QUESTION 9

Question

9.1 In the diagram below, 𝑂 is the centre of the circle and 𝑃, 𝑆 and 𝑇 are
points on the circumference of the circle.
Figure 9.1: Circle with centre 𝑂.
Prove the theorem that states that 𝑆𝑂𝑇 = 2 × 𝑆𝑃𝑇. Redraw the diagram
above to show any necessary constructions.
(5)
9.2 In the diagram, 𝑂 is the centre of the circle with 𝐴𝐵||𝐶𝐷. 𝐶𝐷 = 6 𝑐𝑚,
𝐸𝑂 = 3 𝑐𝑚 and 𝐶𝑂 = 5 𝑐𝑚.
Figure 9.2: Circle with centre 𝑂.
Calculate, the distance between 𝐴𝐵 and 𝐶𝐷.
(6)
[11]

Answer 1

To address both parts of this question, let's proceed step by step:

9.1 Proving the theorem \( \angle SOT = 2 \times \angle SPT \)

Diagram:

1. Draw a circle with center \( O \).
2. Mark points \( P \), \( S \), and \( T \) on the circumference of the circle.
3. Draw lines \( OS \) and \( OT \) so they radiate from the center to points \( S \) and \( T \).
4. Draw segments \( SP \) and \( TP \).

Steps of Proof:

• Step 1: Recognize that \( OS \) and \( OT \) are radii of the circle. Thus, \( OS = OT \).

• Step 2: Consider triangle \( OSP \) and triangle \( OTP \):

  • Since \( OS = OT \) (radii of the circle), \( OP = OP \) (common side), this means \( OSP \) is isosceles.
  • Therefore, \( \angle OSP = \angle OSP \).
  • Using the properties of an isosceles triangle, we note that the angles opposite the equal sides are equal.

• Step 3: Using the inscribed angle theorem, we note that the angle \( \angle SPT \) is subtended by arc \( ST \). Hence by the theorem: \[ \angle SOT = 2 \times \angle SPT \]

This concludes the proof that \( \angle SOT = 2 \times \angle SPT \).

9.2 Calculating the distance between \( AB \) and \( CD \)

Given Data:

1. \( CD = 6 \) cm
2. \( EO = 3 \) cm (distance from \( E \) to center \( O \))
3. \( CO = 5 \) cm (radius from center \( O \) to point \( C \))

Understand the structure:

• Since \( AB \parallel CD \), both lines are horizontal (assuming a Cartesian plane), meaning we can use their perpendicular distance. The circle’s center \( O \) lies somewhere vertically relative to these lines.

Calculation Steps:

1. The line \( CD \) is at distance \( CO \) from the center \( O \).
2. The distance \( EO \) from \( E \) to \( O \) vertically aligns with the radius \( CO \).

Using the Pythagorean theorem in triangle \( EOC \): \[ EC^2 + EO^2 = CO^2 \] Substituting the values: \[ EC^2 + 3^2 = 5^2 \] \[ EC^2 + 9 = 25 \] \[ EC^2 = 16 \] \[ EC = 4 \text{ cm} \]

Distance Calculation:

• The distance (height difference) between parallel lines \( AB \) and \( CD \) is the distance \( EC \).
• Thus the distance between \( AB \) and \( CD = 4 \) cm.

Therefore, the final answers are as follows:

• 9.1: Proven using the properties of circles and isosceles triangles.
• 9.2: The distance between \( AB \) and \( CD \) is 4 cm.