lesson 8.2 properties of chords in a circle
1. draw and label a diagram to illustrate the relationship between
a chord, its perpendicular bisector, and the centre of a circle.
2. point o is the centre of the circle.
determine the values of ( x^circ ) and ( y^circ ).
circle diagram with center o, triangle ocd, angle at d is 48°
3. point o is the centre of the circle; of = 18 cm; and gj = 14 cm.
determine the values of x and y to the nearest tenth of a centimetre where necessary.
circle diagram with center o, diameter of, chord gj, right angle at h
4. a circle has diameter 70 cm.
a chord in the circle is 50 cm long.
how far is the chord from the centre of the circle?
give the answer to the nearest tenth of a centimetre.
5. a circle has diameter 22 cm.
two chords are drawn on opposite sides of the centre of the circle.
one chord is 16 cm long and the other chord is 12 cm long.
a) which chord is closer to the centre of the circle?
b) how much closer to the centre is this chord?
give the answer to the nearest tenth of a centimetre.

Question

Accepted Answer

### Problem 2: Determine $ x^\circ $ and $ y^\circ $
# Explanation:
## Step 1: Identify Triangle Type
The triangle $ OCD $ (or as labeled) is a right triangle? Wait, no, the line from $ O $ to $ C $ is perpendicular to chord $ ED $ (since it's a perpendicular bisector of the chord). So $ OC \perp ED $, meaning $ \angle OCE = 90^\circ $. Wait, the angle at $ D $ is $ 40^\circ $? Wait, the diagram shows a triangle with $ O $, $ C $, $ D $, where $ OC $ is perpendicular to $ ED $, so $ \angle OCD = 90^\circ $. Then, in triangle $ OCD $, we have a right angle at $ C $, angle at $ D $ is $ 40^\circ $, so angle $ x $ (wait, the angle at $ C $ is $ 90^\circ $, angle at $ D $ is $ 40^\circ $, so angle $ y $ (at $ O $): Wait, maybe $ x $ is the right angle? Wait, no, let's re-examine. The chord $ ED $ has $ OC $ as perpendicular bisector, so $ OC \perp ED $, so $ \angle OCE = 90^\circ $, so $ x = 90^\circ $ (since it's a right angle). Then, in triangle $ OCD $, angles sum to $ 180^\circ $, so $ y = 180^\circ - 90^\circ - 40^\circ = 50^\circ $? Wait, maybe the angle at $ D $ is $ 40^\circ $, so:

## Step 1: Recognize Perpendicular Bisector
The line from center $ O $ to chord $ ED $ (at $ C $) is perpendicular, so $ \angle OCE = 90^\circ $, so $ x = 90^\circ $.

## Step 2: Calculate $ y $
In triangle $ OCD $, angles sum to $ 180^\circ $. We know $ \angle OCD = 90^\circ $ and $ \angle ODC = 40^\circ $, so $ y = 180^\circ - 90^\circ - 40^\circ = 50^\circ $.

# Answer:
$ x = 90^\circ $, $ y = 50^\circ $

### Problem 3: $ OF = 18 \, \text{cm} $, $ GJ = 14 \, \text{cm} $, find $ x $ and $ y $
# Explanation:
## Step 1: Identify $ y $ (radius)
$ OF $ is a radius, so $ y = OF = 18 \, \text{cm} $ (since $ OF $ is a radius of the circle).

## Step 2: Find $ x $ (half of chord $ GJ $)
The line $ OH $ (or $ OG $?) is perpendicular to chord $ GJ $, so it bisects $ GJ $. Thus, $ x = \frac{GJ}{2} = \frac{14}{2} = 7 \, \text{cm} $. Wait, but maybe $ x $ is the distance from center to chord? Wait, no, the diagram: $ GJ $ is a chord, $ OH $ is perpendicular to $ GJ $ at $ H $, so $ GH = HJ = \frac{14}{2} = 7 \, \text{cm} $, so $ x = 7 \, \text{cm} $, and $ y $ is the radius, $ OF = 18 \, \text{cm} $, so $ y = 18 \, \text{cm} $.

# Answer:
$ x = 7.0 \, \text{cm} $, $ y = 18.0 \, \text{cm} $

### Problem 4: Chord length 50 cm, diameter 70 cm, find distance from center
# Explanation:
## Step 1: Radius and half-chord
Radius $ r = \frac{70}{2} = 35 \, \text{cm} $. Half of the chord length: $ \frac{50}{2} = 25 \, \text{cm} $.

## Step 2: Apply Pythagorean theorem
Let $ d $ be the distance from center to chord. Then, $ d^2 + 25^2 = 35^2 $. So $ d^2 = 35^2 - 25^2 = 1225 - 625 = 600 $. Thus, $ d = \sqrt{600} \approx 24.5 \, \text{cm} $ (to nearest tenth).

# Answer:
$ \approx 24.5 \, \text{cm} $

### Problem 5a: Which chord is closer?
# Explanation:
For a circle, the longer the chord, the closer it is to the center (since the distance from center to chord is $ d = \sqrt{r^2 - (l/2)^2} $, where $ l $ is chord length; as $ l $ increases, $ d $ decreases). The chords are 16 cm and 12 cm. 16 cm is longer than 12 cm? Wait, no: 16 > 12? Wait, no, 16 cm chord: $ l = 16 $, 12 cm chord: $ l = 12 $. Wait, $ d = \sqrt{r^2 - (l/2)^2} $. Radius $ r = 22/2 = 11 \, \text{cm} $. For 16 cm chord: $ l/2 = 8 $, so $ d_1 = \sqrt{11^2 - 8^2} = \sqrt{121 - 64} = \sqrt{57} \approx 7.55 $. For 12 cm chord: $ l/2 = 6 $, so $ d_2 = \sqrt{11^2 - 6^2} = \sqrt{121 - 36} = \sqrt{85} \approx 9.22 $. Since $ 7.55 < 9.22 $, the 16 cm chord is closer.

# Answer:
The 16 cm long chord is closer.

### Problem 5b: How much closer?
# Explanation:
Calculate $ d_2 - d_1 $. $ d_2 \approx 9.22 $, $ d_1 \approx 7.55 $. So difference is $ 9.22 - 7.55 = 1.67 \approx 1.7 \, \text{cm} $ (to nearest tenth).

# Answer:
$ \approx 1.7 \, \text{cm} $

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QUESTION IMAGE

Question

Explanation:

Problem 2: Determine \( x^\circ \) and \( y^\circ \)

Step 1: Identify Triangle Type

Step 1: Recognize Perpendicular Bisector

Step 2: Calculate \( y \)

Step 1: Identify \( y \) (radius)

Step 2: Find \( x \) (half of chord \( GJ \))

Step 1: Radius and half-chord

Step 2: Apply Pythagorean theorem

Answer:

Problem 3: \( OF = 18 \, \text{cm} \), \( GJ = 14 \, \text{cm} \), find \( x \) and \( y \)