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IAL 2024 Oct Q5

A Level / Edexcel / P1

IAL 2024 Oct Paper · Question 5

题目

Problem

A plot of land OABOAB is in the shape of a sector of a circle with centre OO.

Given

  • OA=OB=5OA=OB=5 km
  • angle AOB=1.2AOB=1.2 radians

(a) find the perimeter of the plot of land.

(2)

A point PP lies on OBOB such that the line APAP divides the plot of land into two regions R1R_1 and R2R_2 as shown in Figure 2.

Figure 2

Given that

area of R1=3×area of R2\begin{align*} \text{area of }R_1=3\times\text{area of }R_2 \end{align*}

(b) show that the area of R2=3.75 km2R_2=3.75\text{ km}^2.

(3)

(c) Find the length of APAP, giving your answer to the nearest 100100 m.

(4)

解答

(a)

解法一

思路

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扇形周长等于两条半径加弧长。弧长公式是 rθr\theta，其中 θ\theta 要用弧度。

答题过程

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The arc length ABAB is

rθ=5(1.2)=6.\begin{align*} r\theta=5(1.2)=6. \end{align*}

So the perimeter is

5+5+6=16.\begin{align*} 5+5+6=16. \end{align*}

Therefore the perimeter is

16 km.\begin{align*} 16\text{ km}. \end{align*}

(b)

解法一

思路

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先求整个扇形面积。若 R1R_1 的面积是 R2R_2 的 33 倍，那么整个扇形面积就是 44 份，其中 R2R_2 占 11 份。

答题过程

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The area of the sector is

12r2θ= 12(52)(1.2)= 15.\begin{align*} \frac12r^2\theta =&\,\frac12(5^2)(1.2)\\ =&\,15. \end{align*}

Since

area of R1=3×area of R2,\begin{align*} \text{area of }R_1=3\times\text{area of }R_2, \end{align*}

the whole sector is made of 44 equal parts of size area of R2\text{area of }R_2.

Therefore

area of R2= 14(15)= 3.75 km2.\begin{align*} \text{area of }R_2 =&\,\frac14(15)\\ =&\,3.75\text{ km}^2. \end{align*}

This is the required result.

(c)

解法一

思路

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R2R_2 就是三角形 OAPOAP，它的两条边 OAOA 和 OPOP 夹角为 1.21.2。先用三角形面积公式求 OPOP，再在三角形 OAPOAP 中用余弦定理求 APAP。

答题过程

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Since R2R_2 is triangle OAPOAP,

12(5)(OP)sin⁡1.2= 3.75OP= 3.752.5sin⁡1.2OP= 1.6093….\begin{align*} \frac12(5)(OP)\sin 1.2=&\,3.75\\ OP=&\,\frac{3.75}{2.5\sin1.2}\\ OP=&\,1.6093\ldots. \end{align*}

Using the cosine rule in triangle OAPOAP,

AP2= 52+(1.6093…)2−2(5)(1.6093…)cos⁡1.2= 21.758….\begin{align*} AP^2 =&\,5^2+(1.6093\ldots)^2\\ &\quad -2(5)(1.6093\ldots)\cos1.2\\ =&\,21.758\ldots. \end{align*}

So

AP=4.6646… km.\begin{align*} AP=4.6646\ldots\text{ km}. \end{align*}

To the nearest 100100 m,

AP=4.7 km.\begin{align*} AP=4.7\text{ km}. \end{align*}

解法二

思路

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也可以从点 PP 向 OAOA 作垂线。三角形 OAPOAP 的面积可以先给出垂直高度，再用直角三角形求 APAP。这种路线避开了余弦定理，但要小心角的位置。

答题过程

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Let XX be the foot of the perpendicular from PP to OAOA.

Since the area of triangle OAPOAP is 3.753.75,

12(5)(PX)= 3.75PX= 1.5.\begin{align*} \frac12(5)(PX)=&\,3.75\\ PX=&\,1.5. \end{align*}

In the right triangle OPXOPX,

tan⁡1.2= PXOXOX= 1.5tan⁡1.2.\begin{align*} \tan1.2=&\,\frac{PX}{OX}\\ OX=&\,\frac{1.5}{\tan1.2}. \end{align*}

Therefore

AX= OA−OX= 5−1.5tan⁡1.2= 4.4164….\begin{align*} AX =&\,OA-OX\\ =&\,5-\frac{1.5}{\tan1.2}\\ =&\,4.4164\ldots. \end{align*}

Now use Pythagoras in triangle APXAPX:

AP2= AX2+PX2= (4.4164…)2+1.52= 21.758….\begin{align*} AP^2 =&\,AX^2+PX^2\\ =&\,(4.4164\ldots)^2+1.5^2\\ =&\,21.758\ldots. \end{align*}

So

AP=4.6646… km.\begin{align*} AP=4.6646\ldots\text{ km}. \end{align*}

To the nearest 100100 m,

AP=4.7 km.\begin{align*} AP=4.7\text{ km}. \end{align*}