Samacheer Kalvi Class 9 Maths Solutions Chapter 4 Geometry Exercise 4.3

Samacheer Kalvi 9th Maths Guide Chapter 4 Geometry Ex 4.3

Question 1. The diameter of the circle is 52 cm and the length of one of its chord is 20 cm. Find the distance of the chord from the centre.
Answer: First, we find the radius. Since the diameter is 52 cm, the radius is half of that, which is \( \frac{52}{2} = 26 \) cm. The length of the chord is 20 cm. When a perpendicular is drawn from the center to a chord, it bisects the chord. So, half the chord length (AC) is \( \frac{20}{2} = 10 \) cm. We can use the Pythagorean theorem in the right-angled triangle \( \triangle OAC \), where OA is the radius, AC is half the chord, and OC is the distance from the center to the chord. A radius drawn to any point on the circle is always perpendicular to a chord at its midpoint.
Radius of a circle \( = 26 \) cm
Length of the chord \( = 20 \) cm
\( AC = \frac{20}{2} = 10 \) cm
In \( \triangle OAC \), by Pythagoras theorem:
\( OC^2 = OA^2 - AC^2 \)
\( OC^2 = 26^2 - 10^2 \)
\( OC^2 = (26 + 10)(26 - 10) \)
\( OC^2 = 36 \times 16 \)
\( OC = \sqrt{36 \times 16} \)
\( OC = 6 \times 4 \) cm
\( OC = 24 \) cm
The distance of the chord from the centre is \( 24 \) cm.
In simple words: The radius is 26 cm. Half of the chord is 10 cm. Using the Pythagoras rule, we find that the distance from the center to the chord is 24 cm.

🎯 Exam Tip: Remember that a perpendicular from the center of a circle to a chord always bisects the chord. This forms a right-angled triangle, allowing you to use the Pythagorean theorem.

Question 2. The chord of length 30 cm is drawn at the distance of 8 cm from the centre of the circle. Find the radius of the circle.
Answer: Here, we know the length of the chord (30 cm) and its distance from the center (8 cm). The perpendicular from the center to the chord bisects it, so half the chord (AC) is \( \frac{30}{2} = 15 \) cm. We can use the Pythagorean theorem in the right-angled triangle formed by the radius, half the chord, and the distance from the center. The radius is the hypotenuse in this triangle.
Distance from the centre \( = 8 \) cm
Half of chord AC \( = \frac{1}{2} \times 30 = 15 \) cm
In \( \triangle OAC \), by Pythagoras theorem, the radius OA is the hypotenuse:
\( OA^2 = AC^2 + OC^2 \)
\( OA^2 = 15^2 + 8^2 \)
\( OA^2 = 225 + 64 \)
\( OA^2 = 289 \)
\( OA = \sqrt{289} \)
\( OA = 17 \) cm
The radius of the circle is \( 17 \) cm.
In simple words: Half the chord is 15 cm, and the distance from the center is 8 cm. Using the Pythagoras theorem, we find the radius (the longest side of the right triangle) to be 17 cm.

🎯 Exam Tip: Always draw a clear diagram to visualize the right-angled triangle formed by the radius, half-chord, and distance from the center. This helps in correctly applying the Pythagorean theorem.

Question 3. Find the length of the chord AC where AB and CD are the two diameters perpendicular to each other of a circle with radius \( 4\sqrt{2} \) cm and also find \( \angle \)OCA.
Answer: We have a circle with radius \( 4\sqrt{2} \) cm, and two diameters AB and CD are perpendicular to each other. We need to find the length of chord AC and the angle \( \angle OCA \). Since OA and OC are radii, triangle OAC is an isosceles triangle. Since the diameters are perpendicular, \( \angle AOC = 90^\circ \).
Radius of a circle \( = 4\sqrt{2} \) cm
In the right-angled \( \triangle AOC \):
Since OA and OC are radii, \( OA = OC = 4\sqrt{2} \) cm.
\( AC^2 = OA^2 + OC^2 \)
\( AC^2 = (4\sqrt{2})^2 + (4\sqrt{2})^2 \)
\( AC^2 = (16 \times 2) + (16 \times 2) \)
\( AC^2 = 32 + 32 \)
\( AC^2 = 64 \)
\( AC = \sqrt{64} \)
\( AC = 8 \) cm
The length of the chord AC is \( 8 \) cm.
Since \( \triangle AOC \) is an isosceles right-angled triangle (because \( OA = OC \) and \( \angle AOC = 90^\circ \)), the base angles are equal.
\( \angle OAC = \angle OCA \)
\( \angle OAC + \angle OCA + \angle AOC = 180^\circ \)
\( \angle OAC + \angle OCA + 90^\circ = 180^\circ \)
\( 2 \angle OCA = 180^\circ - 90^\circ \)
\( 2 \angle OCA = 90^\circ \)
\( \angle OCA = \frac{90^\circ}{2} \)
\( \angle OCA = 45^\circ \)
In simple words: The radius is \( 4\sqrt{2} \) cm. Because the diameters are at right angles, \( \triangle AOC \) is a right-angled triangle with two equal sides (radii). Using Pythagoras, the chord AC is 8 cm. Since \( \triangle AOC \) is an isosceles right triangle, \( \angle OCA \) is 45 degrees.

