GSEB Class 9 Maths Solutions Chapter 10 Circles Exercise 10.4

Question 1. 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.
Answer: Let two circles with centers \( O \) and \( O' \) cross each other at points \( P \) and \( Q \). So, \( OP = OQ = 5 \text{ cm} \), \( O'P = O'Q = 3 \text{ cm} \) and \( OO' = 4 \text{ cm} \). In triangle \( O O' P \), we find that \( OO'^2 + O'P^2 = 4^2 + 3^2 = 16 + 9 = 25 \). This equals \( 5^2 \), which is also \( OP^2 \). Therefore, angle \( O O' P = 90^\circ \) by the converse of the Pythagoras Theorem. We know that \( OO' \) is perpendicular to \( PQ \). If two circles cross each other at two points, the line connecting their centers is the line that cuts their common chord \( PQ \) in half at a right angle. This means \( OO' \) is the same as \( OM \), and \( PQ \) goes through the center \( O' \). The length of the common chord \( PQ \) is \( 2 \times O'P = 2 \times 3 \text{ cm} = 6 \text{ cm} \).
In simple words: To find the common chord's length, we use the radii and the distance between centers to form a right-angled triangle. Applying the Pythagorean theorem helps us calculate the required length.

Exam Tip: Remember that the line joining the centers of two intersecting circles is the perpendicular bisector of their common chord.

Question 2. 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.
Answer:
Given: A circle with center \( O \) has two equal chords, \( AB \) and \( CD \), which cross each other inside the circle at point \( E \).
To Prove: \( AE = DE \) and \( CE = BE \).
Construction: Draw a line segment \( OM \) perpendicular to \( AB \), and a line segment \( ON \) perpendicular to \( CD \). Also, connect point \( O \) to point \( E \).
Proof:
Consider the triangles \( \triangle OME \) and \( \triangle ONE \).
1. \( OM = ON \) (Because equal chords of a circle are always the same distance from the center.)
2. \( OE = OE \) (This is a common side for both triangles.)
Therefore, \( \triangle OME \cong \triangle ONE \) by the RHS (Right angle, Hypotenuse, Side) congruence rule.
This means \( ME = NE \) (These are corresponding parts of congruent triangles, CPCT).
Now, we can write \( AM + ME = DN + NE \).
We know that \( AM = DN \) because \( AM = \frac{1}{2} AB \) and \( DN = \frac{1}{2} CD \), and it is given that \( AB = CD \).
From \( AM + ME = DN + NE \), since \( AM = DN \) and \( ME = NE \), it follows that \( AE = DE \).
Also, \( BE = AB - AE \). Since \( AB = CD \), we can write \( BE = CD - AE \).
Since \( AE = DE \), then \( BE = CD - DE \).
Therefore, \( BE = CE \).
In simple words: When two chords of the same length cross inside a circle, the point where they cross divides them so that the parts of one chord are equal to the matching parts of the other chord. We prove this by showing that two small triangles formed are identical using the RHS rule.

Exam Tip: When proving congruence, clearly state the congruence rule (RHS, SSS, SAS, ASA, AAS) and which parts correspond (CPCT) for full marks.

Question 3. 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.
Answer:
Given: A circle with center \( O \) has two equal chords, \( AB \) and \( CD \), which cross each other inside the circle. Their crossing point is \( E \).
To Prove: The line joining the intersection point \( E \) to the center \( O \) forms equal angles with the chords, meaning \( \angle OEA = \angle OED \).
Construction: Connect point \( O \) to point \( A \) and point \( O \) to point \( D \).
Proof:
Consider the triangles \( \triangle OEA \) and \( \triangle OED \).
1. \( OE = OE \) (This side is common to both triangles.)
2. \( OA = OD \) (These are radii of the same circle, so they are equal in length.)
3. \( AE = DE \) (This was proven in Question 2 above, where equal chords intersect.)
Therefore, \( \triangle OEA \cong \triangle OED \) by the SSS (Side-Side-Side) congruence rule.
This means \( \angle OEA = \angle OED \) (These are corresponding parts of congruent triangles, CPCT).
**Note:** Alternatively, you could draw \( OP \perp AB \) and \( OQ \perp CD \), then show that \( \triangle OPE \cong \triangle OQE \) to prove \( \angle OEP = \angle OEQ \).
In simple words: This question asks us to show that when two equal chords cross, the line from the circle's center to the crossing point creates the same angle with both chords. We prove this by comparing two triangles and using the SSS rule because we already know the chords' segments are equal.

Exam Tip: For angle proofs involving chords, remember that equal chords subtend equal angles at the center and are equidistant from the center. These properties are often key to congruence proofs.

