IB Mathematics SL 3.4 length of an arc area of a sector AI SL Paper 1- Exam Style Questions- New Syllabus
Question
A garden includes a small lawn. The lawn is enclosed by an arc AB of a circle with centre O and radius 6 m, such that \( \angle AOB = 135^\circ \). The straight border of the lawn is defined by chord [AB].
The lawn is shown as the shaded region between arc AB and chord AB.
(a) A footpath is to be laid around the curved side of the lawn. Determine the length of the footpath. [3]
(b) Determine the area of the lawn. [4]
▶️ Answer/Explanation
Markscheme
(a)
The footpath follows the curved side, arc AB, of the circle with radius 6 m and central angle \( \angle AOB = 135^\circ \).
Arc length = \( \dfrac{\theta}{360^\circ} \times 2\pi r \), where \(\theta = 135^\circ, r = 6 m.\)
Calculate: \( \dfrac{135^\circ}{360^\circ} = \dfrac{3}{8} \), circumference = \( 2\pi \times 6 = 12\pi \),
so arc length = \( \dfrac{3}{8} \times 12\pi = \dfrac{36\pi}{8} = \dfrac{9\pi}{2} \).
Using \( \pi \approx 3.14159 \), \( \dfrac{9 \times 3.14159}{2} \approx 14.13715 \) m.
Alternatively, in radians: \( 135^\circ = 135 \times \dfrac{\pi}{180} = \dfrac{3\pi}{4} \),
arc length = \( r \times \theta = 6 \times \dfrac{3\pi}{4} = \dfrac{18\pi}{4} = \dfrac{9\pi}{2} \approx 14.13715 \) m,
rounded to 14.1 m. M1 A1 A1
[3 marks]
The footpath follows the curved side, arc AB, of the circle with radius 6 m and central angle \( \angle AOB = 135^\circ \).
Arc length = \( \dfrac{\theta}{360^\circ} \times 2\pi r \), where \(\theta = 135^\circ, r = 6 m.\)
Calculate: \( \dfrac{135^\circ}{360^\circ} = \dfrac{3}{8} \), circumference = \( 2\pi \times 6 = 12\pi \),
so arc length = \( \dfrac{3}{8} \times 12\pi = \dfrac{36\pi}{8} = \dfrac{9\pi}{2} \).
Using \( \pi \approx 3.14159 \), \( \dfrac{9 \times 3.14159}{2} \approx 14.13715 \) m.
Alternatively, in radians: \( 135^\circ = 135 \times \dfrac{\pi}{180} = \dfrac{3\pi}{4} \),
arc length = \( r \times \theta = 6 \times \dfrac{3\pi}{4} = \dfrac{18\pi}{4} = \dfrac{9\pi}{2} \approx 14.13715 \) m,
rounded to 14.1 m. M1 A1 A1
[3 marks]
(b)
The lawn is the shaded region between arc AB and chord AB, a circular segment.
Area = sector area OAB minus triangle OAB.
Sector area: \( \dfrac{\theta}{360^\circ} \times \pi r^2 = \dfrac{135^\circ}{360^\circ} \times \pi \times 6^2 = \dfrac{3}{8} \times 36\pi = \dfrac{108\pi}{8} = \dfrac{27\pi}{2} \),
\( \dfrac{27\pi}{2} \approx 42.4115 \) m². M1
Triangle OAB: OA = OB = 6 m, \( \angle AOB = 135^\circ \),
area = \( \dfrac{1}{2} \times 6 \times 6 \times \sin 135^\circ = 18 \times \dfrac{\sqrt{2}}{2} = 9\sqrt{2} \approx 12.7279 \) m². M1
Segment area: \( \dfrac{27\pi}{2} – 9\sqrt{2} \approx 42.4115 – 12.7279 = 29.6836 \) m²,
rounded to 29.7 m². A1 A1
[4 marks]
The lawn is the shaded region between arc AB and chord AB, a circular segment.
Area = sector area OAB minus triangle OAB.
Sector area: \( \dfrac{\theta}{360^\circ} \times \pi r^2 = \dfrac{135^\circ}{360^\circ} \times \pi \times 6^2 = \dfrac{3}{8} \times 36\pi = \dfrac{108\pi}{8} = \dfrac{27\pi}{2} \),
\( \dfrac{27\pi}{2} \approx 42.4115 \) m². M1
Triangle OAB: OA = OB = 6 m, \( \angle AOB = 135^\circ \),
area = \( \dfrac{1}{2} \times 6 \times 6 \times \sin 135^\circ = 18 \times \dfrac{\sqrt{2}}{2} = 9\sqrt{2} \approx 12.7279 \) m². M1
Segment area: \( \dfrac{27\pi}{2} – 9\sqrt{2} \approx 42.4115 – 12.7279 = 29.6836 \) m²,
rounded to 29.7 m². A1 A1
[4 marks]
Total Marks: 7