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Worked
Examples
Multiple
Choice
Practice
Questions
Lesson
Notes EXIT
< 𝑄𝑃𝑅 =
1
2
< 𝑅𝑂𝑄
< 𝑂𝑅𝑄 = 32Β°
(π‘π‘Žπ‘ π‘’ π‘Žπ‘›π‘”π‘™π‘’π‘  π‘œπ‘“ π‘Žπ‘› π‘–π‘ π‘œπ‘ π‘π‘’π‘™π‘’π‘  π‘‘π‘Ÿπ‘–π‘Žπ‘›π‘”π‘™π‘’ 𝑂𝑄𝑅 π‘Žπ‘Ÿπ‘’ π‘’π‘žπ‘’π‘Žπ‘™)
< 𝑅𝑂𝑄+ < 𝑂𝑅𝑄 +< 𝑂𝑄𝑅 = 180Β°
(π‘ π‘’π‘š π‘œπ‘“ π‘–π‘›π‘‘π‘’π‘Ÿπ‘–π‘œπ‘Ÿ π‘Žπ‘›π‘”π‘™π‘’π‘  π‘œπ‘“ π‘‘π‘Ÿπ‘–π‘Žπ‘›π‘”π‘™π‘’ 𝑂𝑄𝑅)
< 𝑅𝑂𝑄 + 32Β° + 32Β° = 180Β°
< 𝑅𝑂𝑄 + 64Β° = 180Β°
< 𝑅𝑂𝑄 = 180Β° βˆ’ 64Β°
< 𝑅𝑂𝑄 = 116Β°
∴ < 𝑄𝑃𝑅 =
1
2
116Β°
∴< 𝑄𝑃𝑅 = 58Β°
In the diagram, O is the centre of the circle,
< 𝑂𝑄𝑅 = 32Β° and
< 𝑀𝑃𝑄 = 15Β°.
Calculate:
i) < 𝑄𝑃𝑅,
ii) < 𝑀𝑄𝑂
π‘Šπ΄π‘†π‘†πΆπΈ π½π‘ˆπΏπ‘Œ, 2006.
Show
diagram
More
From π‘‘π‘Ÿπ‘–π‘Žπ‘›π‘”π‘™π‘’ 𝑂𝑀𝑄, < 𝑂𝑀𝑄 +< 𝑀𝑄𝑅 +< 𝑄𝑂𝑀 = 180Β°
But
< 𝑄𝑂𝑀 = 2 < 𝑄𝑃𝑀, π‘Žπ‘›π‘”π‘™π‘’ π‘Žπ‘‘ π‘‘β„Žπ‘’ π‘π‘’π‘›π‘‘π‘Ÿπ‘’ 𝑖𝑠 𝑑𝑀𝑖𝑐𝑒 π‘‘β„Žπ‘’ π‘œπ‘›π‘’ π‘œπ‘› π‘‘β„Žπ‘’
π‘π‘–π‘Ÿπ‘π‘’π‘šπ‘“π‘’π‘Ÿπ‘’π‘›π‘π‘’
β‡’ < 𝑄𝑂𝑀 = 2 15Β°
< 𝑄𝑂𝑀 = 30Β°
β‡’< 𝑂𝑀𝑄 +< 𝑀𝑄𝑂 + 30Β° = 180Β°
Also, < 𝑂𝑀𝑄 =< 𝑀𝑄𝑂, π‘π‘Žπ‘ π‘’ π‘Žπ‘›π‘”π‘™π‘’π‘  π‘œπ‘“ π‘–π‘ π‘œπ‘ π‘π‘’π‘™π‘’π‘  π‘‘π‘Ÿπ‘–π‘Žπ‘›π‘”π‘™π‘’ 𝑂𝑀𝑄
β‡’< 𝑂𝑀𝑄 +< 𝑂𝑀𝑄 + 30Β° = 180Β°
2 < 𝑂𝑀𝑄 + 30Β° = 180Β°
2 < 𝑂𝑀𝑄 = 180Β° βˆ’ 30Β°
2 < 𝑂𝑀𝑄 = 150Β°
< 𝑂𝑀𝑄 =
150Β°
2
∴ < 𝑂𝑀𝑄 = 75Β°
In the diagram, O is the centre of the circle,
< 𝑂𝑄𝑅 = 32Β° and
< 𝑀𝑃𝑄 = 15Β°.
Calculate:
i) < 𝑄𝑃𝑅,
ii) < 𝑀𝑄𝑂
π‘Šπ΄π‘†π‘†πΆπΈ π½π‘ˆπΏπ‘Œ, 2006.
