IB Mathematics SL 3.4 length of an arc area of a sector AI SL Paper 1- Exam Style Questions- New Syllabus
Question
The following diagram shows a semicircle with centre \(O\) and diameter \(PQ\). A rectangle \(OABC\) is also shown, such that \(\,AB=8\,\) and \(\,OA=5\,\).
Find the length of the arc \(BQ\).
▶️Answer/Explanation
Markscheme (with detailed working)
Key facts from the diagram
Since \(O\) is the centre of the semicircle and \(B\) lies on the circle, \(\,OB\) is a radius. In rectangle \(OABC\), the sides are \(|OA|=5\) and \(|AB|=8\), so by Pythagoras \[ OB \;=\; \sqrt{OA^2+AB^2} \;=\; \sqrt{5^2+8^2} \;=\; \sqrt{89}. \] Thus the radius is \(r=\sqrt{89}\). A1
Method 1 (find the central angle \(\angle BOQ\) then arc length)
From the right triangle with legs \(5\) (along \(OA\)) and \(8\) (along \(AB\)), \[ \angle BOA=\arctan\!\left(\frac{8}{5}\right). \] Since \(\angle AOQ=90^\circ\) (radii along the diameter are perpendicular to \(OA\)), the central angle that subtends arc \(BQ\) is \[ \angle BOQ = 90^\circ + \arctan\!\left(\frac{8}{5}\right) \;=\; \frac{\pi}{2} + \arctan\!\left(\frac{8}{5}\right)\ \text{(radians)}. \] Hence the arc length \[ \text{arc }BQ = r\cdot \angle BOQ \;=\; \sqrt{89}\left(\frac{\pi}{2}+\arctan\!\frac{8}{5}\right). \] Numerically, \[ \sqrt{89}\approx 9.43398,\quad \frac{\pi}{2}+\arctan\!\frac{8}{5}\approx 2.58299, \] so \[ \boxed{\text{arc }BQ \approx 24.4}\ (\text{units}). \] A1 M1 A1
Method 2 (subtract a small arc from the semicircle)
The semicircle arc \(PQ\) has length \(\tfrac{1}{2}(2\pi r)=\pi r\). The small central angle \(\angle BOP=\arctan\!\left(\tfrac{5}{8}\right)\), so the small arc \(BP\) has length \(r\,\arctan\!\left(\tfrac{5}{8}\right)\). Therefore \[ \text{arc }BQ = \underbrace{\pi r}_{\text{semicircle}} \;-\; \underbrace{r\,\arctan\!\left(\tfrac{5}{8}\right)}_{\text{arc }BP} \;=\; \sqrt{89}\!\left(\pi-\arctan\!\frac{5}{8}\right) \;\approx\; \boxed{24.4}\ (\text{units}). \] A1 M1
Total Marks: 5
Notes: Award A1 for identifying \(OB\) as a radius and computing \(r=\sqrt{89}\). Full credit for either method with correct angle and arc-length calculation.
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