. Find the area of the two small semi-circles.

Area of two small semi-circles = 2×12×π×r2=π×722 \times \frac{1}{2} \times \pi \times r^2 = \pi \times 7^2
=227×49=154 cm2= \frac{22}{7} \times 49 = 154\text{ cm}^2

2. Find the square of the radius of the big quadrant ($XY^2$).

From the figure, the diameter of the circle YZ=7+7+7+7=28 cmYZ = 7 + 7 + 7 + 7 = 28\text{ cm}.
Since XX is the centre of the quadrant, ∠YXZ=90∘\angle YXZ = 90^\circ and XY=XZXY = XZ.
Using Pythagoras' theorem in △XYZ\triangle XYZ:
XY2+XZ2=YZ2XY^2 + XZ^2 = YZ^2
2XY2=282=7842XY^2 = 28^2 = 784
XY2=392XY^2 = 392

3. Find the area of the big quadrant and the total shaded area.

Area of the big quadrant = 14×π×XY2\frac{1}{4} \times \pi \times XY^2
=14×227×392=308 cm2= \frac{1}{4} \times \frac{22}{7} \times 392 = 308\text{ cm}^2
Total shaded area = 154+308=462 cm2154 + 308 = 462\text{ cm}^2

答案

462 cm2462\text{ cm}^2

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