Lesson Notes By Weeks and Term v5 - Grade 10
Euclidean geometry – Week 10 focus
Download the Lessonotes Mobile South Africa app for faster lesson access on Android and iPhone.
Euclidean geometry – Week 10 focus
Download the Lessonotes Mobile South Africa app for faster lesson access on Android and iPhone.
Subject: Mathematics
Class: Grade 10
Term: 2nd Term
Week: 10
Theme: General lesson support
This page supports the lesson note with a companion video and a short classroom-ready summary.
For class groups and homework, share this lesson page so learners also get the summary, objectives, and full lesson context.
Euclidean geometry is a cornerstone of mathematics, providing the foundation for understanding shapes, sizes, and spatial relationships. In Grade 10, we delve deeper into its principles, building on the basic geometry you learned in earlier grades. This knowledge is crucial not just for passing exams, but for real-world applications like construction, architecture, design, and even understanding maps and spatial awareness in everyday life in South Africa. Think about planning a new house, designing a garden, or even calculating the angles for tiling a bathroom – all these rely on Euclidean geometry principles.
This week focuses on theorems related to circles in Euclidean geometry. These theorems allow us to deduce relationships between angles, lines, and arcs within a circle, leading to powerful problem-solving capabilities. 2.1 Theorem 1: The Perpendicular Bisector Theorem Statement: A line drawn from the centre of a circle perpendicular to a chord bisects the chord. Conversely, the line drawn from the centre of a circle to the midpoint of a chord is perpendicular to the chord.
Explanation: Imagine a circle with centre O. If you draw a line from O that hits a chord (a line segment connecting two points on the circle) at a 90-degree angle (perpendicular), that line will cut the chord exactly in half (bisect it).
The converse is also true: if you connect the centre O to the middle of a chord, that line will always be perpendicular to the chord.
Why it works: Consider two radii drawn from the centre to the endpoints of the chord. These form two congruent right-angled triangles, allowing us to prove that the perpendicular line bisects the chord using congruency theorems.
Example 1: In circle O, AB is a chord and OM is perpendicular to AB at
M. If AB = 8 cm, find the length of A
M. Solution: Since OM is perpendicular to chord AB, it bisects AB (Perpendicular Bisector Theorem).
Therefore, AM = MB = AB/2 = 8/2 = 4 cm.
Example 2: In circle O, AB is a chord, and M is the midpoint of A
B. If the radius of the circle is 5cm and AM = 4cm, find the length of O
M. Solution: Since M is the midpoint of AB, OM is perpendicular to AB (Converse of Perpendicular Bisector Theorem). Triangle OAM is a right-angled triangle.
Using the Pythagorean theorem: OA² = OM² + AM². So, 5² = OM² + 4².
Therefore, OM² = 25 - 16 =
9. Taking the square root, OM = 3 cm. 2.2 Theorem 2: Angle at Centre Theorem Statement: The angle subtended by an arc at the centre of a circle is double the angle subtended by the same arc at the circumference.
Explanation: "Subtended" means "formed by drawing lines from the endpoints of the arc". Imagine an arc on a circle. If you draw lines from the arc's endpoints to the centre of the circle, you create an angle at the centre. If you draw lines from the same arc's endpoints to any point on the circumference of the circle, you create an angle at the circumference. The angle at the centre will always be twice the angle at the circumference.
Why it works: Drawing a radius from the centre to the point on the circumference splits the diagram into isosceles triangles. Through angle chasing, utilizing the properties of isosceles triangles and angles on a straight line, we can prove the theorem.
Example 3: In circle O, arc AB subtends ∠AOB at the centre and ∠ACB at the circumference. If ∠ACB = 35°, find ∠AO
B. Solution: According to the Angle at Centre Theorem, ∠AOB = 2 ∠ACB = 2 * 35° = 70°.
Example 4: In circle O, arc PQ subtends ∠POQ at the centre. Given that ∠POQ = 110°, find ∠PRQ where R is a point on the circumference.
Solution: According to the Angle at Centre Theorem, ∠PRQ = (1/2) ∠POQ = (1/2) * 110° = 55°. 2.3 Theorem 3: Angles in the Same Segment Theorem Statement: Angles in the same segment of a circle are equal.
Explanation: A "segment" of a circle is the area enclosed by a chord and the arc it cuts off. If you choose any two points on the arc of a segment and draw lines from them to a point on the remaining part of the circumference (within the same segment), the angles formed at the circumference will be equal.
Why it works: Each angle in the same segment is half the angle subtended by the same arc at the centre (Angle at Centre Theorem). Since they are both half of the same angle, they must be equal to each other.
Example 5: In circle O, A, B, C, and D are points on the circumference. Angles ∠CAD and ∠CBD are subtended by chord CD within the same segment. If ∠CAD = 42°, find ∠CB
D. Solution: According to the Angles in the Same Segment Theorem, ∠CAD = ∠CB
D. Therefore, ∠CBD = 42°.
Example 6: In circle O, points P, Q, R, and S lie on the circumference. If ∠PSQ = 68 degrees, find ∠PRQ, given that both angles are subtended by chord PQ within the same segment.
Solution: Since angles ∠PSQ and ∠PRQ are in the same segment and subtended by the same chord PQ, they are equal.
Therefore, ∠PRQ = ∠PSQ = 68 degrees. Guided Practice (With Solutions)
Question 1: In circle O, radius OC is perpendicular to chord A
B. If AB = 12cm and OC = 10cm, find the length of the radius O
A. Solution: Draw a diagram. We have a circle with center O, chord AB, and radius OC perpendicular to AB at point M. Since OC is perpendicular to AB, M is the midpoint of AB (Perpendicular Bisector Theorem).
Therefore, AM = AB/2 = 12/2 = 6 cm. Triangle OMA is a right-angled triangle. We want to find OA (the radius).
Using the Pythagorean theorem: OA² = OM² + AM². We know AM = 6cm, but we need to find OM. We know OC = 10cm (radius). Since OC is the radius and M lies on OC, then OM = OC - MC. We are not given MC.