Q1. Can you explain why the converse to Theorem 4 is true, i.e., why does the perpendicular from the centre of a circle to a chord of the circle bisect the chord? (Hint: Use Fig. 5.12. You are told that \( \angle \mathrm{CMA} = \angle \mathrm{CMB} = 90^{\circ} \). You need to show that \( \mathrm{AM} = \mathrm{BM} \).)

Answer:

Given: A circle with centre C, a chord AB, and CM is perpendicular to AB at M.

To prove: M is the midpoint of AB, i.e., AM = BM.

Consider triangles ACM and BCM.

• AC = BC (both are radii of the circle)
• CM = CM (common side)
• \( \angle CMA = \angle CMB = 90° \) (given, CM ⊥ AB)

By the RHS (Right angle-Hypotenuse-Side) congruence criterion, \( \triangle ACM \cong \triangle BCM \).

Therefore, AM = BM (corresponding parts of congruent triangles).

Hence, the perpendicular from the centre of a circle to a chord bisects the chord.

Q2. An isosceles triangle ABC is inscribed in a circle, with \( \mathrm{AB} = \mathrm{AC} \). Show that the altitude from A to BC passes through the centre of the circle.

Answer:

Let the altitude from A to BC meet BC at M.

In \( \triangle ABC \), AB = AC, so it is an isosceles triangle. In an isosceles triangle, the altitude from the vertex angle bisects the base. Therefore, BM = MC.

Since M is the midpoint of chord BC, by Theorem 4, the line joining the centre of the circle to the midpoint of a chord is perpendicular to the chord. So, the line through the centre O and M is perpendicular to BC.

But AM is also perpendicular to BC (as it's the altitude). Through a point M on BC, only one line can be perpendicular to BC. Therefore, AM is the same line as OM, which means the altitude AM passes through the centre O.

Q3. Two parallel chords of lengths 6 cm and 8 cm are on opposite sides of the centre of a circle. If the radius of the circle is 5 cm, find the distance between the midpoints of the chords.

Answer:

Let the circle have centre O. Let the chords be AB = 6 cm and CD = 8 cm.

• Distance from centre to chord AB: \( \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4 \text{ cm} \)
• Distance from centre to chord CD: \( \sqrt{5^2 - 4^2} = \sqrt{25 - 16} = \sqrt{9} = 3 \text{ cm} \)

Since the chords are on opposite sides of the centre, the distance between their midpoints is the sum of their distances from the centre: 4 cm + 3 cm = 7 cm.