I’m Up and Down and Round and Round Class 9 Extra Questions Maths Chapter 5

During revision, students quickly go through Extra Questions for Class 9 Maths and Ganita Manjari Class 9 Maths Chapter 5 I’m Up and Down and Round and Round Important Extra Questions and Answers for clarity.

Class 9 I’m Up and Down and Round and Round Extra Questions

Extra Questions of I’m Up and Down and Round and Round

Class 9 Maths Chapter 5 Extra Questions – I’m Up and Down and Round and Round Extra Questions Class 9

I’m Up and Down and Round and Round Class 9 Short Question Answer

Question 1.
The diameter of a circle is 3.8 cm. Find the length of its radius.
Solution:
Since the diameter of circle is double its radius.
∴ Diameter = 2 × Radius
⇒ \(\frac{1}{2}\) × Diameter = Radius
Radius = \(\frac{1}{2}\) × 3.8 cm
= \(\frac{1}{2} \times \frac{38}{10}\)cm
= \(\frac{19}{10}\) = 1.9 cm

Question 2.
In the adjoining figure, O is the centre of the circle. The chord AB = 10 cm is such that OP ⊥ AB. Find the length of AP.

Solution:
∵ OP ⊥ AB
∴ P is the mid-point of AB.
⇒ AP = \(\frac{1}{2}\) AB
AP = \(\frac{1}{2}\) × 10 cm
= 5 cm

Question 3.
In the adjoining figure, O is the centre of the circle. Find the length of AB.

Solution:
Since chord AB and chord CD subtend equal angles at the centre,
i.e. ∠AOB = ∠COD
∴ Chord AB= Chord CD
⇒ Chord AB = 5 cm [∵ Chord CD = 5 cm]
Thus, the length of chord AB is 5 cm.

Question 4.
In the adjoining figure, O is the centre of the circle and OP = OQ. If AP = 4 cm, then find the length of CD.

Solution:
∵ OP = OQ
∴ Chord AB and chord CD are equidistant from the centre.
⇒ \(\overline{\mathrm{AB}}=\overline{\mathrm{CD}}\)
⇒ \(\frac{1}{2} \overline{\mathrm{AB}}=\frac{1}{2} \overline{\mathrm{CD}}\)
⇒ \(\overline{\mathrm{AP}}=\frac{1}{2} \overline{\mathrm{CD}}\) [P is the mid-point of AB]
⇒ 4 cm = 2_CD
⇒ CD = 2 × 4 cm
⇒ CD = 8 cm
Thus, the required length of CD is 8 cm.

Question 5.
AB and CD are two parallel chords of a circle which are on opposite sides of the centre such that AB = 24 cm and CD = 10 cm and the distance between AB and CD is 17 cm. Find the radius of the circle.
Solution:
∵ Perpendicular from the centre to a chord bisects the chord.

∴ AP = \(\frac{1}{2}\)AB = \(\frac{1}{2}\) x 24 cm = 12 cm
Similarly, CQ = \(\frac{1}{2}\) CD = \(\frac{1}{2}\) x 10 cm = 5 cm
Let OP = x cm ⇒ OQ = (17 – x) cm
Now, in right ∆APO,
x² + 12² = OA² …(1)
Again, in right ∆COQ,
OC² = CQ² + (17 – x)² = 5² + 17² + x² – 34x
= 25 + 289 + x²– 34x = 314 + x² – 34x …(2)
From (1) and (2), we have
x² + 314 – 34x = x² + 122
⇒ x² – x² – 34x = 144 – 314
⇒ -34x = -170
⇒ x = \(\frac{-170}{-34}\) = 5
Now from (1), we have
OA² = 5² + 12² = 25 + 144
⇒ OA² = 169
⇒ OA = 13 cm
Thus, the required radius of the circle = 13 cm.

Question 6.
If O is the centre of the circle, then find the value of x.

Solution:
∵ AB is a diameter.
∴ ∠AOC + ∠COB = 1 80° [Linear pairs]
⇒ 130° + ∠COB =180°
⇒ ∠COB = 180° – 130° = 50°
Now, the arc CB is subtending ∠COB at the centre and ∠CDB at circle outside the arc.
∠CDB = \(\frac{1}{2}\)∠COB
⇒ ∠CDB = \(\frac{1}{2}\) × 50° = 25°
Thus, the measure of x = 25°.

I’m Up and Down and Round and Round Class 9 Long Question Answer

Question 1.
The radius of a circle is 17 cm. A chord of length 30 cm is drawn. Find the distance of the chord from the centre.
Solution:
Length of chord AB = 30 cm.

