Circles — Karnataka (SSLC) Class 9 Mathematics Solutions (Free)

Step 1: Definition of Circle: A circle is the locus of all points in a plane that are at a constant distance (radius) from a fixed point (center). Step 2: Properties: (1) All radii of a circle are equal: Since all points on the circle are equidistant from the center, all radii r are equal. (2) Chord: A line segment with both endpoints on the circle. Examples: diameter (through center), any other line segment between two points on circle. (3) Diameter is the longest chord: Diameter = 2r (passes through center). Any other chord < diameter (doesn't pass through center). Step 3: Given: radius r = 7 cm Step 4: Calculate diameter: Diameter d = 2r = 2 × 7 = 14 cm Step 5: Calculate circumference (perimeter of circle): Circumference C = 2πr C = 2 × π × 7 C = 14π cm C ≈ 14 × 3.14159 C ≈ 43.98 cm or approximately 44 cm Step 6: Alternative circumference formula: C = πd = π × 14 = 14π cm ✓ Answer: Diameter = 14 cm; Circumference = 14π cm ≈ 43.98 cm.

Step 1: Theorem: The angle subtended by an arc at the center of a circle is twice the angle subtended by the same arc at any point on the circle (on the major arc). Angle at center = 2 × Angle at circumference Step 2: Given: Arc AB subtends angle AOB = 60° at center O Point C is on the major arc AB Find: angle ACB (angle at circumference) Step 3: Apply the theorem: angle AOB = 2 × angle ACB 60° = 2 × angle ACB angle ACB = 60°/2 angle ACB = 30° Step 4: Verification: This theorem holds for all positions of point C on the major arc. If C moves along the major arc, angle ACB remains 30°. Step 5: Special case - Angle in a semicircle: If arc AB is a diameter (semicircle), angle at center = 180° Angle at circumference = 180°/2 = 90° This means any angle inscribed in a semicircle is a right angle. Answer: The angle subtended at the circumference (point C) = 30°.

Step 1: Given: Circle with center O and radius r = 5 cm Chord AB at perpendicular distance d = 3 cm from center O Step 2: Let M be the foot of the perpendicular from O to chord AB. Then OM ⊥ AB and OM = 3 cm Step 3: Important property: The perpendicular from the center of a circle to a chord bisects the chord. Therefore: AM = MB = AB/2 Step 4: In right triangle OMA: angle OMA = 90° OA = 5 cm (radius, hypotenuse) OM = 3 cm (perpendicular distance) AM = ? (half of chord) Step 5: By Pythagoras theorem: OA² = OM² + AM² 5² = 3² + AM² 25 = 9 + AM² AM² = 25 - 9 AM² = 16 AM = 4 cm Step 6: Since M bisects the chord: AB = 2 × AM AB = 2 × 4 AB = 8 cm Step 7: Verification using right triangle OMB: OB² = OM² + MB² 5² = 3² + 4² 25 = 9 + 16 25 = 25 ✓ Answer: Length of chord AB = 8 cm.

Step 1: Theorem (Angles in the Same Segment): Angles subtended by the same chord at different points on the same arc of a circle are equal. If chord AB subtends angles at points C and D on the same arc, then angle ACB = angle ADB. Step 2: Given: Circle with center O Chord AB Point C on major arc: angle ACB = 40° Point D also on major arc Find: angle ADB Step 3: Since C and D are both on the major arc (same segment), and they are both subtended by the same chord AB: angle ADB = angle ACB Step 4: Therefore: angle ADB = 40° Step 5: Proof using central angle: Let angle AOB (central angle) = θ Angle at circumference = θ/2 Since both C and D see the same central angle AOB: angle ACB = θ/2 angle ADB = θ/2 Hence angle ACB = angle ADB Step 6: Different segments give different angles: If point E were on the minor arc, angle AEB would be supplementary to angle ACB. angle AEB = 180° - 40° = 140° Answer: Angle ADB = 40° (same as angle ACB, being on the same arc).

