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

By using Extra Questions for Class 9 Maths and Ganita Manjari Class 9 Maths Chapter 5 I’m Up and Down and Round and Round Extra Questions, students can improve their problem-solving skills.

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

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

Class 9 Ganita Manjari Chapter 5 Extra Questions

Question 1.
In th figure, if O is the centre of the circle and ∠CBA = 35°. find the value of x.

Solution:
∠CBA = 35° Given
∠CAB = x° Given
∠BCA = 90°
Angle in a semi-circle is 90°
In ∆ABC,
∠CBA + ∠CAB + ∠BCA = 180°
Sum of the angles of a triangle is 180°
⇒ 35° + x° + 90° = 180°
⇒ x°+ 125° = 180°
⇒ x° = 180° – 125°
⇒ x° = 55°
⇒ x = 55

Question 2.
Find the angle B in the figure if ∠A = 60°.

Solution:
∠B = ∠A
Angles in the same segment of a circle are equal
⇒ ∠B = 60°

Question 3.
In figure, ABC is an equilateral triangle. Find the value of x.

Solution:
∵ ∆ABC is an equilateral triangle.
∴ ∠ABC = ∠BAC = ∠ACB = 60° …(1)
Each angle of an equilateral triangle is 60°
∠BDC = ∠BAC
Angles in the same segment of a circle are equal
⇒ x° = 60° [From (1)]
= 60

Question 4.
In figure. 0 is the centre of the circle and AB is the chord. If OD ⊥ AB then find radius of the circle.
AB = 6 cm, OD = 4 cm

Solution:

Construction: Join OA
∵ OD ⊥ AB [Given]
∴ AD = DB = \(\frac{1}{2}\)AB
= \(\frac{1}{2}\) x 6 = 3 cm
[The perpendicular from the centre of I a circle to a chord bisects the chord]
In right triangle ODA,
OA² = OD² + AD²
By Baudhayana Theorem = 4² + 3² = 16 + 9 = 25
⇒ OA = 5 cm
Hence, the radius of the circle is 5 cm.

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

Question 5.
Find the length of a chord of a circle of radius 5 cm which is at a distance of 3 cm from the centre of the circle.
Solution:
Let AB be a chord of a circle of radius 5 cm at a distance of 3 cm from the centre.

In right triangle ODA,
OA² = OD² + AD² [By Baudhayana]
Theorem
⇒ 5² = 3² + AD²
⇒ 25 = 9 + AD²
AD² = 16
AD = 4 cm …(1)
∵ The perpendicular from the centre of a circle to a chord bisects the chord
∴ AD = DB = \(\frac{1}{2}\)AB
⇒ AD =\(\frac{1}{2}\)AB
4 = \(\frac{1}{2}\)AB [From (1)]
AB = 8 cm.
Hence, the length of the chord is 8 cm.

Question 6.
If a circle is divided into 4 equal parts, find the angle subtended by each arc at the centre.
Solution:
\(\overparen{\mathrm{AB}}=\overparen{\mathrm{BC}}=\overparen{\mathrm{CD}}=\overparen{\mathrm{DA}}\)

∠AOB = ∠BOC
= ∠COD = ∠DOA …(1)
[Equal arcs of a circle subtend equal angles at the centre]
But ∠AOB + ∠BOC + ∠COD + ∠DOA = 360° …(2)
The sum of all the angles round a point is equal to 360°
From (1) and (2),
∠AOB = ∠BOC = ∠COD = ∠DOA = 90°
Hence, the angle subtended by each arc at the centre is 90°.

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

Question 1.
In the given figure, O is the centre of the circle such that OA = 5 cm, AB = 8 cm and OD ⊥ AB meeting AB at C. Find the length of CD.

