CAIE A-Level Mathematics AS 1.4 Circular Measure Questions

$k=\frac{2}{3}$
Marking guidance:

Allow $\mathrm{ACB}=\frac{2 \pi}{3}$.

Perimeter of shaded area $=2 \pi r$

Major sector $O A B=\frac{1}{2} r^{2} \times \frac{4 \pi}{3}$

*M1

Expect $\frac{2}{3} \pi r^{2}$. Finds area of any relevant sector or
triangle. Can be embedded in segment formula.

One or both segments $=[2] \times\left(\frac{1}{2} r^{2} \times \frac{\pi}{3}-\frac{1}{2} r^{2} \sin \frac{\pi}{3}\right)$

*M1

$=[2]\left(r^{2} \frac{\pi}{6}-r^{2} \frac{\sqrt{3}}{4}\right)$
Shaded area $=\frac{2}{3} \pi r^{2}-2\left(\frac{1}{6} \pi r^{2}-\frac{r^{2} \sqrt{3}}{4}\right)$$=\frac{\pi r^{2}}{3}+\frac{r^{2} \sqrt{3}}{2}$
6(c)

Alternative method for Question 6(c)

Sector CAOB $=[2] \times \frac{1}{2} r^{2}$ their $\frac{1}{3} \pi$

*M1

Expect [2] $\times \frac{1}{6} \pi r^{2}$.
Can be embedded in segment formula.

One or both segments $=[2] \times\left(\frac{1}{2} r^{2} \times \frac{\pi}{3}-\frac{1}{2} r^{2} \sin \frac{\pi}{3}\right)$

*M1

$=[2]\left(r^{2} \frac{\pi}{6}-r^{2} \frac{\sqrt{3}}{4}\right)$
Shaded area $=\pi r^{2}-\left\{\frac{1}{3} \pi r^{2}+2\left(r^{2} \frac{\pi}{6}-r^{2} \frac{\sqrt{3}}{4}\right)\right\}$$=\frac{\pi r^{2}}{3}+\frac{r^{2} \sqrt{3}}{2}$
6(c)

Alternative method for Question 6(c)

Area of rhombus AOBC $=[2] \times \frac{1}{2} r^{2} \sin \frac{\pi}{3}$
Expect [2] $\times \frac{\sqrt{3}}{4}$.
Can be embedded in segment formula.

One or both segments $=[2] \times\left(\frac{1}{2} r^{2} \times \frac{\pi}{3}-\frac{1}{2} r^{2} \sin \frac{\pi}{3}\right)$$=[2]\left(r^{2} \frac{\pi}{6}-r^{2} \frac{\sqrt{3}}{4}\right)$
Shaded area $=\pi r^{2}-\left\{\frac{\sqrt{3}}{2} r^{2}-4\left(r^{2} \frac{\pi}{6}-r^{2} \frac{\sqrt{3}}{4}\right)\right\}$$=\frac{\pi r^{2}}{3}+\frac{r^{2} \sqrt{3}}{2}$

[By symmetry] $[6 \times P \hat{A} Q=2 \pi],[P \hat{A} Q=] 2 \pi \div 6$,
Explaining that there are six sectors around the diagram that make up a complete circle.
AG
Alternative method for Question 12(a)
Using area or circumference of circle centre $A \div 6$$\frac{400 \pi}{6}$ or $\frac{40 \pi}{6}$
Justification for dividing by 6 followed by comparison with the sector area or arc length.
AG
Alternative method for Question 12(a)
Explain why $\triangle P A Q$ is an equilateral triangle
Assumption of this scores M0
Using $\triangle P A Q$ is an equilateral triangle $\therefore P \hat{A} Q=\frac{\pi}{3}$
AG
Alternative method for Question 12(a)
Using the internal angle of a regular hexagon $=\frac{2 \pi}{3}$
Or $F \hat{A} O+O \hat{A} B=\frac{2 \pi}{3}$, equilateral triangles
$P \hat{A} Q=2 \pi-\left(\frac{\pi}{2}+\frac{2 \pi}{3}+\frac{\pi}{2}\right)=\frac{\pi}{3}$
AG
12(a)
Alternative method for Question 12(a)
$\operatorname{Sin} \theta=\frac{20}{40}$, with $\theta$ clearly identified
$\theta=\frac{\pi}{6}, 2 \theta=\frac{\pi}{3}=F \hat{A} O$ and by similar triangles $=P \hat{A} Q$
AG

Each straight section of rope has length 40 cm
SOI
Each curved section round each pipe has length $r \theta=20 \times \frac{\pi}{3}$
*M1
Use of $r \theta$ with r=20 and $\theta$ in radians
Total length $=6 \times(($ their 40$)+k \pi)$$6 \times$ (their straight section + their curved section).
Their curved section must be from acceptable use of $r \theta-$ this
could now be numeric.
$240+40 \pi$ or 366 (AWRT) (cm)
Or directly: ( $6 \times$ diameter) + circumference

[Triangle area = ] $\frac{1}{2} \times 40 \times 40 \times \sin \left(\frac{\pi}{3}\right)$ or $\frac{1}{2} \times 40 \times 20 \sqrt{3}$ or
$400 \sqrt{3}$ or 693 (AWRT)
[Total area of hexagon $=6 \times 400 \sqrt{3}=] 2400 \sqrt{3}$
Condone $4800 \frac{\sqrt{3}}{2}$
Alternative method for Question 12(c)
[Trapezium area = ] $\frac{1}{2} \times(40+80) \times 40 \sin \left(\frac{\pi}{3}\right)$ or $1200 \sqrt{3}$ or 2080
(AWRT)
[Total area of hexagon $=2 \times 1200 \sqrt{3}=$ ] $2400 \sqrt{3}$
Condone $4800 \frac{\sqrt{ } 3}{2}$
Alternative method for Question 12(c)
Area of triangle $A B C=400 \sqrt{3}$ or 693 (AWRT) or
$4 \times$ Area of half of triangle $A B C=4 \times 200 \sqrt{3}$ or 1390 (AWRT)
or Area of rectangle $A B D E=1600 \sqrt{3}$ or 2770 (AWRT)
[Total area of hexagon $=2 \times 400 \sqrt{3}+1600 \sqrt{3}=] 2400 \sqrt{3}$
Or $[=4 \times 200 \sqrt{3}+1600=] 2400 \sqrt{3}$
Condone $4800 \frac{\sqrt{ } 3}{2}$
If B0B0, SC B1 can be scored for sight of 4160 (AWRT) as final answer.

Each rectangle area $=40 \times 20(=800)$
SOI, e.g. by sight of 4800
Each sector area $=\frac{1}{2} r^{2} \theta=\frac{1}{2} \times 20^{2} \times \frac{\pi}{3}\left[=\frac{200 \pi}{3}\right]$
SOI.
Total area $=2400 \sqrt{3}+4800+400 \pi$ or $10200\left(\mathrm{~cm}^{2}\right)$ (AWRT)
Or directly: part (c) + 6800+ area circle radius 20 .