Question 4. 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 figure].
Answer:
Construction: Draw a line segment \( OM \) perpendicular to the chord \( BC \).
Proof:
We know that a perpendicular line drawn from the center of a circle to a chord always cuts the chord into two equal halves.
For the larger circle, \( AD \) is a chord. Since \( OM \perp AD \), it means \( M \) is the midpoint of \( AD \).
So, \( AM = DM \) ..........(1)
For the smaller circle, \( BC \) is a chord. Since \( OM \perp BC \), it means \( M \) is the midpoint of \( BC \).
So, \( BM = CM \) ..........(2)
Now, if we subtract equation (2) from equation (1), we get:
\( AM - BM = DM - CM \)
This simplifies to:
\( AB = CD \)
Hence, proved.
In simple words: When a straight line cuts across two circles that share the same center, the parts of the line outside the inner circle are equal in length. We show this by drawing a line from the center perpendicular to the cutting line, which divides both chord segments equally, allowing us to subtract and prove \( AB = CD \).

Exam Tip: Remember the key property that a perpendicular from the center to a chord bisects the chord. This is fundamental for problems involving chords and distances from the center.

Question 5. Three girls Reshma, Salina and Mandip are playing a game by standing on a circle of radius 5 m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salina and between Salina and Mandip is 6 m each, what is the distance between Reshma and Mandip?
Answer: Let R, S, and M denote the positions of Reshma, Salina, and Mandip on the circular park. The radius of the circle is \( OR = OS = OM = 5 \text{ m} \). The problem states that the distance between Reshma and Salina is \( RS = 6 \text{ m} \), and the distance between Salina and Mandip is \( SM = 6 \text{ m} \). We need to determine the distance between Reshma and Mandip, which is \( RM \).
First, let's consider the triangle \( \triangle ORS \). It is an isosceles triangle because \( OR = OS = 5 \text{ m} \).
Let \( K \) be the midpoint of the chord \( RS \). So, \( RK = SK = \frac{RS}{2} = \frac{6}{2} = 3 \text{ m} \). The line segment \( OK \) from the center to the midpoint of the chord is perpendicular to the chord.
Using the Pythagorean theorem in the right-angled triangle \( \triangle ORK \):
\( OK^2 + RK^2 = OR^2 \)
\( OK^2 + 3^2 = 5^2 \)
\( OK^2 + 9 = 25 \)
\( OK^2 = 16 \)
\( OK = 4 \text{ m} \).
Now, let's find the angle \( \angle ROS \) using the cosine rule in \( \triangle ORS \):
\( RS^2 = OR^2 + OS^2 - 2(OR)(OS) \cos(\angle ROS) \)
\( 6^2 = 5^2 + 5^2 - 2(5)(5) \cos(\angle ROS) \)
\( 36 = 25 + 25 - 50 \cos(\angle ROS) \)
\( 36 = 50 - 50 \cos(\angle ROS) \)
\( 50 \cos(\angle ROS) = 50 - 36 \)
\( 50 \cos(\angle ROS) = 14 \)
\( \cos(\angle ROS) = \frac{14}{50} = \frac{7}{25} \).
Similarly, for \( \triangle OSM \), since \( OS = OM = 5 \text{ m} \) and \( SM = 6 \text{ m} \), the triangle \( \triangle OSM \) is congruent to \( \triangle ORS \) (by SSS rule). Thus, \( \angle SOM = \angle ROS \), which means \( \cos(\angle SOM) = \frac{7}{25} \).
The total angle \( \angle ROM = \angle ROS + \angle SOM \).
Therefore, \( \angle ROM = 2 \times \angle ROS \).
We can use the double angle identity for cosine: \( \cos(2\theta) = 2\cos^2\theta - 1 \).
So, \( \cos(\angle ROM) = 2 \left( \frac{7}{25} \right)^2 - 1 \)
\( \cos(\angle ROM) = 2 \left( \frac{49}{625} \right) - 1 = \frac{98}{625} - 1 = \frac{98 - 625}{625} = -\frac{527}{625} \).
Finally, to find \( RM \), we apply the cosine rule to \( \triangle ROM \):
\( RM^2 = OR^2 + OM^2 - 2(OR)(OM) \cos(\angle ROM) \)
\( RM^2 = 5^2 + 5^2 - 2(5)(5) \left( -\frac{527}{625} \right) \)
\( RM^2 = 25 + 25 - 50 \left( -\frac{527}{625} \right) \)
\( RM^2 = 50 + \frac{50 \times 527}{625} \)
\( RM^2 = 50 + \frac{2 \times 527}{25} \)
\( RM^2 = \frac{1250}{25} + \frac{1054}{25} = \frac{2304}{25} \)
\( RM = \sqrt{\frac{2304}{25}} = \frac{48}{5} = 9.6 \text{ m} \).
Therefore, the distance between Reshma and Mandip is \( 9.6 \text{ m} \).
In simple words: We are looking for the distance between two friends (RM) on a circle. We know the circle's radius and the distances between other friends (RS and SM). By using the Pythagorean theorem and the cosine rule, we can find the angles at the center and then calculate the final distance.