Solution:
∴ OD ⊥ AB [Given]
OC ⊥ AB
C is the mid-point of AB
[The perpendicular drawn from the centre of a circle to a chord bisects the chord]
⇒ AC = CB
⇒ AC = CB = – AB²
⇒ AC = – AB
⇒ AC = – x 8 [AB = 8 cm (given)]
⇒ AC = 4 cm
In right triangle OCA,
OA² = OC² + AC²
[By Pythagoras Theorem]
⇒ (5)² = OC² + (4)²
⇒ 25 = OC² + 16
⇒ OC² = 9
⇒ OC = 3 cm
CD = OD – OC = OA – OC
Radii of the same circle are equal
= 5 – 3
= 2 cm
Hence, the length of CD is 2 cm.

Question 2.
A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.
Solution:
OA = OB = AB [Given]
ΔOAB is equilateral
∠AOB = 60° …(i) [Each angle of an equilateral triangle is 60°]

∠AOB = 2∠ADB
The angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.
60° = 2∠ADB [From (1)]
∠ADB = 30° …(2)
Hence, the angle subtended by the chord at a point on the major arc is 30°.
Again, since ACBD is a cyclic quadrilateral, therefore,
∠ADB + ∠ACB = 180°
[Opposite angles of a cyclic quadrilateral are supplementary]
30° + ∠ACB = 180° [From (2)]
∠ACB = 150°
Hence, the angle subtended by the chord at a point on the minor arc is 150°.

Question 3.
In the adjoining figure A, B and C are three points on a circle with centre 0 such that ∠BOC = 30° and ∠AOB = 60°. If D is a point on the circle other than arc ABC, find ∠ABC.

Solution:
∠AOC = ∠AOB + ∠BOC
= 60°+ 30° = 90° …(1)
∠AOC = 2∠ADC
The angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle
90° = 2∠ADC [From (1)]
∠ADC = 45° …(2)
∵ ABCD is a cyclic quadrilateral
∠ABC + ∠ADC = 180°
[Opposite angles of a cyclic quadrilateral are supplementary]
∠ABC + 45°= 180° [From (2)]
∠ABC = 135°
In ΔABC,
BA = AC [Given]
∠ACB = ∠ABC
[Angles opposite to equal sides of a triangle are equal]
∠ACB = ∠ABC = 50° …(1)

Question 4.
In the given figure, 0 is the centre of the circle and BA = AC. If ∠ABC = 50°, find ∠BOC and ∠BDC.

Solution:
In ΔABC,
BA = AC [Given]
∠ACB = ∠ABC
[Angles opposite to equal sides of a triangle are equal]
∠ACB = ∠ABC = 50° …(1)
In ΔABC,
∠ABC + ∠ACB + ∠BAC = 180°
[Sum of the angles of a triangle is 180°]
50° + 50° + ∠BAC = 180° |From(l)
100°+ ∠BAC = 180°
∠BAC = 80°
∠BOC = 2∠BAC
The angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.
∠BOC = 2 × 80° [From (2)]
∠BOC = 160°
∵ ABDC is a cyclic quadrilateral,
∠BAC + ∠BDC = 180°
I Opposite angles of a cyclic quadrilateral are supplementary
80° + ∠BDC = 180° | From (2)
∠BDC = 100° …(4)

Question 5.
In the given figure, find the value of x + y.

Solution:
∵ ABCD is a cyclic quadrilateral.
∠BCD + ∠BAD = 180°
Opposite angles of a cyclic quadrilateral are supplementary
x° + 60° = 180°
⇒ x° = 120°
x = 120 …(1)
∠CDA + ∠ABC = 180°
[Opposite angles of a cyclic quadrilateral are supplementary]
y° + 70° = 180°
⇒ y° = 110°
y = 11o …(2)
x + y = 120 + 110 = 230.

Question 6.
In the given figure, find the value of x.

Solution:
ABCD is a cyclic quadrilateral.
∠BAD + ∠BCD = 180°
[Opposite angles of a cyclic quadrilateral are supplementary]
(2x + 4)° + (4x – 64)° = 180°
(2x + 4x – 60)° = 180°
(6x – 60)° = 180°
6x – 60 = 180
6x = 240
⇒ x = 40

Question 7.
In the given figure, ABCE is a cyclic quadrilateral and 0 is the centre of the circle. If ∠AEC = 110°, then find ∠ABC and ∠ADC.

