CBSE Class 9 Maths Ganita Manjari Part 1 Ch 05 I'm Up and Down and Round and Round Short Answer Questions with Answers

2 Mark Questions

Question 1. What is the least possible radius of a circle through two points A and B?
Answer:
The centre of any circle through A and B lies on the perpendicular bisector of AB, and the radius is the distance from the centre to A.
This distance is smallest when the centre is at the midpoint of AB, so that AB itself is a diameter.
\( \therefore \) Least possible radius = \( \frac{AB}{2} \).
In simple words: The smallest circle through two points has the segment joining them as its diameter.

Exam Tip: Any other centre on the perpendicular bisector is farther from A than the midpoint is.

Question 2. Prove that "Equal chords of a circle subtend equal angles at the centre of the circle".
Answer:
Given: AB and CD are equal chords of a circle with centre O.
To prove: ∠AOB = ∠COD
Proof: In △AOB and △COD:
OA = OC (radii), OB = OD (radii), AB = CD (given)
\( \therefore \) △AOB ≅ △COD (SSS)
\( \therefore \) ∠AOB = ∠COD (CPCT). Hence proved.
In simple words: The two triangles made by the chords and the radii have all three sides equal, so their angles at the centre are equal.

Exam Tip: Write the congruence rule (SSS) and CPCT clearly; each carries marks.

Question 3. ABCD is a cyclic quadrilateral. If ∠B = 75° and ∠D = (3x + 15)°, find x.
Answer:
Opposite angles of a cyclic quadrilateral are supplementary: ∠B + ∠D = 180°.
\( 75 + 3x + 15 = 180 \Rightarrow 3x = 90 \Rightarrow x = 30 \)
\( \therefore x = 30 \) (∠D = 105°; check: 75° + 105° = 180° ✓)
In simple words: ∠B and ∠D are opposite angles, so they add up to 180°.

Exam Tip: Make sure the two angles you add are opposite (B and D), not next to each other.

Question 4. A chord of a circle is 16 cm long and its perpendicular distance from the centre is 6 cm. Find the radius of the circle.
Answer:
The perpendicular from the centre bisects the chord, so half-chord = 8 cm.
By the Baudhāyana–Pythagoras theorem: \( r^2 = 8^2 + 6^2 = 64 + 36 = 100 \)
\( \therefore r = 10 \) cm
In simple words: The radius, half the chord and the distance form a right triangle with sides 6, 8 and 10.

Exam Tip: Always halve the chord first; using 16 instead of 8 is the most common mistake.

Question 5. If two chords of a circle are equally inclined to the diameter passing through their point of intersection, prove that the chords are equal.
Answer:
Given: Chords AB and CD of a circle with centre O intersect at E. The diameter through E makes equal angles with them: ∠OEL = ∠OEM.
Construction: Draw OL ⊥ AB and OM ⊥ CD.
Proof: In △OLE and △OME:
∠OLE = ∠OME = 90°, ∠OEL = ∠OEM (given), OE = OE (common)
\( \therefore \) △OLE ≅ △OME (AAS), so OL = OM (CPCT).
Chords equidistant from the centre are equal, so AB = CD. Hence proved.
In simple words: The chords make equal angles with the diameter, so they are the same distance from the centre, and such chords are equal.

Exam Tip: The key step is showing the perpendicular distances OL and OM are equal.

3 Mark Questions

Question 6. Prove that the perpendicular bisector of a chord passes through the centre of the circle.
Answer:
Given: AB is a chord of a circle with centre O.
Proof: OA = OB (radii), so O is equidistant from A and B.
Every point equidistant from A and B lies on the perpendicular bisector of AB.
\( \therefore \) O lies on the perpendicular bisector of AB; that is, the perpendicular bisector of the chord passes through the centre.
(Alternatively: let M be the midpoint of AB. △OMA ≅ △OMB by SSS, so ∠OMA = ∠OMB = 90°. So OM is the perpendicular bisector of AB, and it passes through O.)
In simple words: The centre is the same distance from both ends of the chord, so it must sit on the chord's perpendicular bisector.