Show
diagram
Flip to see ii
Worked
Example 2
From π‘‘π‘Ÿπ‘–π‘Žπ‘›π‘”π‘™π‘’ 𝑂𝑀𝑄, < 𝑂𝑀𝑄 +< 𝑀𝑄𝑅 +< 𝑄𝑂𝑀 = 180Β°
But
< 𝑄𝑂𝑀 = 2 < 𝑄𝑃𝑀, π‘Žπ‘›π‘”π‘™π‘’ π‘Žπ‘‘ π‘‘β„Žπ‘’ π‘π‘’π‘›π‘‘π‘Ÿπ‘’ 𝑖𝑠 𝑑𝑀𝑖𝑐𝑒 π‘‘β„Žπ‘’ π‘œπ‘›π‘’ π‘œπ‘› π‘‘β„Žπ‘’
π‘π‘–π‘Ÿπ‘π‘’π‘šπ‘“π‘’π‘Ÿπ‘’π‘›π‘π‘’
β‡’ < 𝑄𝑂𝑀 = 2 15Β°
< 𝑄𝑂𝑀 = 30Β°
β‡’< 𝑂𝑀𝑄 +< 𝑀𝑄𝑂 + 30Β° = 180Β°
Also, < 𝑂𝑀𝑄 =< 𝑀𝑄𝑂, π‘π‘Žπ‘ π‘’ π‘Žπ‘›π‘”π‘™π‘’π‘  π‘œπ‘“ π‘–π‘ π‘œπ‘ π‘π‘’π‘™π‘’π‘  π‘‘π‘Ÿπ‘–π‘Žπ‘›π‘”π‘™π‘’ 𝑂𝑀𝑄
β‡’< 𝑂𝑀𝑄 +< 𝑂𝑀𝑄 + 30Β° = 180Β°
2 < 𝑂𝑀𝑄 + 30Β° = 180Β°
2 < 𝑂𝑀𝑄 = 180Β° βˆ’ 30Β°
2 < 𝑂𝑀𝑄 = 150Β°
< 𝑂𝑀𝑄 =
150Β°
2
∴ < 𝑂𝑀𝑄 = 75Β°
In the diagram, 𝑂 is the centre of the
circle PQRS and < 𝑃𝑄𝑅 = 65Β°. Find the
value of angle π‘₯.
π‘Šπ΄π‘†π‘†πΆπΈ π½π‘ˆNE, 2012.
Show
diagram
Find the value of a in the diagram below, if AB
and BC are tangents, and O is the centre of the
circle.
A . 74Β°
B .64Β°
C . 104Β°
D .174Β°
E.124Β°
EXIT
MCQ
In the diagram below, O is the centre of the
circle and AB is a tangent to the circle at A.
Calculate the value of a and b.
B .π‘Ž = 90Β°, 𝑏 = 65Β°
A .π‘Ž = 155Β°, 𝑏 = 90Β°
C π‘Ž = 65Β°, 𝑏 = 90Β°
D .π‘Ž = 90Β°, 𝑏 = 75Β°
E .π‘Ž = 90Β°, 𝑏 = 85Β°
EXIT
MCQ
In the diagram below, P, Q, R and S are four points on a
circle. PR and QS intersect at T, such that
<STR=140Β° and <PRQ=30Β°. Find <SPR.
B 110Β°
A . 280Β°
C .90Β°
D .70Β°
β€’
β€’Β°
E .40Β°
EXIT
MCQ
The diagram shows a circle centre O. A, B and C are
points on the circumference. DBO is a straight line. DA
is a tangent to the circle. Calculate < 𝐡𝐢𝐴.
E .124Β°
A .64Β°
D .62Β°
B .14Β°
C .31Β°
EXIT
MCQ
The diagram shows a circle centre O. A, B and C are
points on the circumference. DBO is a straight line. DA
is a tangent to the circle. Calculate < 𝐴𝑂𝐷.
D.62Β°
A .64Β°
E .124Β°
B .14Β°
C .31Β°
EXIT
MCQ
The diagram shows a circle centre O. A, B and C are points on
the circumference. DBO is a straight line. DA is a tangent to
the circle at A and < 𝐡𝐴𝐷 = 38Β°. Find the value of 𝑧.
C .76Β°
A .38Β°
D .104Β°
B .52Β°
E .164Β° EXIT
MCQ
The diagram shows a circle centre O. A, B and C are points on
the circumference. DBO is a straight line. DA is a tangent to
the circle at A and < 𝐡𝐴𝐷 = 38°. Which theorem can be used to
find 𝑧?