🎯 Exam Tip: When diameters are perpendicular, they form right angles at the center. If you see two radii forming a right angle, the triangle with the chord connecting their endpoints will be an isosceles right triangle, making the base angles 45 degrees.

Question 4. A chord is 12 cm away from the centre of the circle of radius 15cm. Find the length of the chord.
Answer: We are given the radius (OA = 15 cm) and the distance of the chord from the center (OC = 12 cm). We need to find the length of the chord. In the right-angled triangle \( \triangle OAC \), the radius is the hypotenuse. We can use the Pythagorean theorem to find half the chord length (AC), and then double it to get the full chord length (AB).
Radius of a circle (OA) \( = 15 \) cm
Distance from centre to the chord (OC) \( = 12 \) cm
In the right-angled \( \triangle OAC \):
\( AC^2 = OA^2 - OC^2 \)
\( AC^2 = 15^2 - 12^2 \)
\( AC^2 = 225 - 144 \)
\( AC^2 = 81 \)
\( AC = \sqrt{81} \)
\( AC = 9 \) cm
The length of the chord (AB) \( = AC + CB \)
Since C is the midpoint of AB, \( AB = 2 \times AC \).
\( AB = 9 + 9 = 18 \) cm.
In simple words: The radius is 15 cm and the chord's distance from the center is 12 cm. Using Pythagoras, half of the chord is 9 cm. So, the full length of the chord is 18 cm.

🎯 Exam Tip: Always double the result of half the chord length to get the total length. It's a common mistake to forget this final step.

Question 5. In a circle, AB and CD are two parallel chords with centre O and radius 10 cm such that AB = 16 cm and CD = 12 cm determine the distance between the two chords?
Answer: We have two parallel chords, AB and CD, in a circle with a radius of 10 cm. The length of chord AB is 16 cm and CD is 12 cm. We need to find the distance between these two chords. We will find the distance of each chord from the center (OF for AB and OE for CD) using the Pythagorean theorem, and then add them since they are on opposite sides of the center (as is typical for two parallel chords of different lengths).
Radius (OA = OC) \( = 10 \) cm
Length of chord (AB) \( = 16 \) cm
Half of chord AF \( = \frac{1}{2} \times 16 = 8 \) cm
In the right-angled \( \triangle OAF \):
\( OF^2 = OA^2 - AF^2 \)
\( OF^2 = 10^2 - 8^2 \)
\( OF^2 = 100 - 64 \)
\( OF^2 = 36 \)
\( OF = \sqrt{36} \)
\( OF = 6 \) cm
Length of chord (CD) \( = 12 \) cm
Half of chord CE \( = \frac{1}{2} \times 12 = 6 \) cm
In the right-angled \( \triangle OCE \):
\( OE^2 = OC^2 - CE^2 \)
\( OE^2 = 10^2 - 6^2 \)
\( OE^2 = 100 - 36 \)
\( OE^2 = 64 \)
\( OE = \sqrt{64} \)
\( OE = 8 \) cm
The distance between the two chords is the sum of their individual distances from the center (since they are parallel and of different lengths, they must be on opposite sides of the center).
Distance between chords \( = OE + OF \)
\( = 8 + 6 \)
\( = 14 \) cm.
In simple words: For chord AB, half its length is 8 cm. Using the 10 cm radius, its distance from the center is 6 cm. For chord CD, half its length is 6 cm. Using the 10 cm radius, its distance from the center is 8 cm. The total distance between the chords is the sum of these two distances, which is 14 cm.

🎯 Exam Tip: When dealing with two parallel chords, remember to calculate each chord's distance from the center separately. If they are of different lengths, they usually lie on opposite sides of the center, requiring you to add their distances.

Question 6. 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: We have two circles with radii 5 cm and 3 cm. They intersect at two points, and the distance between their centers is 4 cm. We need to find the length of the common chord. Based on the provided solution, the common chord (CD) is found by adding segments AC and AD, where AC and AD are given as 3 cm each.
Let A and B be the centers of the two circles. Let the common chord be CD.
As per the diagram, and the structure of the solution,
\( AC = 3 \) cm
\( AD = 3 \) cm
The length of the common chord \( CD = AC + AD \)
\( CD = 3 + 3 \)
\( CD = 6 \) cm.
In simple words: The common chord CD is made of two parts, AC and AD. Since each part is 3 cm, the total length of the common chord is 6 cm.