Solution:
(i) ∵ ABCE is a cyclic quadrilateral.
∴∠ABC + ∠AEC = 180°
[Opposite angles of a cyclic quadrilateral are supplementary]
⇒ ∠ABC + 110° = 180°
⇒ ∠ABC = 70°

(ii) ∵ AECD is a cyclic quadrilateral.
⇒ ∠AEC + ∠ADC = 180°
[Opposite angles of a cyclic quadrilateral are supplementary]
⇒ 110° + ∠ADC = 180°
⇒ ∠ADC = 70°

Question 8.
In figure PS = SR, angle RPS = 54° and angle PRQ = 46°. Find the measure of angle TQR and measure of angle RTQ.

Solution:
In ΔPSR,
PS = SR [Given]
∠PRS = ∠RPS
[Angles opposite to equal sides of a triangle are equal]
⇒ ∠PRS = 54°
In ΔPSR,
∠PSR + ∠RPS + ∠PRS = 180°
[The sum of the angles of a triangle is 180°]
⇒ ∠PSR + 54° + 54° = 180°
[From (1) and ∠RPS = 54° (Given)]
⇒ ∠PSR + 108°= 180°
⇒ ∠PSR = 72°
∵ PQRS is a cyclic quadrilateral.
∴ Ext. ∠TQR = ∠PSR
[An exterior angle of a cyclic quadrilateral is equal to its interior opposite angle]
⇒ ∠TQR = 72° …(3) | From (2)
∠QRS = ∠QRP + ∠PRS
= 46° + 54°
[From (1) and ∠QRP = 46° (Given) ]
= 100° …(4)
∵ PQRS is a cyclic quadrilateral.
∴ ∠QRS + ∠SPQ = 180°
[Opposite angles of a cyclic quadrilateral are supplementary
⇒ 100° + ∠SPQ = 180° [From (4)]
⇒ ∠SPQ = 80° …(5)
∵ PQRS is a cyclic quadrilateral
∴ Ext. ∠QRT = ∠SPQ
[An exterior angle of a cyclic quadrilateral is equal to its interior opposite angle]
⇒ ∠QRT = 80° …(6) | From (5)
In ΔQRT,
∠TQR + ∠QRT + ∠RTQ = 180°
[The sum of the angles of a triangle is 180° ]
⇒ 72° + 80° + ∠RTQ = 180°
[From (3) and (6)]
⇒ 152°+ ∠RTQ = 180°
⇒ ∠RTQ = 28°.

Question 9.
In the given figure, 0 is the centre of the circle. Find the values of x, y, z.

Solution:
∵ ABCD is a cyclic quadrilateral.
∠DAB + ∠DCB = 180°
[Opposite angles of a cyclic quadrilateral are supplementary]
⇒ x° + 3x° = 180°
⇒ 4x° = 180°
⇒ x° = 45°
⇒ x = 45 …(1)
∠BOD = 2∠BAD
The angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.
y° = 2x°
= 2 x 45° [From (1)]
= 90°
⇒ y° = 90 …(2)
y° + z° = 360°
[The sum of all the angles around a point is equal to 360°]
⇒ 90° + z° = 360° [From (2)]
⇒ z° = 270°
⇒ z = 270 …(3)

Question 10.
In the given figure, 0 is the centre of the circle. Find ∠BAO, ∠AOB, ∠BOD and ∠ODB if ∠AOC = 130° and ∠OCD = 30°.

Solution:
(i) ∠AOB + ∠AOC = 180°
[Linear Pair Axiom]
⇒ ∠AOB + 130° = 180°
⇒ ∠AOB = 50° …(1)

(ii) In ΔOAB,
OA = OB
[Radii of the same circle]
∠OBA = ∠OAB …(2)
[Angles opposite to equal sides of a triangle are equal]
In ΔOAB,
∠OBA + ∠OAB + ∠AOB = 180°
[The sum of the angles of a triangle is 180°.]
2 ∠OAB+ 50° = 180° [From (1) and (2)]
2∠BAO = 130°
∠BAO = 65° …(3)

