Let PQ and RS be two parallel chords of a given circle of radius 6 cm, lying on the same side of the center. If the chords subtends angles of `72^(@)`and `144^(@)` at the center and the distance between the chords is d, then `d^(2)` is equal to

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Geometry
We have a circle with radius \( R = 6 \) cm and two parallel chords \( PQ \) and \( RS \) that subtend angles of \( 72^\circ \) and \( 144^\circ \) at the center \( O \) of the circle. We need to find the distance \( d \) between these two chords and then calculate \( d^2 \).

### Step 2: Calculate the Angles at the Center
- The angle subtended by chord \( PQ \) at the center \( O \) is \( 72^\circ \).
- The angle subtended by chord \( RS \) at the center \( O \) is \( 144^\circ \).

### Step 3: Determine the Angles for the Triangles
- For chord \( PQ \):
- The angle \( \angle POQ = 72^\circ \).
- The angle \( \angle POM = \frac{72^\circ}{2} = 36^\circ \) (where \( M \) is the foot of the perpendicular from \( O \) to \( PQ \)).
- Thus, \( \angle OPM = 90^\circ - 36^\circ = 54^\circ \).

- For chord \( RS \):
- The angle \( \angle ROS = 144^\circ \).
- The angle \( \angle NOR = \frac{144^\circ}{2} = 72^\circ \) (where \( N \) is the foot of the perpendicular from \( O \) to \( RS \)).
- Thus, \( \angle ORN = 90^\circ - 72^\circ = 18^\circ \).

### Step 4: Use Sine to Find Distances
- In triangle \( OPQ \):
\[ OM = OP \cdot \sin(54^\circ) = R \cdot \sin(54^\circ) \]

- In triangle \( ORS \):
\[ ON = OR \cdot \sin(18^\circ) = R \cdot \sin(18^\circ) \]

### Step 5: Calculate the Distance \( d \)
The distance \( d \) between the two chords is given by:
\[ d = OM - ON = R \cdot \sin(54^\circ) - R \cdot \sin(18^\circ) \]
Factoring out \( R \):
\[ d = R \left( \sin(54^\circ) - \sin(18^\circ) \right) \]

### Step 6: Substitute the Radius
Substituting \( R = 6 \) cm:
\[ d = 6 \left( \sin(54^\circ) - \sin(18^\circ) \right) \]

### Step 7: Calculate \( d^2 \)
To find \( d^2 \):
\[ d^2 = \left( 6 \left( \sin(54^\circ) - \sin(18^\circ) \right) \right)^2 = 36 \left( \sin(54^\circ) - \sin(18^\circ) \right)^2 \]

### Step 8: Use Sine Values
Using known sine values:
- \( \sin(54^\circ) = \frac{\sqrt{5} + 1}{4} \)
- \( \sin(18^\circ) = \frac{\sqrt{5} - 1}{4} \)

Substituting these values:
\[ d = 6 \left( \frac{\sqrt{5} + 1}{4} - \frac{\sqrt{5} - 1}{4} \right) \]
\[ = 6 \left( \frac{2}{4} \right) = 6 \cdot \frac{1}{2} = 3 \]

Thus,
\[ d^2 = 3^2 = 9 \]

### Final Answer
\[ d^2 = 9 \]