Since, OP ⊥ AB
P is the mid-point of AB.
⇒ AP = \(\frac{1}{2}\)AB = \(\frac{1}{2}\) × 30 cm = 15 cm
Now, in right ∆APO,
AO² = AP² + OP²
17² = 15² + OP²
OP² = 17² – 15²
= (17 – 15)(17 + 15)
= 2 × 32 = 64
OP = \(\sqrt{64}\) = 8 cm
∴ The distance of the chord AB from the centre O is 8 cm.

Question 2.
Find the length of a chord which is at a distance of 4 cm from the centre of a circle of radius 5 cm.
Solution:
∵ The perpendicular distance, OP = 4 cm

∴ In right ∆APO,
AO² = AP² + OP²
5² = AP² + 4²
AP² = 5² – 4² = (5 – 4)(5 + 4)
= 1 × 9 = 9
AP = √9 = 3 cm
Since the perpendicular from the centre to a chord of a circle divides the chord into two equal parts.
AP = \(\frac{1}{2}\)AB²
AB = 2AP
AB = 2 × 3 cm = 6 cm
Thus, the required length of chord AB is 6 cm.

I’m Up and Down and Round and Round Class 9 Case Based Questions

Question 1.
ICC World Cup Final Match is being played between INDIA and NEW ZEALAND. Toss is won by NEW ZEALAND and his captain decided to Bowl first. He places the three fielders at boundary line at points A, B, and D. Position of bowler is at point P.
Captain also keeps himself at the boundary on the point C. Angle of captain from the fielder B and centre point ‘O’ is of 40° measure.
In the given figure, O is the centre of the circular stadium.

(i) What angle do the bowler ‘P’ and the fielder ‘C’ subtend at the centre point ‘O’?
(ii) What angle fielders ‘D’ and ‘C’ make at the centre point ‘O’?
(iii) At what angle is fielder B from fielders ‘D’ and ‘C’?
OR
At what angle is fielder ‘A’ from the fielder ‘B’ and bowler ‘P’?
Solution:
(i) O is the centre of the circular stadium and ∠BCO = 40°.
In right ∆OCP, we have
∠POC = 180° – (∠OPC + ∠PCO)
⇒ ∠POC =180°- (90° + 40°) = 50°

(ii) Now, ∠AOD = 90° [Given]
∠AOD + ∠DOP = 180° [Angles of a linear pair]
∴ ∠DOP = 180° – ∠AOD = 180° – 90° = 90°
Now, ∠COD = 90° – ∠POC
= 90° – 50°
= 40°

(iii) Since the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle,
∴ ∠CBD = \(\frac{1}{2}\)∠COD
⇒ ∠CBD = \(\frac{1}{2}\) × 40° = 20°
OR
Also, ∠ABD = \(\frac{1}{2}\)∠AOD = \(\frac{1}{2}\) × 90° = 45°
Now, in A ABP, we have
∠BAP + (45° + ∠CBD) + 90°= 180°
⇒ ∠BAP + 45° + 20° + 90° = 180°
⇒ ∠BAP = 180° – 155° = 25°
Hence, ∠BAP = 25°

I’m Up and Down and Round and Round Class 9 Competency Based Questions

Question 1.
AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, the distance of AB from the centre of the circle is
(a) 17 cm
(b) 15 cm
(c) 4 cm
(d) 8 cm
Solution:
Draw OP ⊥ AB.

As perpendicular from the centre to a chord bisects the chord, so
AP = \(\frac{1}{2}\) × AB
\(\frac{1}{2}\) × 30= 15 cm
Radius OA = \(\frac{1}{2}\) × 34 = 17 cm
In right ΔOPA, we have
OP = \(\sqrt{O A^2-A P^2}\)
= \(\sqrt{(17)^2-(15)^2}\)
= \(\sqrt{289-225}=\sqrt{64}\)
= 8 cm
Hence, (d) is the correct answer.

Question 2.
In the given figure, if ∠ACB = 40°, ∠DPB = 120°, find the value of x.
(a) 45°
(b) 20°
(c) 30°
(d) 120°
Solution:
∠ACB = ∠ADB
[∵ Angles in the same segment of a circle are equal]
∴ ∠ADB = 40° [∵ ∠ACB = 40°]
Now in ΔBPD,
x + ∠DPB + ∠PDB = 180°
x+ 120°+ 40°= 180°
[∵ ∠PDB = ∠ADB = 40°]
⇒ x = 180° -160° = 20°
Hence, (b) is the correct answer.

Question 3.
Two circles of radii 5 cm and 3 cm and centres A and B, touch internally. If the perpendicular bisector of segment AB meets the bigger circle in P and Q, find the length of PQ.
Solution:
When two circles touch each other internally, then

Distance between their centres
= Difference of their radii
⇒ AB = (5 – 3) cm = 2 cm.
Now PQ is the ⊥ bisector of segment AB.
∴ AM = MB = 1 cm
From right-angled ΔAMP, we have
AP² = AM² + MP²
5² = 1² + MP²
MP² = 25 – 1 = 24 cm
MP = \(\sqrt{24}\) cm = 2√6 cm
AM ⊥ PQ
PM = MQ
PQ = 2PM
= 2 × 2√6 cm
PQ = 4√6 cm.