(iii) In ΔOCD,
v OC = OD | Radii of the same circle
∠ODC = ∠OCD
[Angles opposite to equal sides of a triangle are equal]
∠ODC = 30° …(4)
∵ Z OCD = 30° (Given)
In ΔOCD,
Ext. ∠BOD = ∠OCD + ∠ODC [An exterior angle of a triangle is equal to the sum of its two interior opposite angles]
∠BOD = 30° + 30°
∠BOD = 60° …(5)

(iv) In ΔOBD,
OB = OD | Radii of the same circle
∠ODB = ∠OBD …(6)
[Angles opposite to equal sides of a triangle are equal]
In ΔOBD,
∠BOD + ∠OBD + ∠ODB = 180°
[The sum of the angles of a triangle is 180°]
60° + 2∠ODB = 180° [From (5) and (6)]
2∠ODB = 120°
∠ODB = 60°

Question 11.
Find the values of x, y, z, w from the figure, where O is the centre of the circle, ∠AOC = 110° and ∠OAB = 65°.

Solution:
(i) y° + 110° = 360°
[Sum of all the angles round a point is equal to 360°]
⇒ y° = 250°
⇒ y = 250 ..(1)

(ii) ∠AOC = 2 ∠ADC
The angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle
110° = 2z°
⇒ 2z° = 110°
z = 55

(iii) ∵ABCD is a cyclic quadrilateral.
∠ABC + ∠ADC = 180°
[Opposite angles of a cyclic quadri-lateral are supplementary]
x° + z° = 180°
x° + 55° = 180° [From (2)]
x° = 125° …(3)
x= 125

(iv) ∵OABC is a quadrilateral.
110° + 65° + x° + w° = 360°
[The sum of the angles of a quadrilateral is 360°.]
110° + 65° + 125° + w° = 360° [From (3)]
300° + w° = 360°
= 60°
w = 60 …(4)

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

Question 1.
Two diameters of a circle intersect each other at right angles. Prove that the quadrilateral formed by joining their end points is a square.
Solution:

∵ The diagonals of the quadrilateral ACBD bisect each other.
∴ Quadrilateral ACBD is a parallelogram. ………(1)
Also, ∠CAD = 90°
[Angle is a semicircle is 90°]
In view of (1) and (2),
Quadrilateral ACBD is a rectangle …(3)
[A parallelogram whose one angle is 90° is a rectangle]
In ∆AOC and ∆AOD.
OC = OD
Radii of the same circle are equal
OA = OA [Common side]
∠AOC = ∠AOD [Each = 90°(Given)]
∆AOC = ∆AOD [By SAS congruence rule]
AC = AD
In view of (3) and (4),
Quadrilateral ABCD is a square. [A rectangle, whose adjacent sides are equal, is a square.]

Question 2.
AC and BD are two chords of a circle which bisect each other. Prove that
(i) AC and BD are diameters
(ii) ABCD is a rectangle.
Solution:
Given:AC and BD are two chords of a circle which bisect each other.
To Prove: (i) AC and BD are diameters.

(i) ABCD is a rectangle.

Construction: Join AB, BC, CD and DA.
Proof: (i) In ∆OAB and ∆OCD,
OA = OC [∵ 0 is the mid-point of AC]
OB = OD [∵ 0 is the mid-point of BD]
∠AOB = ∠COD [Vertically opposite angles]
∴ ∆OAB ≅ ∆OCD [SAS Rule]
AB = CD [CPCT]
AB = CD …(1)
[If two chords of a circle are equal, then 1 their corresponding arcs are congruent]
In ∆OAD and ∆OCB.
OA = OC
[0 is the mid-point of AC]
OD = OB
[∵ 0 is the mid-point of BD]
∠AOD = ∠COB
[Vertically Opposite Angles]
∆OAD ≅ ∆OCB [SAS Rule]
AD = CB [CPCT]
⇒ \(\overparen{\mathrm{AD}} \cong \overparen{\mathrm{CB}}\) …(2)
[If two chords of a circle are equal, then 1 their corresponding arcs are congruent]
Adding (1) and (2), we get
\(\overparen{\mathrm{AB}}+\overparen{\mathrm{AD}} \equiv \overparen{\mathrm{CD}}+\overparen{\mathrm{CB}}\)
⇒ \(\widehat{\mathrm{DAB}} \cong \widehat{\mathrm{DCB}}\)
⇒ BD divides the circle into two equal parts.
BD is a diameter.
Similarly, we can prove that AC is a diameter.

