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Circle Theorem Flipbook PDF
Circle Theorem
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Edited by Elsie Lartey
Circle Theorem 1.1 Intro A circle is a plane figure enclosed by a curved line where :every point on the curved line is at an equal distance from a point within called the centre. 1.2 Properties of circles (a)The distance from the centre to the curve is called the radius, r of the circle. OP in the diagram is the radius of the circle. (b) The boundary of a circle is called the circumference. (c)Any straight line passing through the centre and touching the circumference at each end is called the diameter, d. AB in the diagram is the diameter of the circle.
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(d) A tangent to a circle is a straight line which meets the circle in one point only and does not cut the circle when produced. The straight line shown is a tangent to the circle since it touches the circle at point B only. If radius OB is drawn ∠B is a right angle. (e) A chord of a circle is any straight line which divides the circle into two parts and is terminated at each end by the circumference. ST is a chord (f) An arc is a portion of the circumference of a circle. The distance AB is an arc. (g) Segment is the name given to the parts into which a circle is divided by a chord. When the chord is the diameter the segments are known as semicircles.
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1.3 Circle Theorems Circle theorems are properties that show relationships between angles within the geometry of a circle. 1.31 Inscribed Angle Theorems An angle formed within a circle from two chords intersecting at a point on the circumference is an inscribed angle as shown. The word ‘subtend’ The word ‘subtend’ literally means ‘holds under’, and is often used in geometry to describe an angle. Suppose we have an arc or chord AB and a point P not on AB as shown. We say that the angle ∠APB is the angle subtended by arc or chord AB at point P. ♡Notice that lines AP and BP produced from arc AB hold an angle under point
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P which is why we say arc AB subtends an angle at P Angles subtended by the same arc. ♡Angles subtended by the same arc at the circumference are equal. This means that angles in the same segment are equal. In this example the angles are in the major segment. Thus ∠ACB=∠ADB⇒a=b Interactive Learning Activity 1 Click on Activity 1 to access the first activity on Circle Theorem. Task: 1. Carefully drag the point C or D on the circumference and observe the values of the angles. 2. Drag point A or B to vary the length of arc AB observe the new angles. What do you notice?
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Example Find the angles p and q
Solution 𝑝 = 52° (Angles at the circumference in the same segment are equal) 𝑞 = 40° (Angles at the circumference in the same segment are equal) Angles at the centre and the circumference. ♡An angle at the centre of a circle is twice the angle at the circumference subtended by the same arc.
AB is a minor arc
AB is a major arc
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In the diagrams the lines AO and BO produced from arc AB hold the angle a under centre O. In math jargon, arc AB subtends an angle a at centre O. The lines AC and BC produced from arc AB hold the angle b under point C on the circumference. In math jargon, arc AB subtends an angle b at C. This theorem says 𝑎 = 2𝑏 Other examples of the theorem
The examples above are also cases of “angles at the centre and the circumference”. The third example is a special case of the theorem known as Thales’ theorem which says that the angle at the circumference subtended by the diameter is a right angle. Interactive Learning Activity 2
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Click on Activity 2 to access the third activity on Circle Theorem. Task: 1. Carefully drag point A or B on the circumference and observe the values of the angles at the centre O and point C on the circumference. 2. Drag the point A or B till AB becomes a straight line and thus the diameter of the circle. What do you observe? 3. Activity 2.1 Repeat 1. and 2. for Activity 2.1 Examples Find the marked angles: i.
ii.
iii.
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Solution i. 𝑥 = 2 × 50° = 100°(Angle at the centre is twice the angle at the circumference ) ii. 𝑦 = 2 × 40° (Angle at the centre is twice the angle at the circumference) iii. 𝑆𝑈 is a diameter ⇒ ∠𝑆𝑇𝑈 = 90° From △ 𝑆𝑇𝑈 ∠𝑆𝑇𝑈 + ∠𝑇𝑈𝑆 + ∠𝑈𝑆𝑇 = 180° ⇒ 90° + 31° + 𝑧 = 180° ⇒ 𝑧 = 180° − 121° ⇒ 𝑧 = 59° Try 1. Work out the size of each angle marked with a letter. i.
ii.
iii.
