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NCERT Solutions for Class 9 Maths Chapter 11 Circles Ex 11.5 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 11 Circles Ex 11.5.

NCERT Solutions for Class 9 Maths Chapter 11 Circles Ex 11.5

Question 1.
In 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 the arc ABC, find ∠ ADC.

Solution:
∴ ∠AOC = ∠AOB + ∠BOC = 60P + 30° = 90°
∴ Arc ABC makes 90° at the centre of the circle.
∴ ∠ADC = \(\frac { 1 }{ 2 }\) ∠AOC
(∵ The angle subtended by an arc at the centre is double the angle subtended by it any part of the circle.)
= \(\frac { 1 }{ 2 }\) x 90° = 45°

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:
Let BC be chord, which is equal to the radius. Join OB and OC.

Given, BC=OB = OC
∴ ∆OBC is an equilateral triangle.
∠BOC =60°
∴ BAC = \(\frac { 1 }{ 2 }\) ∠BOC
= \(\frac { 1 }{ 2 }\) x 60° = 30°
(∵ The angle subtended by an arc at the centre is double the angle subtended by it any part of the circle.)
Here, ABMC is a cyclic quadrilateral.
∴ ∠BAC + ∠BMC = 180°
(∵ In a cyclic quadrilateral the sum of opposite angles is 180°)
⇒ ∠BMC= 180° – 30° =150°

Question 3.
In figure, ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR.

Solution:
∴ ∠POR = 2∠PQR = 2 x 100° = 200°
(Since, the angle subtended by the centre is double the angle subtended by circumference.)
Since, in ∆OPR, ∠POR = 360° – 200° = 160° .. (i)
Again, ∆ OPR, OP = OR (Radii of the circle)
∴ ∠OPR = ∠ORP (By property of isosceles triangle)
In ∆POR, ∠OPR + ∠ORP + ∠POR = 180° …(ii)
From Eqs. (i) and (ii), we get
∠OPR + ∠OPR + 160° = 180°
∴ 2 ∠OPR = 180° – 160° = 20°
∴ ∠OPR = \(\frac { { 120 }^{ circ } }{ 2 }\) = 10°

Question 4.
In figure, ∠ABC = 69°,∠ACB = 31°, find ∠BDC.

Solution:
∵ ∠BDC = ∠BAC …(i)
(Since, the angles in the same segment are equal)
Now , in ∆ABC
∴ ∠A + ∠B+ ∠C= 180°
⇒ ∠A+ 69°+ 31° = 180°
⇒ ∠A + 100° = 180°
∴ ∠A = 180° – 100° = 80°
⇒ ∠BAC=80°
∴ From Eq.(i)∠BDC = 80°

Question 5.
In figure, A, B and C are four points on a circle. AC and BD intersect at a point E such that ∠ BEC = 130° and ∠ ECD = 20°. Find ∠BAC.

Solution:
∴ ∠AEB = 180° – 130° = 50° (Linear Pair) …(i)
⇒ ∠CED = ∠AEB = 50° (Vertically opposite)
Again ∠ABD = ∠ACD (Since, the angles in the same segment are equal)
∠ABE = ∠ECD
⇒ ∠ABE = 180° …(ii)
∴ In ∆ CDE
∠A+ 20° + 50° = 180° [From Eqs. (i) and (ii)]
∠A + 70° = 180°
∴ ∠A = 180°- 70° =110°
Hence ∠BAC = 110°

Question 6.
ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC = 70°, ∠BAC is 30°, find ∠BCD. Further, if AB = BC, find ∠ECD.
Solution:
Angles in the same segment are equal.

∴ ∠BDC = ∠BAC
∴ ∠BDC = 30°
In ∆ BCD, we have
∴ ∠BDC + ∠DBC + ∠BCD = 180° (Given, ∠DBC = 70° and ∠BDC = 30°)
∴ 30° + 70° + ∠BCD = 180°
∴ ∠BCD= 180°-30°-70° = 80°
If AB = BC, then ∠BCA = ∠BAC= 80° (Angles opposite to equal sides in a triangle are equal)
Now, ∠ECD = ∠BCD – ∠BCA = 80° – 30P = 50° (∵ ∠BCD = 80° and ∠BCA =30°)
Hence, ∠BCD = 80°
and ∠ECD = 50°

Question 7.
If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
Solution:
Given: Diagonals NP and QM of a cyclic quadrilateral are diameters of the circle through the vertices M, P, Q and N of the quadrilateral NQPM.

To prove: Quadrilateral NQPM is a rectangle.
Proof: ∵ ON = OP = OQ = OM (Radii of circle)
Now, ON = OP = \(\frac { 1 }{ 2 }\) NP
and OM = OQ = \(\frac { 1 }{ 2 }\) MQ
∴ NP = MQ
Hence, the diagonals of the quadrilateral MPQN are equal and bisect each other. So, quadrilateral NQPM is a rectangle.

Question 8.
If the non-parallel sides of a trapezium are equal, prove that it is cyclic.
Solution:
Given: Non-parallel sides PS and QR of a trapezium PQRS are equal.

To prove: ABCD is a cyclic trapezium.
Construction: Draw SM ⊥ PQ and RN ⊥ PQ.
Proof In ∆SMP and ∆RNQ, we get
SP = RQ (Given)
∠SMP = ∠RNQ (Each = 90°)
and SM = RN
(∵ Distance between two parallel lines is always equal)
∴ By RHS criterion, we get
∆ SMP ≅ ∆ RNQ
So, ∠P = ∠Q (By CPCT)
and ∠PSM = ∠QRN
Now, ∠PSM = ∠QRN
∴ 90° + ∠PSM = 90° + ∠QRN (Adding both sides 90°)
∴ ∠MSR + ∠PSM = ∠NRS + ∠QRN (∵∠MSR = ∠NRS = 90°)
So, ∠PSR = ∠QRS
i.e., ∠S = ∠R
Thus, ∠P = ∠Q and ∠R = ∠S …(i)
∴ ∠P+ ∠Q+ ∠R+ ∠S = 360° (∵ Sum of the angles of a quadrilateral is 360°)
∴ 2∠S + ∠Q = 360° [From Eq. (i)]
∠S+∠O = 180°
Hence, PQRS is a cyclic trape∠ium.

