Exercise-9.2
Maths - Exercise
Q1. Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.
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 \( \boxed{6\, \text{cm}} \).
Q2. If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.
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.
Q3. If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.
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 \( CD \).
Q4. If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that AB = CD (see Fig. 9.12).
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 \( AB = CD \).
Q5. Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6m each, what is the distance between Reshma and Mandip?
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: \[ \boxed{9.6\,\text{m}} \]
Q6. A circular park of radius 20m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone.
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: \[ \boxed{34.64\,\text{m}} \]