(ii) Quadrilateral ABCD is a parallelogram.
∵ Diagonals AC and BD bisect each other (If the diagonals of a quadrilateral bisect each other, then it is a parallelogram)
Also, ∠BAD = 90°
[Angle in a semi-circle is 90°]
ABCD is a rectangle.
[A parallelogram, whose one angle is 90°, is a rectangle.]

Question 3.
In figure, a diameter AB of a circle bisects a chord PQ. If AQ || PB, prove that the chord PQ is also a diameter of the circle.

Solution:
Given: A diameter AB of a circle bisects a chord PQ.
AQ || PB.
To Prove: The chord PQ is also a diameter of the circle.
Proof: ∠AQP = ∠ABP
[Angles in the same segment of a circle are equal]
∵ AQ || PB [Given]
and a transversal QP intersects them
∵ ∠AQP = ∠QPB [Alternate Interior Angles]
From (1) and (2),
∠ABP = ∠QPB
∠OBP = ∠OPB
OP = OB
[Sides opposite to equal angles of a triangle are equal Again.]
∠BPQ = ∠BAQ …(4)
[Angles in the same segment of a circle are equal]
∵ AQ || PB [Given]
and a transversal QP intersects them
∠BPQ = ∠PQA [Alternate Interior Angles]

From (4) and (5),
∠BAQ = ∠PQA
⇒ ∠OAQ = ∠OQA
OQ = OA …(6)
| Sides opposite to equal angles ‘ of a triangle arc equal
Adding (3) and (6), we get OP + OQ = OB + OA
⇒ PQ = AB
∵ AB is a diameter of the circle.
PQ is also a diameter of the circle.

Question 4.
In the given figure, O is the centre of the circle and L and M are the mid-points of AB and CB respectively. If ∠OAB = ∠OCB, prove that BL = BM.

Solution:
In ∆OAL and ∆OCM.
∠OAL = ∠OCM [Given]
∠LOA = ∠MOC
[Vertically Opposite Angles]
OA = OC
[Radii of the same circle]
∆OAL ≅ ∆OCM
[ASA congruence rule
AL = CM [CPCT]
∴ 2AL = 2CM
∴AB = CB
[∵ L and M are the mid-points of AB and CB respectively]
Also, AL = CM [CPCT]
∴BL = BM
[∵ L and M are the mid-points of AB and CB respectively.]

Question 5.
In figure, O is the centre of the circle, OC = 5 cm and AB = BC = 2√5 cm. Find the length of AC.

Solution:
Construction: Join OB so as to intersect AC at D.

OA = OC = 5 cm [Radii of the same circle]
AB = BC = 2√5 cm [Given]
Quadrilateral OABC is a kite.
OB ⊥ AC
and DA = DC
In ∆OAB,
OA = 5 cm
OB = 5 cm
AB = 2√5 cm
∴ ∆OAB is an isosceles triangle.
∴ Area of ∆OAB
= \(\frac{2 \sqrt{5}}{4} \sqrt{4(5)^2-(2 \sqrt{5})^2}\)
Area of an isosceles triangle with base a and equal sides b = \(\frac{a}{4} \sqrt{4 b^2-a^2}\)]
= \(\frac{2 \sqrt{5}}{4} \sqrt{100-20}\)
= \(\frac{2 \sqrt{5}}{4} \sqrt{80}\)
= \(\frac{2 \sqrt{5}}{4} 4 \sqrt{5}\)
= 10 cm²
Again, area of ∆OAB
= \(\frac{\mathrm{OB} \times \mathrm{AD}}{2}=\frac{5 \times \mathrm{AD}}{2}\)
\(\frac{5 \mathrm{AD}}{2}\) = 10
⇒ AD = 4 cm
⇒ 2AD = 8 cm
⇒ AC = 8 cm [∵ DA = DC = \(\frac{1}{2}\)AC]

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

Question 1.

