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Geometrical Properties of Circles

Secondary 3 practice

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Geometrical Properties of Circles practice questions

Secondary 3 topics from the O-Level Mathematics 4052 syllabus. Every question below has a full worked solution.

When a student gets a question wrong in Math Amigo, it does not show the answer straight away. It reads the working, identifies the step that went wrong, and asks a question that points at the correction. The worked solution is there once a genuine attempt has been made.

Practise Geometrical Properties of Circles free

What this topic covers

Worked examples

Example 1

A circle with centre OO has radius 3030 cm. Two equal chords ABAB and CDCD each have length 3636 cm. Find the distance from the centre to each chord.
ABMAB = 36?CDCD = 36O30 cmNot drawn to scale
Show the worked solution
  • Equal chords are equidistant from the centreproperty of equal chords\text{Equal chords are equidistant from the centre} \quad \scriptsize\textit{property of equal chords}
  • Let M be the midpoint of AB\text{Let } M \text{ be the midpoint of } AB
  • OM⊥ABperpendicular from centre bisects chordOM \perp AB \quad \scriptsize\textit{perpendicular from centre bisects chord}
  • AM=AB2=362=18AM = \frac{AB}{2} = \frac{36}{2} = 18 cm
  • OA2=OM2+AM2Pythagoras’ theoremOA^2 = OM^2 + AM^2 \quad \scriptsize\textit{Pythagoras' theorem}
  • 302=OM2+18230^2 = OM^2 + 18^2
  • OM2=302−182OM^2 = 30^2 - 18^2
  • OM2=900−324OM^2 = 900 - 324

Answer: 2424

Example 2

PAPA and PBPB are tangents to a circle with centre OO, touching it at AA and BB. Given that ∠APO=44°\angle APO = 44°, find ∠AOB\angle AOB.
?OABPNot drawn to scale
Show the worked solution
  • PO bisects ∠APBthe line to the centre bisects the angle between the tangentsPO \text{ bisects } \angle APB \quad \scriptsize\textit{the line to the centre bisects the angle between the tangents}
  • ∠APB=2×44°\angle APB = 2 \times 44°
  • =88°= 88°
  • ∠OAP=∠OBP=90°tangent⊥radius\angle OAP = \angle OBP = 90° \quad \scriptsize\textit{tangent} \perp \textit{radius}
  • In quadrilateral OAPB:∠AOB=360°−90°−90°−88°\text{In quadrilateral } OAPB: \angle AOB = 360° - 90° - 90° - 88°
  • ∠AOB=92°\angle AOB = 92°

Answer: 9292

Example 3

A, B and C are points on a circle with centre O. AOB is a diameter. ∠BAC = 22°22°. Find ∠BOC.
22°?OABCNot drawn to scale
Show the worked solution
  • ∠BOC=2×∠BAC∠ at centre = 2 × ∠ at circumference, arc BC\angle BOC = 2 \times \angle BAC \quad \scriptsize\angle\text{ at centre = 2 }{\times}\text{ }\angle\text{ at circumference, arc BC}
  • ∠BOC=2×22°\angle BOC = 2 \times 22°
  • =44°= 44°

Answer: 4444

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Which syllabus does this follow?

O-Level Mathematics (4052), for Singapore secondary students. Math Amigo covers Secondary 1 to 4, both O-Level Mathematics and Additional Mathematics.

Will it just show my child the answer?

No. On a wrong answer it looks at the working, points to the step that went wrong, and asks a question that helps the student find the correction. The full worked solution is available after a genuine attempt.

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