A . Angles in alternate
segment
C . Alternate angles
D . Angles in the same
segment
B . Cyclic quadrilaterals
E . Angles about the centre
EXIT
MCQ
The diagram shows a circle centre O. A, B and C are points on
the circumference. DBO is a straight line. DA is a tangent to
the circle at A and < 𝐡𝐴𝐷 = 38Β°. Find the value of π‘₯.
B .52Β°
A .38Β°
D .104Β°
C .76Β°
E .164Β° EXIT
MCQ
The diagram shows a circle centre O. A, B and C are points on
the circumference. DBO is a straight line. DA is a tangent to
the circle at A and < 𝐡𝐴𝐷 = 38Β°. Find the value of 𝑀 βˆ’ 𝑦.
D .270Β°
A .38Β°
B .90Β°
C .60Β°
E .180Β°
EXIT
MCQ
In the diagram, O is the centre of the circle. BD and CD
are tangents. Which of the following statements is NOT
true about the diagram?
D .< 𝑂𝐡𝐷 =< 𝑂𝐸𝐡
A .< 𝑂𝐡𝐷+ < 𝑂𝐢𝐷 = 180Β°
C .< 𝐢𝑂𝐡+ < 𝐡𝐷𝐢 = 180Β°
B .< 𝑂𝐡𝐸 =< 𝑂𝐸𝐷
C . < 𝑂𝐡𝐷 =< 𝑂𝐢𝐷
Semi-circles
Semi-circles
π‘Ÿ
𝑂 π‘‘π‘–π‘Žπ‘šπ‘’π‘‘π‘’π‘Ÿ
The circle is a locus of points equidistant
from a fixed point called the centre, O.
The radius is π‘Ÿ.
The diameter divides the circle into two
semi-circles. The diameter is a special
chord.
A chord divides the circle into two semi-
circles.
EXIT
MCQ
The tangent to a circle meets the radius at an angle
of 90Β°. The same thing applies to the diameter
EXIT
MCQ
The angle between a chord and a tangent is equal to the
angle created in the alternate segment
EXIT
MCQ
Angle π‘Ž and 𝑏 are equal because they are all created in the same
Segment by the same arc or chord
EXIT
MCQ
Angle at the centre
πœƒ
The angle subtended at the centre is twice the one
subtended at the circumference by an arc.
𝑂
EXIT
MCQ
A quadrilateral that has its four vertices on the circumference of a circle is calle
a cyclic quadrilateral
The opposite angles of a cyclic quadrilateral will always sum up to 180Β°

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  • 2. < 𝑄𝑃𝑅 = 1 2 < 𝑅𝑂𝑄 < 𝑂𝑅𝑄 = 32Β° (π‘π‘Žπ‘ π‘’ π‘Žπ‘›π‘”π‘™π‘’π‘  π‘œπ‘“ π‘Žπ‘› π‘–π‘ π‘œπ‘ π‘π‘’π‘™π‘’π‘  π‘‘π‘Ÿπ‘–π‘Žπ‘›π‘”π‘™π‘’ 𝑂𝑄𝑅 π‘Žπ‘Ÿπ‘’ π‘’π‘žπ‘’π‘Žπ‘™) < 𝑅𝑂𝑄+ < 𝑂𝑅𝑄 +< 𝑂𝑄𝑅 = 180Β° (π‘ π‘’π‘š π‘œπ‘“ π‘–π‘›π‘‘π‘’π‘Ÿπ‘–π‘œπ‘Ÿ π‘Žπ‘›π‘”π‘™π‘’π‘  π‘œπ‘“ π‘‘π‘Ÿπ‘–π‘Žπ‘›π‘”π‘™π‘’ 𝑂𝑄𝑅) < 𝑅𝑂𝑄 + 32Β° + 32Β° = 180Β° < 𝑅𝑂𝑄 + 64Β° = 180Β° < 𝑅𝑂𝑄 = 180Β° βˆ’ 64Β° < 𝑅𝑂𝑄 = 116Β° ∴ < 𝑄𝑃𝑅 = 1 2 116Β° ∴< 𝑄𝑃𝑅 = 58Β° In the diagram, O is the centre of the circle, < 𝑂𝑄𝑅 = 32Β° and < 𝑀𝑃𝑄 = 15Β°. Calculate: i) < 𝑄𝑃𝑅, ii) < 𝑀𝑄𝑂 π‘Šπ΄π‘†π‘†πΆπΈ π½π‘ˆπΏπ‘Œ, 2006. Show diagram More
