CalcGospel 國際數學圖譜
返回

IAL 2023 Oct Q9

A Level / Edexcel / P1

IAL 2023 Oct Paper · Question 9

题目

Problem

Figure 3 shows the plan view of the area being used for a ball-throwing competition.

Figure 3

Competitors must stand within the circle CC and throw a ball as far as possible into the target area, PQRSPQRS, shown shaded in Figure 3.

Given that

  • circle CC has centre OO
  • PP and SS are points on CC
  • OPQRSOOPQRSO is a sector of a circle with centre OO
  • the length of arc PSPS is 0.720.72 m
  • the size of angle POSPOS is 0.60.6 radians

(a) show that OP=1.2OP=1.2 m.

(1)

Given also that

  • the target area, PQRSPQRS, is 9090 m2^2
  • length PQ=xPQ=x metres

(b) show that

5x2+12x−1500=0.\begin{align*} 5x^2+12x-1500=0. \end{align*}
(3)

(c) Hence calculate the total perimeter of the target area, PQRSPQRS, giving your answer to the nearest metre.

(3)

解答

(a)

解法一

思路

展开

弧长公式是 s=rθs=r\theta。这里弧 PSPS 的长度是 0.720.72，圆心角是 0.60.6 radians，所以半径就是 OPOP。

答题过程

展开

Using s=rθs=r\theta,

0.72= OP(0.6)OP= 0.720.6= 1.2.\begin{align*} 0.72=&\,OP(0.6)\\ OP=&\,\frac{0.72}{0.6}\\ =&\,1.2. \end{align*}

Therefore

OP=1.2 m.\begin{align*} OP=1.2\text{ m}. \end{align*}

(b)

解法一

思路

展开

目标区域 PQRSPQRS 是大扇形减去小扇形。小半径是 1.21.2，而 PQ=xPQ=x，所以大半径是 x+1.2x+1.2。

答题过程

展开

The radius of the larger sector is

x+1.2.\begin{align*} x+1.2. \end{align*}

The target area is the larger sector minus the smaller sector:

12(x+1.2)2(0.6)−12(1.2)2(0.6)= 90.\begin{align*} \frac12(x+1.2)^2(0.6)-\frac12(1.2)^2(0.6)=&\,90. \end{align*}

Since 12(0.6)=0.3\frac12(0.6)=0.3,

0.3((x+1.2)2−1.22)= 900.3(x2+2.4x+1.44−1.44)= 900.3x2+0.72x= 90.\begin{align*} 0.3\left((x+1.2)^2-1.2^2\right)=&\,90\\ 0.3\left(x^2+2.4x+1.44-1.44\right)=&\,90\\ 0.3x^2+0.72x=&\,90. \end{align*}

Multiply by 5050:

15x2+36x= 45005x2+12x−1500= 0.\begin{align*} 15x^2+36x=&\,4500\\ 5x^2+12x-1500=&\,0. \end{align*}

(c)

解法一

思路

展开

先解 (b) 的二次方程，取正根作为长度 xx。周长由两条直边 PQPQ、RSRS，内弧 PSPS，以及外弧 QRQR 组成。外弧半径是 x+1.2x+1.2，角度仍是 0.60.6。

答题过程

展开

Solve

5x2+12x−1500=0.\begin{align*} 5x^2+12x-1500=0. \end{align*}

Using the quadratic formula,

x= −12±122−4(5)(−1500)2(5)= −12±3014410= −6±218845= −6±44715.\begin{align*} x =&\,\frac{-12\pm\sqrt{12^2-4(5)(-1500)}}{2(5)}\\ =&\,\frac{-12\pm\sqrt{30144}}{10}\\ =&\,\frac{-6\pm2\sqrt{1884}}{5}\\ =&\,\frac{-6\pm4\sqrt{471}}{5}. \end{align*}

Since xx is a length,

x=−6+44715=16.159….\begin{align*} x=\frac{-6+4\sqrt{471}}{5}=16.159\ldots. \end{align*}

The outer arc QRQR has length

(x+1.2)(0.6).\begin{align*} (x+1.2)(0.6). \end{align*}

So the perimeter of PQRSPQRS is

2x+0.72+0.6(x+1.2)= 2(16.159…)+0.72+0.6(17.359…)= 43.454….\begin{align*} 2x+0.72+0.6(x+1.2) =&\,2(16.159\ldots)+0.72+0.6(17.359\ldots)\\ =&\,43.454\ldots. \end{align*}

Therefore the perimeter is

43 m\begin{align*} 43\text{ m} \end{align*}

to the nearest metre.