Question 4.
Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see Fig.). Prove that ∠ACP = ∠QCD.

Solution:
∵ Angles in the same segment of a circle are equal
∴ ∠ACP = ∠ABP …(1)
Also ∠QCD = ∠QBD …(2)
But ∠ABP = ∠QBD [Vertically Opp. ∠s] …(3)
Now from (1), (2), (3), we get,
∠ACP = ∠QCD.
Hence proved.

Question 5.
Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that the angles of the ΔDEF are 90°, 90° – \(\frac{A}{2}\), 90° – \(\frac{B}{2}\) and 90° – \(\frac{C}{2}\) respectively.
Solution:
Since AD, BE and CF are the bisectors of ∠A, ∠B and ∠C respectively,

∠1 = ∠2 = \(\frac{1}{2}\)∠A
∠3 = ∠4 = \(\frac{1}{2}\)∠B
∠5 = ∠6 = \(\frac{1}{2}\)∠C
Now ∠ADE = ∠3 …(1)
[∵ Angles in the same segment of a circle are equal]
Also ∠ADF = ∠6 [Same reason] …(2)
Adding (1) and (2), we get ∠ADE + ∠ADF = ∠3 + ∠6
⇒ ∠D = \(\frac{1}{2}\) ∠B + \(\frac{1}{2}\) ∠C
[∵ ∠3 = \(\frac{1}{2}\) ∠B and ∠6 = \(\frac{1}{2}\) ∠C]
⇒ ∠D = \(\frac{1}{2}\) (∠B + ∠C) = \(\frac{1}{2}\) (180° – ∠A)
[∵ ∠A + ∠B + ∠C = 180°]
⇒ ∠D = 90° – \(\frac{1}{2}\) ∠A
Similarly, we can prove that
∠E = 90° – \(\frac{1}{2}\) ∠B
and ∠F = 90° – \(\frac{1}{2}\) ∠C.

I’m Up and Down and Round and Round Class 9 Extra Questions for Practice

Multiple Choice Questions

Question 1.
In a circle AB is diameter and C is a point on the circle, Which of following is the measure of ZACB?
(a) 60°
(b) 90°
(c) 180°
(d) None of these

Question 2.
In a circle with centre O, radius OA = 5 cm and chord AB = 8 cm. If OC ⊥ AB, then which of the following is length of AC?
(a) 2 cm
(b) 3 cm
(c) 4 cm
(d) 5 cm

Question 3.
A, B and C are three distinct points on a circle with centre O. If ∠ABC = 20° then which of the following is the measure of ∠AOC (provided O and B are on the same side of AC)?
(a) 10°
(b) 20°
(c) 40°
(d) 60°

Assertion Reason Questions

Two statements are given, one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer from the options (a), (b), (c), and (d) given below.
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(c) Assertion (A) is true, but Reason (R) is false.
(d) Assertion (A) is false, but Reason (R) is true.

Question 1.
Assertion (A): AB and CD are two chords of a circle with centre at O intersecting each other at point R as shown in the figure then ∠BRQ = ∠CRQ.
Reason (R): Equal chord are always equidistant from the centre of the circle.

Question 2.
Assertion (A): Angle in the same segment are equal.
Reason (R): Angle in a semi circle is a right angle.

Short Answer Type Questions

Question 1.
In the given figure, ∠ABC = 45°, prove that OA ⊥OC.

Question 2.
In the given figure. 0 is the centre of the circle. 0M ⊥ BC, OL ⊥ AB, ON ⊥ AC and OM = ON = OL. Is ∆ABC equilateral? Give reason.

Question 3.
A, B and C are three distinct points on a circle with centre O. If AOB is diameter and AC BC, then find the measure of ∠BAC.

Question 4.
If BM and CN are the perpendiculars drawn on the sides AC and AB of the triangle ABC, prove that the points B, C, M and N are concyclic.

Question 5.
If a pair of opposite sides of a cyclic quadrilateral are equal, prove that its diagonals are also equal.

Question 6.
Prove that chords of a circle are at equal distance from the centre.

Long Answer Type Questions

Question 1.
Prove that the sum of the angles in the four segments exterior to a cyclic quadrilateral is equal to 6 right angles.

Question 2.
AB is a diameter of the circle with centre O and chord CD is equal to radius OC (as shown in figure). AC and BD produced to meet at P. Prove that ∠APB = 60°.

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