Lesson Notes By Weeks and Term v5 - Grade 11
Euclidean geometry (circles) – Week 4 focus
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Euclidean geometry (circles) – Week 4 focus
Download the Lessonotes Mobile South Africa app for faster lesson access on Android and iPhone.
Subject: Mathematics
Class: Grade 11
Term: 3rd Term
Week: 4
Theme: General lesson support
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Euclidean geometry, particularly the study of circles, is a cornerstone of mathematical understanding. It builds upon previously learned geometric principles and introduces new concepts that are crucial for further mathematical study and practical applications. In a South African context, understanding circles can be applied to numerous real-world scenarios, from designing circular gardens or rondavels to calculating areas and circumferences in construction or farming contexts. It also lays the foundation for understanding trigonometry, which is essential in surveying and navigation. This week, we will delve into the theorems related to angles subtended by chords and tangents of circles.
This week focuses on four crucial circle theorems: Angle at Centre Theorem: The angle subtended by an arc or chord at the centre of a circle is twice the angle it subtends at the circumference on the same side of the chord.
Explanation: Imagine a chord AB within a circle with centre O. The angle AOB (at the centre) will always be twice the angle ACB, where C is any point on the circumference on the same side of the chord A
B. Why: This theorem arises from the properties of isosceles triangles formed within the circle (radii are equal). Drawing radii from the centre to the points on the circumference forms two isosceles triangles, which allows us to relate the angles using the angle sum property of triangles.
How: Always remember to identify the central angle and the angle at the circumference subtended by the same arc or chord. The central angle is always twice the angle at the circumference.
Reason for proof: Let angle ACB = x. Draw CO and extend to meet the circumference at D. Angle AOC = 2 Angle ABC and Angle BOC = 2 * Angle BAC. Proof of these two statements use properties of isosceles triangles (OA = OC, OB = OC) and exterior angle theorem. Adding the two statements, the overall angle subtended at the center is twice the angle at the circumference.
Example 1: In circle with centre O, angle AOB = 100°. Find angle ACB, where C lies on the major arc A
B. Solution: Angle ACB = 1/2 Angle AOB = 1/2 * 100° = 50°.
Reason: Angle at centre = 2 angle at circumference.
Angles in the Same Segment Theorem: Angles subtended by the same chord (or arc) in the same segment of a circle are equal.
Explanation: Consider a chord AB within a circle. If C and D are two points on the circumference lying in the same segment (on the same side of the chord), then angle ACB will be equal to angle AD
B. Why: This theorem stems directly from the Angle at Centre Theorem. Since the angle at the centre subtended by chord AB is fixed, and this angle is twice any angle on the circumference subtended by the same chord, all angles in the same segment must be equal.
How: Identify the chord and the segment. Ensure both angles being compared are subtended by the same chord and lie in the same segment.
Example 2: In a circle, chord PQ subtends angles PRQ and PSQ at points R and S respectively, lying on the same segment. If angle PRQ = 35°, find angle PS
Q. Solution: Angle PSQ = Angle PRQ = 35°.
Reason: Angles in the same segment are equal.
Cyclic Quadrilateral Theorem: The opposite angles of a cyclic quadrilateral are supplementary (add up to 180°).
Explanation: A cyclic quadrilateral is a four-sided figure where all four vertices lie on the circumference of a circle. In such a quadrilateral, the sum of any pair of opposite angles is always 180°.
Why: The sum of the angles at the center subtended by two opposite sides of the cyclic quadrilateral is 360°(entire circle). Each of those angles is twice the corresponding angle at the circumference.
Therefore the sum of the angles at the circumference (opposite angles) is half of 360° ie. 180°.
How: Identify the cyclic quadrilateral. Add the measures of opposite angles to verify they sum to 180°. If one angle is unknown, subtract the known opposite angle from 180° to find it.
Example 3: ABCD is a cyclic quadrilateral. If angle ABC = 80°, find angle AD
C. Solution: Angle ADC = 180° - Angle ABC = 180° - 80° = 100°.
Reason: Opposite angles of a cyclic quadrilateral are supplementary. Tangent-Chord Theorem (Alternate Segment Theorem): The angle between a tangent and a chord at the point of contact is equal to the angle in the alternate segment.
Explanation: If a tangent is drawn to a circle at a point on the circumference, and a chord is drawn from that same point, then the angle between the tangent and the chord is equal to the angle subtended by that chord in the alternate segment (the segment on the opposite side of the chord from the tangent).
Why: This is proven using the properties of a radius drawn to the point of tangency (which is perpendicular to the tangent) and the angle sum property of a triangle.
How: Identify the tangent, the chord at the point of contact, and the alternate segment. The angle between the tangent and the chord will be equal to the angle in the alternate segment.
Example 4: AT is a tangent to a circle at point A. Chord AB is drawn. Angle TAB = 60°. Find angle ACB, where C lies on the circumference in the alternate segment.
Solution: Angle ACB = Angle TAB = 60°.
Reason: Angle between tangent and chord equals angle in alternate segment. Guided Practice (With Solutions)
Question 1: In the circle with centre O, angle AOB = 120°. Calculate the size of angle ACB, where C is a point on the circumference.
Solution: Angle AOB = 120° (Given) Angle ACB = ½ Angle AOB (Angle at centre = 2 * angle at circumference) Angle ACB = ½ 120° = 60°
Commentary: This question directly applies the angle at centre theorem.