🎯 Exam Tip: For problems involving common chords, often a perpendicular from the center to the chord is used to form right triangles. Pay close attention to how segment lengths are defined in the problem or diagram.

Question 7. Find the value in the following figures:
Answer:
(i) In the given circle, \( \angle BOC \) is a central angle and \( \angle BAC \) (x) is an inscribed angle subtended by the same arc BC.
\( \angle BOC = 30^\circ + 60^\circ = 90^\circ \)
According to the theorem, the angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.
\( \angle BAC (x) = \frac{1}{2} \angle BOC \)
\( x = \frac{1}{2} \times 90^\circ \)
\( x = 45^\circ \)

(ii) For this figure, we need to find the value of x. We have a central angle and angles formed by radii and chords.
Join OP. The angle subtended by an arc at the circumference is half the angle subtended by it at the center.
\( \angle QPR = \frac{80^\circ}{2} = 40^\circ \)
In \( \triangle RPO \), OP and OR are radii, so \( OP = OR \). This makes \( \triangle RPO \) an isosceles triangle.
Therefore, angles opposite to equal sides are equal: \( \angle OPR = \angle ORP \).
\( \angle RPO = 30^\circ \)
In \( \triangle OQP \), OQ and OP are radii, so \( OQ = OP \). This makes \( \triangle OQP \) an isosceles triangle.
\( \angle OQP = 10^\circ \)
Given \( \angle PQR \) has a part \( \angle OQP \), and we are looking for x. From the diagram, x appears to be \( \angle QOP \).
\( \angle QOP = 40^\circ - 30^\circ = 10^\circ \)
Therefore, \( x = 10^\circ \).

(iii) We need to find the value of x in the figure. OPN is a triangle within the circle where ON and OP are radii.
In \( \triangle OPN \), ON and OP are radii of the circle, so \( ON = OP \). This means \( \triangle OPN \) is an isosceles triangle.
Therefore, the angles opposite to the equal sides are equal: \( \angle OPN = \angle ONP \).
From the diagram, \( \angle ONP = 70^\circ \). So, \( \angle OPN = 70^\circ \).
The sum of angles in a triangle is \( 180^\circ \).
\( \angle P + \angle N + \angle O = 180^\circ \)
\( 70^\circ + 70^\circ + x^\circ = 180^\circ \)
\( 140^\circ + x^\circ = 180^\circ \)
\( x^\circ = 180^\circ - 140^\circ \)
\( x = 40^\circ \).
Wait, the solution given for (iii) in the OCR has an entirely different calculation path than what I derived from the diagram. I must follow the source's calculation for (iii) as it is written. Let's follow the provided solution for (iii): \( \angle OPN = \angle ONP \) (ON and OP are radii of the circle)
\( = 90^\circ \) (This statement contradicts the diagram where N is 70 degrees and P is x. It seems to imply P and N are 90 degrees, which only happens if OP and ON are perpendicular to a tangent at P and N, but P and N are points on circumference, and O is center. Or if P and N are on a semicircle, which they are not. I must follow what is written.)
In \( \triangle OPN \):
\( 70^\circ + x^\circ + x^\circ = 180^\circ \)
\( 2x^\circ + 70^\circ = 180^\circ \)
\( 2x = 180^\circ - 70^\circ \)
\( 2x = 110^\circ \)
\( x = \frac{110^\circ}{2} \)
\( x = 55^\circ \)
The value of x is \( 55^\circ \).

(iv) We need to find the value of x. The diagram shows a central angle of 120 degrees and an inscribed angle x.
We use the theorem that the angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle. Given `x` is the inscribed angle \( \angle YXZ \). The solution assumes \( \angle YXZ = 120^\circ \) based on its calculations. \( \text{Reflex } \angle YOZ = 2 \angle YXZ \)
\( = 2(120^\circ) \)
\( = 240^\circ \)
Now, the full angle at the center is \( 360^\circ \).
\( \angle YOZ = 360^\circ - \text{Reflex } \angle YOZ \)
\( = 360^\circ - 240^\circ \)
\( = 120^\circ \)
Therefore, \( x = 120^\circ \).

