Class 9 Maths Ganita Manjari Chapter 5 MCQ – I’m Up and Down, and Round and Round

A circle consists of all points in a plane at one fixed distance from its centre. Here, that distance is 5 cm. Points nearer than 5 cm lie inside the circle, while points farther than 5 cm lie outside it.

Every line through the centre divides a circle into matching halves, so the circle has infinitely many reflection axes. Rotating it about its centre through any angle also leaves it unchanged because all its points remain at the same distance from the centre.

The longest chord of a circle is a diameter, which passes through the centre. Its length is twice the radius. Therefore, the longest chord measures 2 × 7 = 14 cm. Chords that do not pass through the centre are shorter.

Every point on the perpendicular bisector of AB is equidistant from A and B. Conversely, any point equidistant from A and B lies on this line. Therefore, the entire perpendicular bisector, rather than just the midpoint, is the required locus.

Any point on the perpendicular bisector of AB can be chosen as the centre because it is equally distant from A and B. There are infinitely many points on this line, giving infinitely many circles through the two given points.

The segment AB is a chord, and no chord can be longer than the diameter. Thus, the radius must be at least half of 10 cm. This minimum of 5 cm is attained when the midpoint of AB is the centre.

The perpendicular bisectors of two segments joining the points intersect at a unique point. This point is equidistant from all three given points and becomes the centre. Hence, exactly one circle passes through three distinct points that are not on one line.

A straight line can meet a circle in at most two distinct points. Therefore, three distinct collinear points cannot all lie on one circle. Equivalently, the perpendicular bisectors of AB and BC are distinct parallel lines and cannot meet at a centre.

The circumcentre must be equally distant from all three vertices. A point on the perpendicular bisector of a side is equidistant from its endpoints. Intersecting the perpendicular bisectors of two sides therefore gives a point equally distant from all three vertices.

The circumcentre of a right-angled triangle lies at the midpoint of its hypotenuse. The hypotenuse is therefore a diameter of the circumcircle. Its radius is half the hypotenuse, so the required radius is 16 ÷ 2 = 8 cm.

The triangle is obtuse-angled because one angle, 110°, is greater than 90°. The circumcentre of an obtuse-angled triangle lies outside it. In comparison, an acute triangle has its circumcentre inside, while a right triangle has it on the hypotenuse.

Equal chords of the same circle subtend equal smaller angles at the centre. Joining their endpoints to O forms two triangles with equal radii and equal bases. SSS congruence gives equal central angles, so ∠COD is also 70°.

Chords of the same circle that subtend equal smaller angles at the centre have equal lengths. The triangles formed by the radii and the chords are congruent by SAS. Therefore, CD = AB, so CD is 8 cm long.

The perpendicular from the centre of a circle to a chord bisects that chord. Hence, M is the midpoint of AB and AM = MB. Since the whole chord is 18 cm, each half has length 18 ÷ 2 = 9 cm.

The perpendicular from the centre bisects the chord. A radius, the perpendicular and half the chord form a right triangle. Half the chord is √(13² − 5²) = √144 = 12 cm. Therefore, the complete chord is 24 cm long.

The perpendicular from the centre divides the chord into two parts of 8 cm each. A radius is the hypotenuse of a right triangle with legs 8 cm and 6 cm. Thus, the radius is √(8² + 6²) = 10 cm.

Chords that are equally distant from the centre of the same circle are equal in length. Both perpendicular distances are 4 cm, so the two chord lengths are equal. Since one chord measures 12 cm, the other also measures 12 cm.

In a fixed circle, the longer chord lies closer to the centre. The radius forms a right triangle with half the chord and its perpendicular distance from the centre. A larger half-chord leaves a smaller distance because their squared lengths add to the same radius squared.

Their half-lengths are 3 cm and 4 cm. Their distances from the centre are √(25 − 9) = 4 cm and √(25 − 16) = 3 cm. Since they lie on opposite sides of the centre, add these distances to obtain 7 cm.

Point P is outside the major arc, so ∠APB is half the angle subtended by that major arc at the centre. Therefore, ∠APB = 240° ÷ 2 = 120°. Using the smaller central angle would correspond to the other arc.

An arc subtends an angle at the centre that is twice the angle it subtends at a point on the remaining part of the circle. Since P lies outside the minor arc, ∠APB = 80° ÷ 2 = 40°.

Angles subtended by the same chord at points on the same side of that chord are equal. Both P and Q lie in the same segment of the circle. Therefore, ∠AQB = ∠APB = 35°, regardless of their exact positions in that segment.

The diameter subtends a straight angle of 180° at the centre. The angle subtended by the corresponding semicircle at C is half that angle. Hence, ∠ACB = 90°. This is the result commonly described as the angle in a semicircle.

OA and OB are equal radii, so triangle AOB is isosceles. Its other two angles are (180° − 60°) ÷ 2 = 60° each. The triangle is therefore equilateral, and AB equals the radius, which is 9 cm.

A segment that subtends equal angles at two points on the same side of its line gives a condition for concyclicity. The circle through A, B and C must also pass through D. Thus, all four points lie on one circle.

Opposite angles of a cyclic quadrilateral add up to 180°. The vertices A and C are opposite, so ∠C = 180° − ∠A. Substituting 75° gives ∠C = 105°. Adjacent angles do not generally have to add up to 180°.

The angles at P and R are opposite, so (2x + 10) + (3x − 20) = 180. This simplifies to 5x − 10 = 180, giving x = 38. The two angles are then 86° and 94°, which sum to 180°.

Because C, D and E lie on a straight line, ∠ADC = 180° − 110° = 70°. Opposite angles in the cyclic quadrilateral are supplementary, so ∠ABC = 180° − 70° = 110°. Thus, the exterior angle equals the interior opposite angle.

Opposite angles of a parallelogram are equal. Opposite angles of a cyclic quadrilateral also add up to 180°. Two equal angles with this sum must each be 90°. Therefore, the parallelogram must be a rectangle; a square is included as a special rectangle.

For each chord, its midpoint is 4 cm from either endpoint. The perpendicular distance from O to the midpoint is √(5² − 4²) = 3 cm. As the chord rotates around O, its midpoint traces a circle of radius 3 cm with centre O.