Exam Tip: State the locus property: points equidistant from A and B lie on the perpendicular bisector of AB.

Question 7. Quadrilateral PQRS is inscribed in a circle. If ∠P = (2x + 10)° and ∠R = (3x − 20)°, find the value of x and the measures of ∠P and ∠R.
Answer:
P and R are opposite angles of a cyclic quadrilateral, so ∠P + ∠R = 180°.
\( (2x + 10) + (3x - 20) = 180 \Rightarrow 5x - 10 = 180 \Rightarrow x = 38 \)
∠P = 2(38) + 10 = 86°; ∠R = 3(38) − 20 = 94°
Check: 86° + 94° = 180° ✓
In simple words: Opposite angles in a cyclic quadrilateral add up to 180°, which gives an equation in x.

Exam Tip: Always substitute x back and check that the two angles add up to 180°.

Question 8. In a circle with centre O, the central angle AOB is 60°. If the radius of the circle is 12 cm, what is the length of the chord AB?
Answer:
In △OAB, OA = OB = 12 cm (radii), so it is isosceles with ∠AOB = 60°.
Base angles: \( \angle OAB = \angle OBA = \frac{180^\circ - 60^\circ}{2} = 60^\circ \).
All angles are 60°, so △OAB is equilateral.
\( \therefore AB = OA = 12 \) cm
In simple words: Two radii with a 60° angle between them form an equilateral triangle, so the chord equals the radius.

Exam Tip: A 60° central angle always gives a chord equal to the radius; this is worth remembering.

Question 9. Explain why the following statement is true: If the perpendicular distance of a chord from the centre is d and the radius is r, then the chord length is \( 2\sqrt{r^2 - d^2} \).
Answer:
Let AB be the chord, O the centre and M the foot of the perpendicular from O to AB, so OM = d and OA = r.
The perpendicular from the centre bisects the chord, so \( AM = \frac{AB}{2} \).
In right △OMA: \( OA^2 = OM^2 + AM^2 \Rightarrow r^2 = d^2 + AM^2 \Rightarrow AM = \sqrt{r^2 - d^2} \)
\( \therefore AB = 2AM = 2\sqrt{r^2 - d^2} \)
In simple words: The radius, the distance d and half the chord form a right triangle; find half the chord and double it.

Exam Tip: Mention both facts used: the perpendicular bisects the chord, and the Baudhāyana–Pythagoras theorem.

Question 10. In a circle with centre O, chords AB and AC are congruent. Explain why this statement is true: "The centre of the circle lies on the angle bisector of ∠BAC".
Answer:
Join OA, OB and OC.
In △OAB and △OAC: OA = OA (common), OB = OC (radii), AB = AC (given)
\( \therefore \) △OAB ≅ △OAC (SSS), so ∠OAB = ∠OAC (CPCT).
So AO divides ∠BAC into two equal parts; that is, the centre O lies on the bisector of ∠BAC.
In simple words: The two triangles formed by the equal chords and the radii are identical, so AO splits the angle at A equally.

Exam Tip: Congruent triangles are the standard way to show two angles are equal.

Question 11. The distance of a chord AB of a circle from the centre is 12 cm and the length of the chord AB is 10 cm. Find the diameter of the circle.
Answer:
Let OD ⊥ AB, so OD = 12 cm and D is the midpoint of AB: AD = 5 cm.
In right △AOD: \( OA^2 = AD^2 + OD^2 = 25 + 144 = 169 \Rightarrow OA = 13 \) cm
\( \therefore \) Diameter = 2 × 13 = 26 cm
In simple words: Half the chord (5) and the distance (12) give a radius of 13, so the diameter is 26 cm.

Exam Tip: The question asks for the diameter; do not stop at the radius.