Step 1: Theorem (Intersecting Chords): When two chords intersect inside a circle, the products of their segments are equal. If chords AB and CD intersect at P, then: AP × PB = CP × PD Step 2: Given: Chord AB with intersection point P AP = 6 cm PB = 4 cm Chord CD with intersection point P CP = 3 cm PD = ? Step 3: Apply the intersecting chords theorem: AP × PB = CP × PD 6 × 4 = 3 × PD 24 = 3 × PD Step 4: Solve for PD: PD = 24/3 PD = 8 cm Step 5: Verification: Product of segments of chord AB = AP × PB = 6 × 4 = 24 Product of segments of chord CD = CP × PD = 3 × 8 = 24 ✓ Step 6: Therefore: Chord AB = AP + PB = 6 + 4 = 10 cm Chord CD = CP + PD = 3 + 8 = 11 cm Answer: Length of PD = 8 cm.

Step 1: Angle-in-Segment Theorem Verification: Angles subtended by a chord at points on the major arc are equal. Angles subtended by a chord at points on the minor arc are equal. Angles in the major arc + angles in the minor arc = 180° Step 2: Given: Chord AB Angle at point C on major arc = 60° Angle at point D on minor arc = 120° Step 3: Verify the relationship: angle ACB + angle ADB = 60° + 120° = 180° ✓ Step 4: Explanation using central angle: Let the central angle AOB = θ For point C on major arc: angle ACB = θ/2 = 60° Therefore: θ = 120° Step 5: For point D on minor arc: The angle subtended at a point on the minor arc is: angle ADB = (360° - θ)/2 = (360° - 120°)/2 = 240°/2 = 120° Step 6: Alternatively: angle ACB + angle ADB = θ/2 + (360° - θ)/2 = [θ + 360° - θ]/2 = 360°/2 = 180° ✓ Step 7: This confirms: Angles subtended by a chord at points on opposite arcs are supplementary (sum to 180°). Answer: The theorem is verified: angles in major arc (60°) + angles in minor arc (120°) = 180°.

Step 1: Given: Circle with center O and radius r = 6 cm Central angle subtended by arc AB = 90° Step 2: Arc Length Formula: Arc length = (θ/360°) × 2πr Where θ is the central angle in degrees. Step 3: Calculate arc length: Arc AB = (90°/360°) × 2π × 6 Arc AB = (1/4) × 2π × 6 Arc AB = (1/4) × 12π Arc AB = 3π cm Arc AB ≈ 3 × 3.14159 ≈ 9.42 cm Step 4: Sector Area Formula: Area of sector = (θ/360°) × πr² Where θ is the central angle in degrees. Step 5: Calculate sector area: Area of sector AOB = (90°/360°) × π × 6² Area of sector AOB = (1/4) × π × 36 Area of sector AOB = 9π cm² Area of sector AOB ≈ 9 × 3.14159 ≈ 28.27 cm² Step 6: Verification: Full circle: arc = 2π × 6 = 12π cm, area = π × 6² = 36π cm² Quarter circle: arc = 12π/4 = 3π cm ✓, area = 36π/4 = 9π cm² ✓ Answer: Arc length AB = 3π cm ≈ 9.42 cm; Area of sector AOB = 9π cm² ≈ 28.27 cm².

Step 1: Cyclic Quadrilateral: A quadrilateral whose all four vertices lie on a circle. Step 2: Property of Cyclic Quadrilateral: The opposite angles of a cyclic quadrilateral are supplementary. angle A + angle C = 180° angle B + angle D = 180° Step 3: Given: angle A = 75° angle C = 105° Step 4: Verify the sum of given angles: angle A + angle C = 75° + 105° = 180° ✓ This confirms ABCD is a cyclic quadrilateral (though the problem states it). Step 5: Since opposite angles sum to 180°: angle A + angle C = 180° ✓ angle B + angle D = 180° ... (1) Step 6: We need another condition to find B and D individually. However, from the property of cyclic quadrilaterals and the sum of all angles = 360°: angle A + angle B + angle C + angle D = 360° 75° + angle B + 105° + angle D = 360° angle B + angle D = 360° - 180° angle B + angle D = 180° ✓ Step 7: Without additional information, B and D cannot be uniquely determined individually. But we know: angle B + angle D = 180° angle B = 180° - angle D angle D = 180° - angle B Step 8: Example: If angle B = 85°, then angle D = 95°. Verification: 75° + 85° + 105° + 95° = 360° ✓ Answer: angle B + angle D = 180°. Opposite angles are supplementary: angle A + angle C = 180° and angle B + angle D = 180°.