Edexcel GCSE Maths (1MA1), Higher tier · Geometry and measures
Practise Circle theorems. 8 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy3 marks
\(A\), \(B\) and \(C\) are points on the circumference of a circle, centre \(O\). \(AB\) is a diameter of the circle. Diagram NOT accurately drawn
[object Object]
Work out the size of the angle marked \(x\). Give reasons for your answer.[3]
Show the answer and mark scheme
Answer: \(58^\circ\) °
M1 for angle \(ACB = 90^\circ\) or \(180 - 90 - 32\)
A1 for 58
C1 for the angle in a semicircle is 90° (and angles in a triangle add up to 180°)
Worked solution: Angle \(ACB = 90^\circ\) (the angle in a semicircle is 90°). \(x = 180 - 90 - 32 = 58\) (angles in a triangle add up to 180°).
Question 2Medium2 marks
\(A\), \(B\) and \(C\) are points on a circle, centre \(O\). \(B\) is on the major arc \(AC\). Angle \(AOC = 130^\circ\). Maya says, 'Angle \(ABC = 130^\circ\), because angles subtended by the same arc are equal.'
(a) Explain Maya's mistake.[1]
(b) Work out the size of angle \(ABC\).[1]
Show the answer and mark scheme
(a)Answer: Angle \(AOC\) is at the centre and angle \(ABC\) is at the circumference: the angle at the centre is twice the angle at the circumference.
C1 for explaining that angle \(AOC\) is at the centre, so it is twice angle \(ABC\) (angles in the same segment are only equal when both are at the circumference)
Worked solution: The 'same arc' rule is for two angles at the circumference. Here one angle is at the centre, so it is twice the angle at the circumference.
(b)Answer: 65°
B1 for 65
Worked solution: \(130 \div 2 = 65^\circ\).
Question 3Hard4 marks
\(A\), \(B\) and \(C\) are points on the circumference of a circle, centre \(O\). The line \(CO\) is extended to the point \(D\). Diagram NOT accurately drawn
[object Object]
Prove that angle \(AOB\) is twice angle \(ACB\). Give a reason for each stage of your working.[4]
M1 for angle \(OAC\) = angle \(OCA\) (\(= x\)) because \(OA = OC\) (radii)
M1 for angle \(AOD = 2x\) (the exterior angle of triangle \(AOC\), or \(180 - (180 - 2x)\))
M1 for angle \(BOD = 2y\) using the same method in triangle \(BOC\)
C1 for a complete proof with all reasons (radii are equal, base angles of an isosceles triangle are equal, angles in a triangle add up to 180°, angles on a straight line add up to 180°), concluding angle \(AOB = 2x + 2y = 2(x + y) = 2\) × angle \(ACB\)
Worked solution: Let angle \(OCA = x\) and angle \(OCB = y\), so angle \(ACB = x + y\). \(OA = OC\) (radii), so angle \(OAC = x\) (base angles of an isosceles triangle are equal). Angle \(AOC = 180 - 2x\) (angles in a triangle add up to 180°), so angle \(AOD = 2x\) (angles on a straight line add up to 180°). Similarly, \(OB = OC\) gives angle \(BOD = 2y\). Angle \(AOB = 2x + 2y = 2(x + y) = 2\) × angle \(ACB\).