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1.32 Alternate segment theorem The left-hand diagram below shows two angles ∠P and ∠Q lying in the same segment of a circle – we have proven that these two angles are equal. In the next two diagrams, the angle ∠BQU remains equal to ∠P as the point Q moves around the arc closer and closer to A. In the last diagram, Q coincides with A, and AU is a tangent.
The alternate segment theorem says that ∠BAU will still equal ∠P ♡The angle between the tangent and the chord at the point of contact is equal to the angle in the alternate segment. Here ∠BAU is ‘an angle between a chord and a tangent. The word ‘alternate’ means ‘other’ – the chord AB divides the circle into two segments,
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major and minor. So the angle between the chord and the tangent in the minor segment is equal to the angle subtended by the chord in the major segment and vice versa.
∠𝐵𝐴𝑈 = ∠𝑃
Note: To easily spot this property of a circle, look out for a triangle in the interior of the circle with one of its vertices resting on the point of contact of the tangent to the circle.
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Example Find the missing angles 𝑥,𝑦,𝑧 Solution 𝑦 = 90° (Angle subtended at the circumference by diameter) Angles in a triangle add up to 180° 90° + 30° + 𝑧 = 180° ⇒ 𝑧 = 180° − 120° ⇒ 𝑧 = 60° 𝑥 = 𝑧 = 60° (Alternate segment) Try Find the size of the angles marked with a letter. 1. i.
ii.
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1.33 Cyclic Quadrilateral A cyclic quadrilateral is a quadrilateral with all four vertices on the circumference of a circle. The opposite angles of a cyclic quadrilateral The distinctive property of a cyclic quadrilateral is that its opposite angles are supplementary. 𝑥 + 𝑦 = 180° 𝑝 + 𝑞 = 180° Examples Calculate the angles a and b Solution Opposite angles of a cyclic quadrilateral add up to 180° 𝑎 + 60° = 180° 𝑎 = 180° − 60°
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𝑎 = 120° 𝑏 + 140° = 180° 𝑏 = 180° − 140° 𝑏 = 40° Try Find the size of the angles marked with a letter. i.
ii.
Exterior angles of a cyclic quadrilateral An exterior angle of a cyclic quadrilateral is equal to the opposite interior angle.
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Example Find the angles a, b and c. Solution An exterior angle of a cyclic quadrilateral is equal to the opposite interior angle ⇒ 𝑏 = 105° 𝑎 + 𝑏 = 180°(Angles on a straight line) ⇒ 𝑎 = 180° − 105° ⇒ 𝑎 = 75° 𝑐 + 86° = 180 (Opposite angles in a cyclic quadrilateral are supplementary) 𝑐 = 180° − 86° 𝑐 = 94° Circle theorems involving the tangent of a circle . 1. Tangents from a common external point have equal length . |𝐶𝐵| = |𝐷𝐵|
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2. The angle between a tangent and a line through the centre is 90° 𝑎 + 𝑏 = 90° Example Calculate the angles g and h.
Since |𝐴𝑃| = |𝐵𝑃| (tangents from a common external point) △ 𝐴𝐵𝑃 is isosceles. ⇒ 𝑔 + 𝑔 + 56° = 180° ⇒ 2𝑔 = 180° − 56° ⇒ 2𝑔 = 124° ⇒ 𝑔 = 62° The tangent to a circle at a point is perpendicular to the radius at that point. ⇒ ∠𝐶𝐵𝑃 = 90°
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⇒ ℎ + 𝑔 = 90° ⇒ ℎ = 90° − 62° ⇒ ℎ = 28° Try Work out the size of the angles marked with a letter. i.
ii.
Edited by Elsie Lartey
Edited by Elsie Lartey
Edited by Elsie Lartey
Edited by Elsie Lartey
Edited by Elsie Lartey