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

5 Mark Questions

Question 1. Two parallel chords of lengths 6 cm and 8 cm are on opposite sides of the centre of a circle. If the radius of the circle is 5 cm, find the distance between the midpoints of the chords.
Answer:
Let AB = 6 cm and CD = 8 cm, with OM ⊥ AB and ON ⊥ CD. The perpendiculars bisect the chords: AM = 3 cm and CN = 4 cm.
\( OM = \sqrt{5^2 - 3^2} = \sqrt{16} = 4 \) cm
\( ON = \sqrt{5^2 - 4^2} = \sqrt{9} = 3 \) cm
Since the chords are parallel and on opposite sides of O, M, O and N lie on one straight line.
Distance between midpoints = OM + ON = 4 + 3 = 7 cm
In simple words: The 6 cm chord is 4 cm from the centre and the 8 cm chord is 3 cm away on the other side, so they are 7 cm apart.

Exam Tip: If the chords were on the same side of the centre, you would subtract instead: 4 − 3 = 1 cm.

Question 2. Prove that "The angle subtended by an arc at the centre of the circle is double the angle subtended by the arc at any point on the circle outside the arc."
Answer:
Given: Arc AB subtends ∠AOB at the centre O and ∠ACB at a point C on the remaining part of the circle.
To prove: ∠AOB = 2∠ACB
Construction: Join CO and extend it to a point D.
Proof: In △OAC, OA = OC (radii), so ∠OAC = ∠OCA.
The exterior angle ∠AOD = ∠OAC + ∠OCA = 2∠OCA … (1)
Similarly, in △OBC, OB = OC, so ∠BOD = 2∠OCB … (2)
Adding (1) and (2): ∠AOD + ∠BOD = 2(∠OCA + ∠OCB)
\( \therefore \) ∠AOB = 2∠ACB. Hence proved.
In simple words: Draw the diameter through C; each small isosceles triangle shows the outer angle at O is twice the angle at C.

Exam Tip: State the exterior angle property clearly; it is the heart of this proof.

Question 3. How would you use the following figure to justify the statement that the sum of the opposite angles of a cyclic quadrilateral is 180°?

Answer:
ABCD is a cyclic quadrilateral with centre O. Join OA, OB, OC and OD.
OA = OB = OC = OD (radii), so △OAB, △OBC, △OCD and △ODA are all isosceles.
Let the equal base angles be: ∠OAB = ∠OBA = p, ∠OBC = ∠OCB = q, ∠OCD = ∠ODC = u, ∠ODA = ∠OAD = v.
Then ∠A = p + v, ∠B = p + q, ∠C = q + u and ∠D = u + v.
Angle sum of the quadrilateral: ∠A + ∠B + ∠C + ∠D = 360° ⇒ 2(p + q + u + v) = 360° ⇒ p + q + u + v = 180°.
So ∠A + ∠C = (p + v) + (q + u) = 180° and ∠B + ∠D = (p + q) + (u + v) = 180°.
Hence, the opposite angles of a cyclic quadrilateral are supplementary.
In simple words: Each vertex angle splits into two base angles of isosceles triangles, and every opposite pair ends up using all four of them once.

Exam Tip: Label the four equal-angle pairs with letters before writing the angle sums.

Question 4. A quadrilateral MNOP is inscribed in a circle. If MN is a diameter, what can you say about ∠MOP and ∠MNP? Explain your reasoning.
Answer:
∠MOP = ∠MNP.
Both angles are subtended by the same chord MP, at the points O and N on the circle.
Since the vertices are in the order M, N, O, P around the circle, N and O lie on the same arc cut off by MP, so these are angles in the same segment.
Angles in the same segment of a circle are equal, so ∠MOP = ∠MNP.
Also, because MN is a diameter, ∠MON = ∠MPN = 90° (angles in a semicircle).
In simple words: Both angles look at the same chord MP from the same side, so they are equal.

Exam Tip: Here O is a vertex of the quadrilateral, not the centre; read the question carefully.

