Lesson Notes By Weeks and Term v5 - Grade 10
Euclidean geometry – Week 8 focus
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Euclidean geometry – Week 8 focus
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Subject: Mathematics
Class: Grade 10
Term: 3rd Term
Week: 8
Theme: General lesson support
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Euclidean geometry is the study of shapes, sizes, relative positions of figures, and the properties of space. It's named after the ancient Greek mathematician Euclid, and it forms the foundation of much of what we understand about the world around us. This week, we'll be focusing on understanding and applying the theorem of the line drawn from the centre of a circle perpendicular to a chord bisects the chord. This theorem, along with its converse, is fundamental for solving geometric problems and understanding relationships within circles.
Theorem 1: The line drawn from the centre of a circle perpendicular to a chord bisects the chord.
Statement: If a line is drawn from the centre of a circle to a chord such that the line is perpendicular to the chord, then that line bisects the chord.
Explanation: "Bisect" means to cut into two equal parts. So, this theorem states that if you have a circle and draw a line from the very centre of the circle to a chord, and that line forms a 90-degree angle with the chord (is perpendicular), then the line will cut the chord exactly in half.
Diagram: ``` O / \ / \ / \ A-------M-------B \ / \ / \ / C ``` Where: `O` is the centre of the circle. `AB` is the chord. `OM` is perpendicular to `AB` (OM ⊥ AB).
Therefore, `AM = MB`.
Proof: Given: Circle with centre O, and OM ⊥ A
B. To Prove: AM = M
B. Construction: Join OA and OB (radii).
Proof: In ΔOMA and ΔOMB: OA = OB (Radii of the same circle) OM = OM (Common side) ∠OMA = ∠OMB = 90° (Given OM ⊥ AB) Therefore, ΔOMA ≡ ΔOMB (RHS - Right angle, Hypotenuse, Side congruency) Therefore, AM = MB (Corresponding sides of congruent triangles are equal).
Example 1: In circle with centre O, AB is a chord. OM ⊥ AB, AB = 8cm. Calculate A
M. Solution: Since OM ⊥ AB, AM = MB (Line from centre ⊥ chord) AB = 8cm AM = (1/2) AB = (1/2) * 8cm = 4cm Example 2: In circle with centre O, chord PQ = 12cm, and radius OP = 10cm. OR ⊥ P
Q. Calculate the length of O
R. Diagram: ``` O / \ / \ 10cm / \ P-------R-------Q \ / 6cm \ / \ / ``` Where: O is the centre of the circle. PQ is the chord of length 12cm. OP is a radius of length 10cm. OR is perpendicular to P
Q. Solution: PR = (1/2) PQ = (1/2) * 12cm = 6cm (Line from centre ⊥ chord) In right-angled triangle ΔOPR: OP² = OR² + PR² (Pythagoras) 10² = OR² + 6² 100 = OR² + 36 OR² = 100 - 36 = 64 OR = √64 = 8cm Theorem 2: The line drawn from the centre of a circle to the midpoint of a chord is perpendicular to the chord (Converse of the previous theorem).
Statement: If a line is drawn from the centre of a circle to the midpoint of a chord, then the line is perpendicular to the chord.
Explanation: This is the reverse of the first theorem. If you draw a line from the centre of the circle to the point where the chord is cut in half (the midpoint), then that line will always form a 90-degree angle with the chord.
Diagram: (Same diagram as before, but with different given information) ``` O / \ / \ / \ A-------M-------B \ / \ / \ / C ``` Where: `O` is the centre of the circle. `AB` is the chord. `AM = MB` (M is the midpoint of AB).
Therefore, `OM ⊥ AB`.
Proof: Given: Circle with centre O, and AM = M
B. To Prove: OM ⊥ A
B. Construction: Join OA and OB (radii).
Proof: In ΔOMA and ΔOMB: OA = OB (Radii of the same circle) OM = OM (Common side) AM = MB (Given) Therefore, ΔOMA ≡ ΔOMB (SSS - Side, Side, Side congruency) Therefore, ∠OMA = ∠OMB (Corresponding angles of congruent triangles are equal). ∠OMA + ∠OMB = 180° (Angles on a straight line).
Therefore, ∠OMA = ∠OMB = 90°.
Therefore, OM ⊥ A
B. Example 3: In a circle centre O, AB is a chord and M is the midpoint of A
B. If ∠OAM = 30°, determine ∠OM
A. Solution: OM ⊥ AB (Line from centre to midpoint of chord) ∠OMA = 90° Guided Practice (With Solutions)
Question 1: In a circle with centre O, chord CD has a length of 16cm. OM is drawn from the centre O to CD such that OM ⊥ C
D. Determine the length of C
M. Solution: CM = MD (Line from centre ⊥ chord) CD = 16cm CM = (1/2) CD = (1/2) * 16cm = 8cm
Commentary: This question directly applies the first theorem. We recognize that since OM is perpendicular to CD, it bisects C
D. Therefore, CM is half the length of C
D. Question 2: In a circle with centre O, radius OA = 13cm and chord AB = 24cm. M is the midpoint of A
B. Calculate the length of O
M. Solution: AM = (1/2) AB = (1/2) * 24cm = 12cm (M is the midpoint of AB) OM ⊥ AB (Line from centre to midpoint of chord) In right-angled triangle ΔOMA: OA² = OM² + AM² (Pythagoras) 13² = OM² + 12² 169 = OM² + 144 OM² = 169 - 144 = 25 OM = √25 = 5cm
Commentary: This question combines the converse of the theorem with Pythagoras' theorem. First, we use the fact that M is the midpoint to deduce that OM is perpendicular to A
B. Then, we use the Pythagorean theorem to find the unknown length O
M. Question 3: Circle with center O. Chord XY is 30cm long. The perpendicular distance from O to XY is 8cm. Calculate the length of the radius O
X. Solution: Let M be the point where the perpendicular from O meets XY. Then OM = 8cm. XM = (1/2) XY = (1/2) * 30cm = 15cm (Line from centre ⊥ chord bisects chord) ΔOMX is a right-angled triangle. OX² = OM² + XM² (Pythagoras) OX² = 8² + 15² OX² = 64 + 225 = 289 OX = √289 = 17cm
Commentary: Again, we're using the first theorem and combining it with the Pythagorean theorem. Recognizing the right-angled triangle and knowing two sides allows us to calculate the third (the radius). Independent Practice (Questions Only) In a circle with centre O, chord AB = 10cm and the radius of the circle is 13cm.