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Exam questions · Maths · Circle Theorems

Angles at the Centre and in a Semicircle

  • 6 exam questions
  • 19 marks
  • 9 quick checks
  1. 1 Calculate [2 marks]

    The diagram is not drawn to scale. \(A\), \(B\) and \(C\) are points on a circle, centre \(O\). Angle \(ACB = 36^\circ\). Calculate angle \(AOB\), giving a reason for your answer. [2 marks]

    A circle diagram showing the angle at the centre and the angle at the circumference.
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    Model answer

    Angle \(AOB = 2 \times 36 = 72^\circ\), because the angle at the centre is twice the angle at the circumference.

    Mark scheme

    • \(72\) — B1
    • The angle at the centre is twice the angle at the circumference — B1
  2. 2 Calculate [3 marks]

    The diagram is not drawn to scale. \(AB\) is a diameter of a circle and \(C\) is a point on the circle. Angle \(BAC = 27^\circ\). Calculate angle \(ABC\), giving a reason for each step. [3 marks]

    A circle diagram showing a triangle in a semicircle.
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    Model answer

    Angle \(ACB = 90^\circ\), because the angle in a semicircle is a right angle. Then \(ABC = 180 - 90 - 27 = 63^\circ\).

    Mark scheme

    • Angle \(ACB = 90^\circ\) — B1
    • The angle in a semicircle is a right angle — B1
    • \(63\) — B1
  3. 3 Calculate [3 marks]

    \(A\), \(B\) and \(C\) are points on a circle, centre \(O\), with \(C\) on the major arc \(AB\). Angle \(AOB = 7x - 6\) and angle \(ACB = 3x + 4\). Calculate the value of \(x\). [3 marks]

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    Model answer

    The angle at the centre is twice the angle at the circumference, so \(7x - 6 = 2(3x + 4) = 6x + 8\). Then \(x = 14\).

    Mark scheme

    • \(7x - 6 = 2(3x + 4)\) — M1
    • \(7x - 6 = 6x + 8\) — M1
    • \(14\) — A1
  4. 4 Calculate [4 marks]

    \(A\), \(B\) and \(C\) are points on a circle, centre \(O\), with \(C\) on the major arc \(AB\). Angle \(OAB = 22^\circ\). Calculate angle \(ACB\), giving reasons for your answer. [4 marks]

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    Model answer

    \(OA = OB\) because they are radii, so angle \(OBA = 22^\circ\). Angle \(AOB = 180 - 22 - 22 = 136^\circ\). The angle at the circumference is half the angle at the centre, so \(ACB = 68^\circ\).

    Mark scheme

    • \(180 - 22 - 22\) or angle \(OBA = 22^\circ\) — M1
    • \(136\) — A1
    • \(68\) — A1
    • Radii make an isosceles triangle, and the angle at the centre is twice the angle at the circumference — B1
  5. 5 Calculate [4 marks]

    \(AB\) is a diameter of a circle and \(C\) is a point on the circle. \(AC = 5\) cm and \(BC = 12\) cm. Calculate the radius of the circle, giving a reason for your answer. [4 marks]

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    Model answer

    Angle \(ACB = 90^\circ\), because the angle in a semicircle is a right angle. \(AB^2 = 5^2 + 12^2 = 25 + 144 = 169\), so \(AB = 13\) cm and the radius is \(6.5\) cm.

    Mark scheme

    • The angle in a semicircle is a right angle — B1
    • \(5^2 + 12^2\) or \(25 + 144\) — M1
    • \(AB = 13\) — A1
    • \(6.5\) — A1
  6. 6 Calculate [3 marks]

    The diagram is not drawn to scale. \(A\), \(B\) and \(C\) are points on a circle, centre \(O\). Angle \(AOB = 96^\circ\), and \(C\) is on the minor arc \(AB\). Calculate angle \(ACB\), giving a reason for your answer. [3 marks]

    A circle diagram showing the angle at the circumference on the minor arc.
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    Model answer

    The reflex angle \(AOB = 360 - 96 = 264^\circ\). The angle at the circumference is half of this, so \(ACB = 132^\circ\).

