Given:

• Radius of first circle, \( r_1 = 5 \text{ cm} \)
• Radius of second circle, \( r_2 = 3 \text{ cm} \)
• Distance between centers, \( d = 4 \text{ cm} \)

To Find: Length of the common chord \( PQ \) of the two circles.

Solution:

Let the centers of the circles be \( O \) and \( O' \), and the common chord \( PQ \) intersect the line segment \( OO' \) at point \( M \). Since \( PQ \) is the common chord, it is perpendicular to \( OO' \) at \( M \).

Using the right triangles formed, we apply the following steps:

The distance \( OM \) from center \( O \) to \( M \) is given by:

\[ OM = \frac{r_1^2 - r_2^2 + d^2}{2d} \]

Substituting values:

\[\begin{aligned} OM &= \frac{5^2 - 3^2 + 4^2}{2 \times 4} \\\\&= \frac{25 - 9 + 16}{8} \\\\&= \frac{32}{8} \\\\&= 4 \text{ cm} \end{aligned}\]

Length of the perpendicular from center \( O \) to chord \( PQ \) is:

\[\begin{aligned} PM &= \sqrt{r_1^2 - OM^2} \\&= \sqrt{5^2 - 4^2} \\&= \sqrt{25 - 16} \\&= \sqrt{9} \\&= 3 \text{ cm} \end{aligned}\]

As \( PQ \) is the chord bisected at \( M \), the full length is:

\[ PQ = 2 \times PM = 2 \times 3 = 6 \text{ cm} \]

Therefore, the length of the common chord is \( \boxed{6\, \text{cm}} \).

Solution:

To Prove: If two equal chords \( AB \) and \( CD \) intersect at point \( Q \) inside the circle, then the segments satisfy \[ QD = QB \] where \( QD \) and \( QB \) are segments on the chords.

Construction: Draw perpendiculars from the center \( O \) of the circle to both chords, meeting them at points \( S \) and \( T \) respectively.

Proof:

• In triangles \( \triangle SQO \) and \( \triangle TRO \): \[ OQ = OQ \quad (\text{common side}) \]
• Since the chords \( SQ \) and \( TR \) are equal, the perpendicular distances from the center to these chords are equal: \[ OS = OT \]
• The angles at \( S \) and \( T \) are right angles, because the radius is perpendicular to the chord: \[ \angle S = \angle T = 90^\circ \]

By the RHS (Right angle-Hypotenuse-Side) congruence criterion:

\[ \triangle SQO \cong \triangle TRO \]

Therefore, by CPCT (Corresponding Parts of Congruent Triangles):

\[ SQ = TQ \]

Because \( Q \) is the intersection point of the chords, the segments satisfy:

\[ SB = TB \]

Adding equal segments on both sides yields:

\[ \begin{aligned}SB + SQ &= TB + TQ \\\implies QD &= QB\end{aligned} \]

Hence, the segments of one chord are equal to the corresponding segments of the other chord.

Solution:

Given: Two equal chords \( AB \) and \( CD \) of a circle intersect at point \( P \).

To Prove:

\[ \angle QPO = \angle RPO \] where points \( Q \) and \( R \) lie on chords \( AB \) and \( CD \), respectively, such that \( PQ \) and \( PR \) are segments from \( P \) to the ends of the chords.

Proof:

Consider triangles \( \triangle PQO \) and \( \triangle PRO \), where \( O \) is the center of the circle.

• \( PO = PO \) (Common side)
• Since \( AB \) and \( CD \) are equal chords, their perpendicular distances from the center \( O \) are equal:
  \[ OQ = OR \]
• Also, the radii \( OQ \) and \( OR \) are perpendicular to chords \( AB \) and \( CD \) respectively:
  \[ \angle PQO = \angle PRO = 90^\circ \]

By the RHS (Right angle-Hypotenuse-Side) congruence criterion:

\[ \triangle PQO \cong \triangle PRO \]

Therefore, corresponding angles are equal by CPCT:

\[ \therefore \angle QPO = \angle RPO \]

Hence, the line joining the intersection point \( P \) to the center \( O \) makes equal angles with the two equal chords \( AB \) and \( CD \).

