Home / IB Mathematics AHL 3.7 area of sector, length of arc AI HL Paper 1- Exam Style Questions

IB Mathematics AHL 3.7 area of sector, length of arc AI HL Paper 1- Exam Style Questions

IB Mathematics AHL 3.7 area of sector, length of arc AI HL 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 \( OA = 5 \).

Diagram of a semicircle with centre O, diameter PQ, and rectangle OABC

Find the length of the arc \( BQ \).

▶️ Answer/Explanation
Markscheme

Recognition that OB is a radius (M1)
\( r = \sqrt{5^2 + 8^2} = \sqrt{89} \) (A1)
EITHER (finding angle BOQ)
Correct calculation for finding BOA: \( \angle BOA = \arctan\left(\frac{8}{5}\right) \) or \( \tan \angle BOA = \frac{8}{5} \) (A1)
Expressing BOQ as 90° + BOA: \( \angle BOQ = 90^\circ + \arctan\left(\frac{8}{5}\right) \) or \( \angle BOQ = \frac{\pi}{2} + \arctan\left(\frac{8}{5}\right) \) (A1)
\( \angle BOQ \approx 147.994^\circ \) or 2.58299 radians (A1)
Substituting their radius and angle BOQ correctly into arc length formula:
\( \text{arc BQ} = \frac{90^\circ + \arctan\left(\frac{8}{5}\right)}{360^\circ} \times 2\pi \sqrt{89} \) or
\( \text{arc BQ} = \left( \frac{\pi}{2} + \arctan\left(\frac{8}{5}\right) \right) \times \sqrt{89} \) (M1)
\( \text{arc BQ} \approx 24.4 \) m (24.3679…) (A1)
OR (finding angle BOP)
Correct calculation for finding angle BOP: \( \angle BOP = \arctan\left(\frac{5}{8}\right) \) or \( \tan \angle BOP = \frac{5}{8} \) (A1)
Substituting their radius and BOP correctly into arc length formula:
\( \text{arc BP} = \frac{\arctan\left(\frac{5}{8}\right)}{360^\circ} \times 2\pi \sqrt{89} \) (M1)
Subtracting their arc BP from arc PQ:
\( \text{arc BQ} = \pi \sqrt{89} – \frac{\arctan\left(\frac{5}{8}\right)}{360^\circ} \times 2\pi \sqrt{89} \) (M1)
\( \text{arc BQ} \approx 24.4 \) m (24.3679…) (A1)
Result: 24.4 m (24.3679…) [5]

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