    Mark scheme

    • \(360 - 96 = 264\) — M1
    • \(132\) — A1
    • The angle at the centre is twice the angle at the circumference — B1

Quick check

  1. 1

    The angle at the centre of a circle is \(84^\circ\). What is the angle at the circumference made by the same arc?

    1. A\(168^\circ\)
    2. B\(42^\circ\)
    3. C\(84^\circ\)
    4. D\(96^\circ\)
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    B: \(42^\circ\)

    The angle at the centre is twice the angle at the circumference, so the angle at the circumference is \(84 \div 2 = 42^\circ\).

  2. 2

    What is the size of an angle in a semicircle?

    1. A\(90^\circ\)
    2. B\(180^\circ\)
    3. C\(45^\circ\)
    4. D\(60^\circ\)
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    A: \(90^\circ\)

    The angle at the centre on a diameter is \(180^\circ\), so the angle at the circumference is half of it, \(90^\circ\).

  3. 3

    Which reason fits this statement? \(\angle AOB = 2 \times \angle ACB\)

    1. AAngles in the same segment are equal
    2. BThe angle in a semicircle is \(90^\circ\)
    3. COpposite angles of a cyclic quadrilateral add up to \(180^\circ\)
    4. DThe angle at the centre is twice the angle at the circumference
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    D: The angle at the centre is twice the angle at the circumference

    The angle at \(O\), the centre, is double the angle at \(C\), on the circumference.

  4. 4

    \(AB\) is a diameter and \(C\) is on the circle. Angle \(CAB = 35^\circ\). What is angle \(CBA\)?

    1. A\(35^\circ\)
    2. B\(90^\circ\)
    3. C\(55^\circ\)
    4. D\(145^\circ\)
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    C: \(55^\circ\)

    Angle \(ACB = 90^\circ\) in a semicircle, so \(CBA = 180 - 90 - 35 = 55^\circ\).

  5. 5

    The angle at the circumference is \(3x\) and the angle at the centre on the same arc is \(5x + 20\). What is \(x\)?

    1. A\(10\)
    2. B\(20\)
    3. C\(4\)
    4. D\(40\)
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    B: \(20\)

    \(5x + 20 = 2 \times 3x = 6x\), so \(x = 20\).

  6. 6

    In triangle \(OAB\), \(O\) is the centre and \(\angle OAB = 28^\circ\). What is angle \(AOB\)?

    1. A\(124^\circ\)
    2. B\(62^\circ\)
    3. C\(56^\circ\)
    4. D\(152^\circ\)
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    A: \(124^\circ\)

    \(OA = OB\) are radii, so \(\angle OBA = 28^\circ\) and \(\angle AOB = 180 - 28 - 28 = 124^\circ\).

  7. 7

    \(AB\) is a diameter of length 10 cm and \(AC = 6\) cm. What is \(BC\)?

    1. A4 cm
    2. B\(\sqrt{136}\) cm
    3. C16 cm
    4. D8 cm
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    D: 8 cm

    Angle \(ACB = 90^\circ\) in a semicircle, so \(BC^2 = 10^2 - 6^2 = 64\) and \(BC = 8\).

  8. 8

    Two points \(A\) and \(B\) are on a circle with centre \(O\), and \(\angle AOB = 130^\circ\). \(C\) is on the minor arc \(AB\). What is angle \(ACB\)?

    1. A\(65^\circ\)
    2. B\(130^\circ\)
    3. C\(115^\circ\)
    4. D\(230^\circ\)
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    C: \(115^\circ\)

    The reflex angle at the centre is \(360 - 130 = 230^\circ\), so \(ACB = 230 \div 2 = 115^\circ\).

  9. 9

    Which statement is true for every triangle with a diameter as one side and its third corner on the circle?

    1. AIt is isosceles
    2. BIt is right-angled
    3. CIt is equilateral
    4. DIt has an angle of \(60^\circ\)
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    B: It is right-angled

    The angle opposite the diameter is in a semicircle, so it is always \(90^\circ\).