Solution:

Given: Two concentric circles with center \( O \). A line intersects the outer circle at points \( A \) and \( D \), and the inner circle at points \( B \) and \( C \) (see Fig. 9.12).

To Prove:

\[ AB = CD \]

Construction: Draw a perpendicular \( OP \) from the center \( O \) to the line \( AD \), meeting it at point \( P \).

Proof:

The perpendicular from the center of a circle to a chord bisects the chord. Therefore:

• Since \( OP \perp AD \), and \( AD \) is a chord of the outer circle: \[ AP = PD \tag{i} \]
• Since \( OP \perp BC \), and \( BC \) is a chord of the inner circle: \[ BP = PC \tag{ii} \]

Now, subtracting equation (ii) from equation (i):

\[ AP - BP = PD - PC \]

This simplifies to:

\[ AB = CD \]

Hence proved that the segments \( AB = CD \).

Solution:

Given: Circle of radius \( 5\,\text{m} \).
\( RS \) and \( SM \) are equal chords of length \( 6\,\text{m} \).
\( O \) is the center, \( R, S, M \) are points on the circle.
We need the distance \( RM \) between Reshma and Mandip.

Draw perpendiculars from \( O \) to chords \( RS \) that bisect these chords.

• Let the midpoint of \( RS \) be \( P \), then \( RP = PS = 3\,\text{m} \) (half the chord).
• Since \( O \) is the center, \( OR = OS = 5\,\text{m} \) (radius).
• Using Pythagoras' theorem in right triangle \( ORP \):

\[ \begin{aligned} OP &= \sqrt{OR^2 - RP^2}\\ &= \sqrt{25 - 9}\\ &= \sqrt{16} \\&= 4\,\text{m} \end{aligned}\]

Area of triangle \( ORS \) using base \( RS = 6\,\text{m} \) and height \( OP = 4\,\text{m} \): \[\begin{aligned} \text{Area} &= \frac{1}{2} \times 6 \times 4 \\&= 12\,\text{m}^2 \end{aligned}\]

Area of triangle \( ORM \) using base \( OR = 5\,\text{m} \) and height \( a \) (to be found): \[\begin{aligned} \frac{1}{2} \times 5 \times a &= 12 \\5a&= 24 \\\implies a &= \frac{24}{5} \\&= 4.8\,\text{m} \end{aligned}\]

Since the triangle is isosceles with the maximum height from \( O \) to \( RM \), and the configuration is such that the height is bisected, the required distance: \[\begin{aligned} RM &= 2a \\&= 2 \times 4.8 \\&= 9.6\,\text{m} \end{aligned}\]

Therefore, the distance between Reshma and Mandip is: \[ \boxed{9.6\,\text{m}} \]

Solution:

Given: The radius of the circular park is \( 20\,\text{m} \).
Boys Ankur, Syed, and David sit at equal distances on the circumference.

• Let the vertices of the equilateral triangle on the circle be \( A \), \( S \), and \( D \).
• The center \( O \) is the centroid of the equilateral triangle.
• Let the side of the equilateral triangle be \( 2a \).

The centroid divides the median in ratio \( 2:1 \), so: \[ OA : AP = 2:1 \] where \( OA \) is the radius and \( AP \) is one-third of the median.

The distance from the center to a side (the centroid to side) is: \[ OP = \frac{1}{3} \times \text{median} \]

Therefore, \[ OA = 2 \times OP \] \[ 20 = 2x \implies x = 10\,\text{m} \] Height of \(\triangle ASP\) \[20+10=30\,m\]

Using Pythagoras Theorem

In \(\triangle ASP \): \[ \begin{aligned} (2a)^2 &= a^2 + (30)^2\\ 4a^2 &= a^2 + 900\\ 4a^2 - a^2 &= 900\\ 3a^2 &= 800\\ a^2 &= \frac{900}{3}\\ a &= \sqrt{300} \\&= 10\sqrt{3} \\&\approx 17.32\,\text{m}\\ 2a &= 2 \times 17.32 \\&\approx 34.64\,\text{m} \end{aligned} \]

Length of String

Each side of the equilateral triangle is the length of string between each pair: \[ \text{Length of each string} = 2a \approx 34.64\,\text{m} \]

Therefore, the length of the string of each telephone is: \[ \boxed{34.64\,\text{m}} \]