  • 3. From π‘‘π‘Ÿπ‘–π‘Žπ‘›π‘”π‘™π‘’ 𝑂𝑀𝑄, < 𝑂𝑀𝑄 +< 𝑀𝑄𝑅 +< 𝑄𝑂𝑀 = 180Β° But < 𝑄𝑂𝑀 = 2 < 𝑄𝑃𝑀, π‘Žπ‘›π‘”π‘™π‘’ π‘Žπ‘‘ π‘‘β„Žπ‘’ π‘π‘’π‘›π‘‘π‘Ÿπ‘’ 𝑖𝑠 𝑑𝑀𝑖𝑐𝑒 π‘‘β„Žπ‘’ π‘œπ‘›π‘’ π‘œπ‘› π‘‘β„Žπ‘’ π‘π‘–π‘Ÿπ‘π‘’π‘šπ‘“π‘’π‘Ÿπ‘’π‘›π‘π‘’ β‡’ < 𝑄𝑂𝑀 = 2 15Β° < 𝑄𝑂𝑀 = 30Β° β‡’< 𝑂𝑀𝑄 +< 𝑀𝑄𝑂 + 30Β° = 180Β° Also, < 𝑂𝑀𝑄 =< 𝑀𝑄𝑂, π‘π‘Žπ‘ π‘’ π‘Žπ‘›π‘”π‘™π‘’π‘  π‘œπ‘“ π‘–π‘ π‘œπ‘ π‘π‘’π‘™π‘’π‘  π‘‘π‘Ÿπ‘–π‘Žπ‘›π‘”π‘™π‘’ 𝑂𝑀𝑄 β‡’< 𝑂𝑀𝑄 +< 𝑂𝑀𝑄 + 30Β° = 180Β° 2 < 𝑂𝑀𝑄 + 30Β° = 180Β° 2 < 𝑂𝑀𝑄 = 180Β° βˆ’ 30Β° 2 < 𝑂𝑀𝑄 = 150Β° < 𝑂𝑀𝑄 = 150Β° 2 ∴ < 𝑂𝑀𝑄 = 75Β° In the diagram, O is the centre of the circle, < 𝑂𝑄𝑅 = 32Β° and < 𝑀𝑃𝑄 = 15Β°. Calculate: i) < 𝑄𝑃𝑅, ii) < 𝑀𝑄𝑂 π‘Šπ΄π‘†π‘†πΆπΈ π½π‘ˆπΏπ‘Œ, 2006. Show diagram Flip to see ii Worked Example 2
  • 4. From π‘‘π‘Ÿπ‘–π‘Žπ‘›π‘”π‘™π‘’ 𝑂𝑀𝑄, < 𝑂𝑀𝑄 +< 𝑀𝑄𝑅 +< 𝑄𝑂𝑀 = 180Β° But < 𝑄𝑂𝑀 = 2 < 𝑄𝑃𝑀, π‘Žπ‘›π‘”π‘™π‘’ π‘Žπ‘‘ π‘‘β„Žπ‘’ π‘π‘’π‘›π‘‘π‘Ÿπ‘’ 𝑖𝑠 𝑑𝑀𝑖𝑐𝑒 π‘‘β„Žπ‘’ π‘œπ‘›π‘’ π‘œπ‘› π‘‘β„Žπ‘’ π‘π‘–π‘Ÿπ‘π‘’π‘šπ‘“π‘’π‘Ÿπ‘’π‘›π‘π‘’ β‡’ < 𝑄𝑂𝑀 = 2 15Β° < 𝑄𝑂𝑀 = 30Β° β‡’< 𝑂𝑀𝑄 +< 𝑀𝑄𝑂 + 30Β° = 180Β° Also, < 𝑂𝑀𝑄 =< 𝑀𝑄𝑂, π‘π‘Žπ‘ π‘’ π‘Žπ‘›π‘”π‘™π‘’π‘  π‘œπ‘“ π‘–π‘ π‘œπ‘ π‘π‘’π‘™π‘’π‘  π‘‘π‘Ÿπ‘–π‘Žπ‘›π‘”π‘™π‘’ 𝑂𝑀𝑄 β‡’< 𝑂𝑀𝑄 +< 𝑂𝑀𝑄 + 30Β° = 180Β° 2 < 𝑂𝑀𝑄 + 30Β° = 180Β° 2 < 𝑂𝑀𝑄 = 180Β° βˆ’ 30Β° 2 < 𝑂𝑀𝑄 = 150Β° < 𝑂𝑀𝑄 = 150Β° 2 ∴ < 𝑂𝑀𝑄 = 75Β° In the diagram, 𝑂 is the centre of the circle PQRS and < 𝑃𝑄𝑅 = 65Β°. Find the value of angle π‘₯. π‘Šπ΄π‘†π‘†πΆπΈ π½π‘ˆNE, 2012. Show diagram
  • 5. Find the value of a in the diagram below, if AB and BC are tangents, and O is the centre of the circle. A . 74Β° B .64Β° C . 104Β° D .174Β° E.124Β° EXIT MCQ
  • 6. In the diagram below, O is the centre of the circle and AB is a tangent to the circle at A. Calculate the value of a and b. B .π‘Ž = 90Β°, 𝑏 = 65Β° A .π‘Ž = 155Β°, 𝑏 = 90Β° C π‘Ž = 65Β°, 𝑏 = 90Β° D .π‘Ž = 90Β°, 𝑏 = 75Β° E .π‘Ž = 90Β°, 𝑏 = 85Β° EXIT MCQ
  • 7. In the diagram below, P, Q, R and S are four points on a circle. PR and QS intersect at T, such that <STR=140Β° and <PRQ=30Β°. Find <SPR. B 110Β° A . 280Β° C .90Β° D .70Β° β€’ β€’Β° E .40Β° EXIT MCQ
  • 8. The diagram shows a circle centre O. A, B and C are points on the circumference. DBO is a straight line. DA is a tangent to the circle. Calculate < 𝐡𝐢𝐴. E .124Β° A .64Β° D .62Β° B .14Β° C .31Β° EXIT MCQ
  • 9. The diagram shows a circle centre O. A, B and C are points on the circumference. DBO is a straight line. DA is a tangent to the circle. Calculate < 𝐴𝑂𝐷. D.62Β° A .64Β° E .124Β° B .14Β° C .31Β° EXIT MCQ