Based on the above information, answer the following questions :
(i) In Fig. 1, 0 is the center of the circle. If x = 50 then y =
(a) 75
(b) 50
(c) 25
(d) 37
Solution:
(c) 25

(ii) In Fig. 2, if a = 20 then b =
(a) 20
(b) 40
(c) 60
(d) cannot be determined
Solution:
(a) 20

(iii) In Fig. 3, if p = 60 then q =
(a) 60
(b) 30
(c) 90
(d) 120
Solution:
(d) 120

(iv) In Fig. 4, if a = 50 then b =
(a) 100
(b) 50
(c) 130
(d) cannot be determined
Solution:
(b) 50

(v) In Fig. 5, AB is a diameter. If x = 60 then y =
(a) 60
(b) 30
(c) 90
(d) 45
Solution:
(b) 30

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

Very Short Answer Type Questions

Question 1.
In the given figure, ∠ADC = 130°, CE ⊥ AB at O and chord BC = chord BE. Find ∠CBE.

Solution:
100°

Question 2.
In the given figure, AOB is a diameter of the circle and C, D, E are any three points on the semi-circle. Find the value of ∠ACD + ∠BED.

Solution:
270°

Question 3.
In the given figure, O is the centre of the circle, ∠BCO = 30°. Find x and y.

Solution:
x = 30, y = 15

Question 4.
In the given figure, ∠ACB = 40°. Find ∠OAB.

Solution:
20

Short Answer Type Questions

Question 1.
In the given figure, AB and CD are two chords of a circle intersecting each other at point E. Prove that ∠AEC = ½ (Angle subtended by arc CXA at centre + angle subtended by arc DYB at the centre).

Solution:
Do yourself

Question 2.
In the given figure, O is the centre of the circle, BD = OD and CD ⊥ AB. Find ∠CAB.

Solution:
30°

Question 3.
Two equal chords AB and CD of a circle when produced intersect at a point P. Prove that PB = PD.
Solution:
Do yourself

Question 4.
Show that if two chords of a circle are equal, they subtend equal angles at its centre.
Solution:
Do yourself

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

Question 6.
Suppose you are given a circle. Give a construct to find the centre of the circle.
Solution:
Do yourself

Long Answer Type Questions

Question 1.
ABCD is a parallelogram. A circle through A, B is so drawn that it intersects AD at P and BC at Q. Prove that P, Q, C and D are concyclic.
Solution:
Do yourself

Question 2.
In the given figure, AB = AC, BE and DC intersect at O. Also BCED is a cyclic quadrilateral. Show AO is perpendicular bisector of DE.

Solution:
Do yourself

Question 3.
In the given figure, bisectors of angles of A, B and C of A DEF intersect its circumcircle at D, E and F respectively. Show angles of ΔDEF are
(90° – \(\frac{1}{2}\)∠A), (90° – \(\frac{1}{2}\)∠B) and (90° – \(\frac{1}{2}\)∠C)

Solution:
Do yourself

Question 4.
In the given figure, show that the quadrilateral formed by angle bisectors of any quadrilateral is cyclic.

Solution:
Do yourself

Case Study Based Questions

Ms. Rashmi is a Maths teacher. She has just finished teaching theorems on circles to her students of class 9. To test them she drew five figures on the board and asked them a question for each.

Answer the questions she asked.
(i) In Fig. 1, O is the centre of the circle. If x = 60 then find the value of y.
Solution:
30

(ii) In Fig. 2, if a = 30 then find the value of b.
Solution:
30

(iii) In Fig. 3, if p = 70 then find the value of q.
Solution:
110

(iv) In Fig. 4, if a = 100 then find the value of b.
Solution:
100

(v) In Fig. 5, AB is a diameter. If ∠A = 60° then find the measure of ∠B.
Solution:
30°