Ganita Manjari Class 9 Maths Chapter 5: I'm Up and Down, and Round and Round — NCERT Solutions

Explanation: Join the radii CA and CG. In right triangles CEA and CHG, angle CEA = angle CHG = 90°, the hypotenuses CA and CG are equal (radii of the same circle), and CE = CH (given). By the RHS congruence rule, triangle CEA is congruent to triangle CHG, so AE = GH. Since the perpendicular from the centre bisects a chord, AB = 2AE and GF = 2GH. Therefore AB = 2AE = 2GH = GF, which proves that chords equidistant from the centre are equal in length.

Explanation: In the figure, BA is a diameter passing through the centre C, and D is a point on the circle. The angle subtended by BA at the centre is angle BCA = 180°, since B, C and A lie on one straight line. By the central angle theorem, the angle subtended by an arc at the centre is twice the angle subtended by it at any point on the remaining part of the circle. Hence angle BDA = (1/2) x angle BCA = (1/2) x 180° = 90°, so a diameter always subtends a right angle at any point on the circumference.

Explanation: Join the radii CA and CB. In triangles CMA and CMB, angle CMA = angle CMB = 90° (given, since CM is perpendicular to chord AB), the hypotenuses CA and CB are equal (radii of the same circle), and side CM is common to both triangles. By the RHS congruence rule, triangle CMA is congruent to triangle CMB. Therefore AM = BM by CPCT, which shows that the perpendicular drawn from the centre to a chord must bisect the chord.

Explanation: Let ABCD be a quadrilateral inscribed in a circle with centre O; join OB and OD. Arc BCD subtends angle BOD at the centre and angle BAD at point A on the remaining part of the circle, so angle BOD = 2 x angle A. Similarly, arc BAD subtends the reflex angle BOD at the centre and angle BCD at C, so reflex angle BOD = 2 x angle C. Adding these, angle BOD + reflex angle BOD = 360° gives 2(angle A + angle C) = 360°, so angle A + angle C = 180°. Since all four angles of a quadrilateral sum to 360°, the other pair also satisfies angle B + angle D = 360° - 180° = 180°.

Explanation: On measuring with a protractor, the angles subtended by arc AKB at P, Q and R come out equal: angle APB = angle AQB = angle ARB. This is the observation — an arc subtends equal angles at all points on the remaining part of the circle (angles in the same segment are equal). Comparing with the central angle, each of these angles measures exactly half of angle AOB, so the relationship is angle AOB = 2 x angle APB. Repeating the activity with a different arc (including a major arc, where the central angle is reflex) gives the same two results, confirming the central angle theorem.

Explanation: Let A, B, C be three non-collinear points. Every point equidistant from A and B lies on the perpendicular bisector of AB, and every point equidistant from B and C lies on the perpendicular bisector of BC. Since A, B, C are not collinear, AB and BC are not parallel, so their perpendicular bisectors are also not parallel and intersect at exactly one point O; this point satisfies OA = OB = OC. The circle with centre O and radius OA therefore passes through all three points. For uniqueness, the centre of any circle through A, B and C must be equidistant from all three, so it must lie on both perpendicular bisectors and hence coincide with O, with radius OA — so exactly one such circle exists.

Explanation: Place the protractor at the centre O. The central angle AOB corresponding to arc AKB measures about 100°; since this is less than 180°, arc AKB is a minor arc. The central angle corresponding to arc CLD is the reflex angle COD (shown shaded at O in the figure), which measures about 200°; since this is more than 180°, arc CLD is a major arc. The rule used is: an arc is minor when its central angle is less than 180° and major when its central angle is a reflex angle, i.e. greater than 180°.

Explanation: Points A, D, C and B all lie on the circle, so ADCB is a cyclic quadrilateral in which angle ADC = 100° and angle ABC = x are a pair of opposite angles. Opposite angles of a cyclic quadrilateral are supplementary, so angle ADC + angle ABC = 180°. Therefore x = 180° - 100° = 80°. Equivalently, angles ADC and ABC lie in opposite segments of the chord AC, so they must add up to 180°.

Explanation: Every circle through A and B has its centre on the perpendicular bisector of AB. As the centre moves farther away from AB along this bisector, the radius OA keeps increasing, so the circle becomes larger. Curvature is inversely related to the radius (a bigger circle bends less sharply), so the curvature decreases and the arc through A and B looks flatter and flatter, approaching a straight line. Conversely, moving the centre towards the midpoint of AB makes the radius smaller and the curvature greater.

Explanation: Circles appear widely in nature because many natural processes act equally in all directions from a central point. Common examples include the discs of the Sun and the full Moon, circular ripples spreading on still water, the iris and pupil of the human eye, and the annual growth rings seen in a cut tree trunk. Other examples are a rainbow (a full circle when seen from high above), water droplets and soap bubbles seen face-on, the cross-sections of fruits such as oranges and lemons, and round flower heads such as sunflowers. Any such set of naturally round objects is an acceptable answer.

Explanation: The perpendicular from the centre bisects a chord, so the half-lengths of the chords are 3 cm and 4 cm. For the 6 cm chord, its distance from the centre is sqrt(5² - 3²) = sqrt(25 - 9) = sqrt(16) = 4 cm, using Pythagoras with the radius as hypotenuse. For the 8 cm chord, the distance is sqrt(5² - 4²) = sqrt(25 - 16) = sqrt(9) = 3 cm. Since the chords are parallel and on opposite sides of the centre, the distance between them is the sum 4 + 3 = 7 cm.

Explanation: Let the chords be AB and CD in a circle with centre O, so the two triangles are OAB and OCD. Here OA = OC and OB = OD, because all four are radii of the same circle (this is also why each triangle is isosceles). The bases are equal, AB = CD, since the chords are given to be equal. By the SSS congruence rule, triangle OAB is congruent to triangle OCD. In particular, by CPCT angle AOB = angle COD, showing that equal chords subtend equal angles at the centre.

Explanation: (i) No. Points on the circle on the same side of AB lie in the same segment, and angles in the same segment of a circle are equal, so angle AXB = angle AYB always holds. (ii) Not necessarily: if X and Y lie on opposite sides of AB, the angles lie in opposite segments and are supplementary, so they can still be equal when each is 90°, which happens exactly when AB is a diameter; for any chord that is not a diameter, equal angles do force X and Y to the same side. (iii) Yes, when X and Y lie on the same side of line AB: by the converse of the angles-in-the-same-segment theorem, two points on the same side of AB subtending equal angles at AB must be concyclic with A and B, so the circle through A, B and X necessarily passes through Y as well.

Explanation: The longest chord of a circle is its diameter. For any chord AB, joining its endpoints to the centre O gives AB <= OA + OB = 2r by the triangle inequality, with equality exactly when the chord passes through the centre. So the longest chord has length 2 x 5 = 10 units. A smallest chord does not exist: a chord must meet the circle at two distinct points, so its length is always greater than 0, yet by taking the two points closer and closer together the chord can be made shorter than any given positive length. Since no positive length can be the minimum, there is no smallest chord.