题目
Problem
Figure 1 shows a sketch of the entrance to a tunnel.
Figure 1
The shape of the entrance consists of a sector BCDF, of a circle centre F, joined to two congruent (identical) triangles ABF and EDF.
Given that
AFE is a straight line
AF=FE=6.4 m
FB=FD=6.2 m
angle BFD=2.275 radians
(a) Show that angle AFB=0.433 radians to 3 decimal places.
(1)
(b) Find the perimeter of the entrance to the tunnel, ABCDEFA, in metres, to one decimal place.
(4)
(c) Find the cross-sectional area of the entrance to the tunnel, ABCDEFA, in m2, to one decimal place.
(4)
解答
(a)
解法一
思路
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AFE 是直线,所以围绕点 F 的上半部分总角度是 π。中间扇形角是 2.275,两侧三角形全等,所以两侧角相等。
答题过程
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Since AFE is a straight line and the two triangles are congruent,
∠AFB===2π−2.2750.433296…0.433 radians
to 3 decimal places.
(b)
解法一
思路
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周长由底边 AF+FE、两条斜边 AB,DE 和弧长 BCD 组成。弧长用 rθ;斜边 AB 可在三角形 ABF 中用余弦定理求出。
答题过程
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The arc length BCD is
rθ==6.2(2.275)14.105.
Using the cosine rule in triangle ABF,
AB2==6.42+6.22−2(6.4)(6.2)cos(0.433296…)7.367….
So
AB=2.7142….
The perimeter is
2AB+arc BCD+AF+FE==2(2.7142…)+14.105+12.832.333….
Therefore the perimeter is
32.3 m.
(c)
解法一
思路
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总面积等于中间扇形面积加两侧全等三角形面积。扇形面积用 21r2θ,三角形面积用 21absinC。
答题过程
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The area of sector BCDF is
21r2θ==21(6.2)2(2.275)43.7255.
The area of triangle ABF is
21(6.4)(6.2)sin(0.433296…)=8.325….
There are two congruent triangles, so the total area is
43.7255+2(8.325…)=60.376….
Therefore the cross-sectional area is
60.4 m2.