Case Study Based Questions

Question 5. Four friends Rima, Sita, Mohan and Sohan are sitting on the circumference of a circular park. Their locations are marked by points A, P, Q and R. Rohit joins them and sits at the centre of the circular park, so he is equidistant from all the other friends. His position is marked as O. They are sitting in such a way that ∠PQR = 110°.
(i) What is the measure of reflex ∠POR? Give reason.
(ii) What is the measure of ∠PAR? Give reason.
(iii) Find ∠OPR. Give reason.
Answer:
(i) The angle subtended by an arc at the centre is double the angle it subtends at a point on the remaining part of the circle. Arc PAR subtends ∠PQR at Q.
Reflex ∠POR = 2 × ∠PQR = 2 × 110° = 220°
(ii) APQR is a cyclic quadrilateral, and ∠PAR and ∠PQR are opposite angles.
∠PAR = 180° − 110° = 70°
(iii) ∠POR (non-reflex) = 360° − 220° = 140°. In △POR, OP = OR (radii), so ∠OPR = ∠ORP.
∠OPR = \( \frac{180^\circ - 140^\circ}{2} \) = 20°
In simple words: The centre angle is twice the circle angle (220°), opposite angles make 180° (70°), and the isosceles triangle POR gives 20°.

Exam Tip: Be careful to use the reflex angle in (i) and the ordinary angle 140° in (iii).

Question 6. Hina and Vinni are best friends. They usually go to the park in the evening to play. One day, Tarun, Hina and Vinni were sitting on a merry-go-round at A, B and C respectively and passing a ball to each other. The seats were at equal distances from each other, and the radius of the base circle is 50 cm.
(a) Which type of triangle is formed by them?
(b) If the distance between Vinni and Hina is 2x cm, find the value of CP.
(c) Find the distance between A and P in terms of x.
(d) Find the value of x.
(e) If the distance between any two of them is m√3 cm, find m.
Answer:
(a) The seats are equally spaced, so AB = BC = CA: △ABC is an equilateral triangle.
(b) OP ⊥ BC, and the perpendicular from the centre bisects the chord, so CP = \( \frac{1}{2} \times 2x = \) x cm.
(c) In right △ACP: AC = 2x, so \( AP^2 = (2x)^2 - x^2 = 3x^2 \Rightarrow AP = \) \( \sqrt{3}x \) cm.
(d) OP = AP − OA = \( \sqrt{3}x - 50 \). In right △OCP: \( 50^2 = (\sqrt{3}x - 50)^2 + x^2 \)
\( 2500 = 3x^2 - 100\sqrt{3}x + 2500 + x^2 \Rightarrow 4x^2 = 100\sqrt{3}x \Rightarrow x = 25\sqrt{3} \) (x ≠ 0)
(e) Side = 2x = \( 50\sqrt{3} \) cm = \( m\sqrt{3} \), so m = 50.
In simple words: The three friends form an equilateral triangle with side 50√3 cm inside a circle of radius 50 cm.

Exam Tip: For an equilateral triangle in a circle of radius r, the side is always r√3; use this to check your answer.

Question 7. A newly constructed sports stadium has a perfectly circular running track with its central hub at point O. The radius of this circular setup is 10 metres. Four athletes A, B, C and D are stationed on the track boundary, forming a cyclic quadrilateral ABCD. The opposite angles are ∠DAB = (2x + 15)° and ∠BCD = (3x − 10)°. Two underground water pipes EF and GH cross beneath the field as chords of the circle. Pipe EF is 16 m long and lies 6 m from O. Pipe GH lies 8 m from O. An arc AB subtends ∠AOB = 80° at O.
(i) What is the value of x determined from the positions of athletes A and C?
(ii) Find the length of the underground water pipe GH.
(iii) If another athlete E stands at any point on the major arc AB, what would be the measure of ∠AEB?
(iv) Which of the following theorems explains why pipe EF is closer to the centre than pipe GH?
(a) Equal chords of a circle subtend equal angles at the centre.
(b) The longer chord is closer to the centre than the shorter chord.
(c) Chords equidistant from the centre are equal in length.
(d) The perpendicular from the centre to a chord bisects the chord.
Answer:
(i) Opposite angles of a cyclic quadrilateral are supplementary: \( (2x + 15) + (3x - 10) = 180 \Rightarrow 5x = 175 \Rightarrow \) x = 35 (∠A = 85°, ∠C = 95°).
(ii) Half of GH = \( \sqrt{10^2 - 8^2} = \sqrt{36} = 6 \) m, so GH = 12 m.
(Check with EF: half of EF = \( \sqrt{100 - 36} = 8 \) m, so EF = 16 m ✓)
(iii) ∠AEB = \( \frac{1}{2} \) ∠AOB = \( \frac{80^\circ}{2} \) = 40°
(iv) (b) EF (16 m) is longer than GH (12 m), and the longer chord is nearer the centre (6 m < 8 m).
In simple words: x is 35, pipe GH is 12 m, E sees arc AB at 40°, and the longer pipe EF sits closer to the centre.

Exam Tip: Always check given data: here the 16 m pipe at 6 m confirms the radius is 10 m.