Exam Tip: For problems involving distances on a circle, remember that all points on the boundary are equidistant from the center. Using the cosine rule in triangles formed by radii and chords is often an effective strategy.

Question 6. A circular park of radius 20 m 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.
Answer: Let the three boys Ankur, Syed, and David be represented by points A, B, and C on the boundary of the circular park. Since they are sitting at equal distances, \( \triangle ABC \) forms an equilateral triangle inscribed in the circle. Let \( O \) be the center of the park (circle), and its radius is \( R = 20 \text{ m} \). The length of the string of each phone will be the side length of the equilateral triangle, which is \( AB = BC = AC \).
**Method 1: Using Area of Triangle**
Let \( D \) be the midpoint of \( BC \). So \( BD = DC \). Let \( BD = x \text{ m} \).
The line segment \( OD \) is perpendicular to \( BC \). Consider the right-angled triangle \( \triangle ODB \).
\( OB \) is the radius of the circle, so \( OB = 20 \text{ m} \).
Using the Pythagorean theorem in \( \triangle ODB \):
\( OB^2 = OD^2 + BD^2 \)
\( 20^2 = OD^2 + x^2 \)
\( 400 = OD^2 + x^2 \)
Therefore, \( OD^2 = 400 - x^2 \), which means \( OD = \sqrt{400 - x^2} \).
The side length of the equilateral triangle \( ABC \) is \( BC = 2 \times BD = 2x \text{ m} \).
The area of an equilateral triangle with side 's' is \( \frac{\sqrt{3}}{4} s^2 \).
So, Area\( (\triangle ABC) = \frac{\sqrt{3}}{4} (2x)^2 = \frac{\sqrt{3}}{4} (4x^2) = \sqrt{3} x^2 \) ..........(1)
Alternatively, the area of \( \triangle ABC \) can also be expressed as the sum of the areas of three smaller triangles: \( \triangle OAB \), \( \triangle OBC \), and \( \triangle OAC \). Since \( \triangle ABC \) is equilateral and \( O \) is the circumcenter, these three smaller triangles are congruent.
So, Area\( (\triangle ABC) = 3 \times \text{Area}(\triangle OBC) \).
Area\( (\triangle OBC) = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times BC \times OD \).
Area\( (\triangle ABC) = 3 \times \frac{1}{2} \times (2x) \times \sqrt{400 - x^2} = 3x \sqrt{400 - x^2} \) ..........(2)
Now, equate the two expressions for Area\( (\triangle ABC) \) from (1) and (2):
\( \sqrt{3} x^2 = 3x \sqrt{400 - x^2} \)
Divide both sides by \( x \) (since \( x \neq 0 \)):
\( \sqrt{3} x = 3 \sqrt{400 - x^2} \)
Divide by \( \sqrt{3} \):
\( x = \sqrt{3} \sqrt{400 - x^2} \)
Squaring both sides:
\( x^2 = 3 (400 - x^2) \)
\( x^2 = 1200 - 3x^2 \)
\( x^2 + 3x^2 = 1200 \)
\( 4x^2 = 1200 \)
\( x^2 = \frac{1200}{4} = 300 \)
\( x = \sqrt{300} = \sqrt{100 \times 3} = 10\sqrt{3} \text{ m} \).
Since \( BC = 2x \), the length of the string of each phone is \( BC = 2 \times 10\sqrt{3} = 20\sqrt{3} \text{ m} \).
**Method 2: Using Properties of Centroid (Alternative Method)**
In an equilateral triangle, the centroid, orthocenter, incenter, and circumcenter all coincide. So, \( O \) is also the centroid.
The centroid divides the median in the ratio 2:1. Let \( AD \) be a median (altitude in an equilateral triangle).
So, \( AO : OD = 2 : 1 \).
Since \( AO \) is the radius \( R = 20 \text{ m} \), we have \( OD = \frac{1}{2} AO = \frac{1}{2} \times 20 = 10 \text{ m} \).
Now consider the right-angled triangle \( \triangle ODB \).
\( OB^2 = OD^2 + BD^2 \)
\( 20^2 = 10^2 + BD^2 \)
\( 400 = 100 + BD^2 \)
\( BD^2 = 300 \)
\( BD = \sqrt{300} = 10\sqrt{3} \text{ m} \).
The side length of the equilateral triangle \( BC = 2 \times BD = 2 \times 10\sqrt{3} = 20\sqrt{3} \text{ m} \).
Thus, the length of the string of each phone is \( 20\sqrt{3} \text{ m} \).
In simple words: Three boys are equally spaced on a circular park, forming an equilateral triangle. We need to find the length of the sides of this triangle. By using the circle's radius and properties of triangles (like the Pythagorean theorem or centroid properties), we can calculate the side length.

Exam Tip: For problems involving inscribed equilateral triangles, remember that the center of the circle is also the centroid of the triangle, which divides the medians in a 2:1 ratio. This shortcut can simplify calculations significantly.