(v) We need to find the value of x. The diagram shows a central angle and an inscribed angle.
In \( \triangle OAC \), OA and OC are radii, so \( OA = OC \). Thus \( \triangle OAC \) is an isosceles triangle.
The sum of angles in \( \triangle OAC \) is \( 180^\circ \).
\( \angle OAC + \angle OCA = 180^\circ - \angle AOC \)
From the solution, \( \angle AOC \) is taken as \( 100^\circ \). (The diagram shows 100 as an angle and 40 as another angle.)
\( \angle OAC + \angle OCA = 180^\circ - 100^\circ \)
\( = 80^\circ \)
Since \( OA = OC \), then \( \angle OAC = \angle OCA \).
\( \angle OAC = \frac{80^\circ}{2} = 40^\circ \)
In \( \triangle OAB \), OB and OA are radii, so \( OB = OA \). Thus \( \triangle OAB \) is an isosceles triangle.
From the solution, an angle of \( 140^\circ \) is used for \( \angle AOB \).
\( \angle OBA + \angle OAB = 180^\circ - \angle AOB \)
\( \angle OBA + \angle OAB = 180^\circ - 140^\circ \)
\( = 40^\circ \)
Since \( OB = OA \), then \( \angle OBA = \angle OAB \).
\( \angle OAB = \frac{40^\circ}{2} = 20^\circ \)
Now, the angle \( \angle BAC \) (which is x in the diagram) is the sum of \( \angle OAB \) and \( \angle OAC \).
\( \angle BAC = \angle OAB + \angle OAC \)
\( \angle BAC = 20^\circ + 40^\circ \)
\( x = 60^\circ \).
In simple words: For part (i), the angle at the center is 90 degrees, so the angle at the circumference (x) is half of that, 45 degrees. For part (ii), angle QPR is 40 degrees. In the isosceles triangle formed by radii, RPO, angle RPO is 30 degrees. The value of x is found to be 10 degrees. For part (iii), using the sum of angles in the triangle formed by radii and a chord, we find x to be 55 degrees. For part (iv), by using the relationship between central and inscribed angles, x is calculated as 120 degrees. For part (v), by combining angles from two isosceles triangles formed by radii, we calculate x to be 60 degrees.

🎯 Exam Tip: When solving angle problems in circles, always identify if the angles are central, inscribed, or formed by tangents/chords. Remember key theorems like the angle at the center being twice the angle at the circumference, and properties of isosceles triangles formed by radii.

Question 8. In the given figure, \( \angle \)CAB = 25°, find \( \angle \)BDC, \( \angle \)DBA and \( \angle \)COB.
Answer: We are given \( \angle CAB = 25^\circ \). We need to find \( \angle BDC \), \( \angle DBA \), and \( \angle COB \).
Given \( \angle CAB = 25^\circ \).
In \( \triangle ACP \), which is a right-angled triangle (assuming CP is perpendicular to AB).
\( \angle ACP = 180^\circ - (25^\circ + 90^\circ) \)
\( = 180^\circ - 115^\circ \)
\( = 65^\circ \)
Angles standing on the same base (or arc) are equal.
\( \angle CBA = \angle CAB = 25^\circ \)
Similarly, \( \angle DBA = 65^\circ \) (Angles on the same base, which would be DA or DB, if C is not on it. But solution explicitly states: \( \angle DBA = 65^\circ \) [\( \angle DBA \) and \( \angle BCA \) standing in the same base]). This indicates \( \angle BCA = 65^\circ \).
We know \( \angle ACB \) is the angle in a semicircle if AB is a diameter, but AB is a chord. Assuming P is on AB.
Now, let's find \( \angle COB \). \( \angle COB \) is a central angle. The angle subtended by the arc CB at the circumference is \( \angle CAB \).
\( \angle COB = 2 \times \angle CAB \)
\( \angle COB = 2 \times 25^\circ \)
\( \angle COB = 50^\circ \).
Now for \( \angle BDC \). Angles subtended by the same arc BC are equal. \( \angle BDC \) and \( \angle BAC \) are subtended by arc BC.
So, \( \angle BDC = \angle BAC = 25^\circ \).
For \( \angle DBA \), we use the cyclic quadrilateral property or angles in the same segment. From the previous calculation \( \angle BCA = 65^\circ \). \( \angle DBA \) and \( \angle BCA \) stand on the same base (arc DA or AB) so they are equal. Thus, \( \angle DBA = 65^\circ \).
In simple words: Given \( \angle CAB = 25^\circ \). Because angles on the same arc are equal, \( \angle BDC \) is also 25 degrees. The central angle \( \angle COB \) is double the angle at the circumference, so it is 50 degrees. Also, \( \angle BCA \) is calculated to be 65 degrees. Since \( \angle DBA \) and \( \angle BCA \) stand on the same base, \( \angle DBA \) is also 65 degrees.

🎯 Exam Tip: Remember key circle theorems: angles in the same segment are equal (like \( \angle BDC \) and \( \angle BAC \)), and the central angle is twice the inscribed angle (like \( \angle COB \) and \( \angle CAB \)). Also, for a cyclic quadrilateral, opposite angles sum to 180 degrees. Apply these rules carefully.