  • 10. The diagram shows a circle centre O. A, B and C are points on the circumference. DBO is a straight line. DA is a tangent to the circle at A and < 𝐡𝐴𝐷 = 38Β°. Find the value of 𝑧. C .76Β° A .38Β° D .104Β° B .52Β° E .164Β° EXIT MCQ
  • 11. The diagram shows a circle centre O. A, B and C are points on the circumference. DBO is a straight line. DA is a tangent to the circle at A and < 𝐡𝐴𝐷 = 38Β°. Which theorem can be used to find 𝑧? A . Angles in alternate segment C . Alternate angles D . Angles in the same segment B . Cyclic quadrilaterals E . Angles about the centre EXIT MCQ
  • 12. The diagram shows a circle centre O. A, B and C are points on the circumference. DBO is a straight line. DA is a tangent to the circle at A and < 𝐡𝐴𝐷 = 38Β°. Find the value of π‘₯. B .52Β° A .38Β° D .104Β° C .76Β° E .164Β° EXIT MCQ
  • 13. The diagram shows a circle centre O. A, B and C are points on the circumference. DBO is a straight line. DA is a tangent to the circle at A and < 𝐡𝐴𝐷 = 38Β°. Find the value of 𝑀 βˆ’ 𝑦. D .270Β° A .38Β° B .90Β° C .60Β° E .180Β° EXIT MCQ
  • 14. In the diagram, O is the centre of the circle. BD and CD are tangents. Which of the following statements is NOT true about the diagram? D .< 𝑂𝐡𝐷 =< 𝑂𝐸𝐡 A .< 𝑂𝐡𝐷+ < 𝑂𝐢𝐷 = 180Β° C .< 𝐢𝑂𝐡+ < 𝐡𝐷𝐢 = 180Β° B .< 𝑂𝐡𝐸 =< 𝑂𝐸𝐷 C . < 𝑂𝐡𝐷 =< 𝑂𝐢𝐷
  • 15. Semi-circles Semi-circles π‘Ÿ 𝑂 π‘‘π‘–π‘Žπ‘šπ‘’π‘‘π‘’π‘Ÿ The circle is a locus of points equidistant from a fixed point called the centre, O. The radius is π‘Ÿ. The diameter divides the circle into two semi-circles. The diameter is a special chord. A chord divides the circle into two semi- circles. EXIT MCQ
  • 16. The tangent to a circle meets the radius at an angle of 90Β°. The same thing applies to the diameter EXIT MCQ
  • 17. The angle between a chord and a tangent is equal to the angle created in the alternate segment EXIT MCQ
  • 18. Angle π‘Ž and 𝑏 are equal because they are all created in the same Segment by the same arc or chord EXIT MCQ
  • 19. Angle at the centre πœƒ The angle subtended at the centre is twice the one subtended at the circumference by an arc. 𝑂 EXIT MCQ
  • 20. A quadrilateral that has its four vertices on the circumference of a circle is calle a cyclic quadrilateral The opposite angles of a cyclic quadrilateral will always sum up to 180Β°