Question 9.
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 figure). Prove that ∠ ACP = ∠QCD.

Solution:
Given: 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.
To prove: ∠ACP = ∠QCD
Proof: In circle I, ∠ACP = ∠ABP (Angles in the same segment) …(i)
In circle II, ∠QCD = ∠QBD{Angles in the same segment)…(ii)
∠ABP = ∠QBD (Vertically opposite angles)
From Eqs. (i) and (ii), we get ∠ACP = ∠QCD

Question 10.
If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.
Solution:
Given: Two circles are drawn with sides AC and AB of AABC as diameters . Both circles intersect each other at D.
To prove: D lies on BC.
Construction: Join AD.
Proof: Since, AC and AB are the diameters of the two circles.
∠ADB = 90° ( ∴ Angles in a semi-circle) …(i)
and ∠ADC = 90° (Angles in a semi-circle) …(ii)

On adding Eqs. (i) and (ii), we get
∠ADB + ∠ADC = 90° + 90° = 180°
Hence, BCD is a straight line.
So, D lies on BC.

Question 11.
ABC and ADC are two right angled triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD.
Solution:
Since, ∆ADC and ∆ABC are right angled triangles with common hypotenuse.

Draw a circle with AC as diameter passing through B and D. Join BD.
∵ Angles in the same segment are equal.
∴ ∠CBD = ∠CAD

Question 12.
Prove that a cyclic parallelogram is a rectangle.
Solution:
Given: PQRS is a parallelogram inscribed in a circle.
To prove: PQRS is a rectangle.

Proof: Since, PQRS is a cyclic quadrilateral.
∴ ∠P+∠R = 180°
(∵ Sum of opposite angles in a cyclic quadrilateral is 180°) …(i)
But ∠P = ∠R (∵ In a || gm opposite angles are equal) …(ii)
From Eqs. (i) and (ii), we get
∠P = ∠R = 90°
Similarly, ∠Q = ∠S = 90
∴ Each angle of PQRS is 90°.
Hence, PQRS is a rectangle.
We hope the NCERT Solutions for Class 9 Maths Chapter 11 Circles Ex 11.5, help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 11 Circles Ex 11.5, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 9 Maths Chapter 11 Circles Ex 11.3 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 11 Circles Ex 11.3.

NCERT Solutions for Class 9 Maths Chapter 11 Circles Ex 11.3

Question 1.
Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?
Solution:
Different pairs of circles are
(i) Two points common

(ii) One point is common

(iii) No point is common

(iv) No point is common

(v) One point is common

From figures, it is obvious that these pairs many have 0 or 1 or 2 points in common.
Hence, a pair of circles cannot intersect each other at more than two points.

Question 2.
Suppose you are given a circle. Give a construction to find its centre.
Solution:
Steps of construction

Taking three points P,Q and R on the circle.
Join PQ and QR,
Draw MQ and NS, respectively the perpendicular bisectors of PQ and RQ, which intersect each other at O.
Hence, O is the centre of the circle.

Question 3.
If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.
Solution:
Given: Two circles with centres O and O’ intersect at two points M and N so that MN is the common chord of the two circles and OO’ is the line segment joining the centres of the two circles. Let OO’ intersect MN at P.
To prove: OO’ is the perpendicular bisector of MN.

Construction: Draw line segments OM, ON, O’M and O’N.
Proof In ∆ OMO’ and ONO’, we get
OM = ON (Radii of the same circle)
O’M = O’N (Radii of the same circle)
OO’ = OO’ (Common)
∴ By SSS criterion, we get
∆ OMO’ ≅ ONO’
So, ∠ MOO’ = ∠ N00′ (By CPCT)
∴ ∠ MOP = ∠ NOP …(i)
(∵ ∠ MOO’ = ∠ MOP and ∠ NOO’ = ∠ NOP)
In ∆ MOP and ∆ NOP, we get
OM = ON (Radii of the same circle)
∠ MOP = ∠NOP [ From Eq(i)]
and OM = OM (Common)
∴ By SAS criterion, we get
∆ MOP ≅ ∆NOP
So, MP = NP (By CPCT)
and ∠ MPO = ∠ NPO
But ∠ MPO + ∠NPO = 180° ( ∵MPN is a straight line)
∴ 2 ∠ MPO = 180° ( ∵ ∠ MPO = ∠ NPO)
⇒ ∠ MPO = 90°
So, MP = PN
and ∠ MPO = ∠ NPO = 90°
Hence, OO’ is the perpendicular bisector of MN.

We hope the NCERT Solutions for Class 9 Maths Chapter 11 Circles Ex 11.3, help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 11 Circles Ex 11.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 9 Maths Chapter 11 Circles Ex 11.2 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 11 Circles Ex 11.2.

NCERT Solutions for Class 9 Maths Chapter 11 Circles Ex 11.2

Question 1.
Recall that two circles are congruent, if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres
Solution:
Given MN and PQ are two equal chords of two congruent circles with centre at O and O’.
To prove ∠ MON = ∠ PO’Q

Proof In ∆ MON and ∆ PO’Q, we have
MO = PO’ (Radii of congruent circles)
NO = QO’ (Radii of congruent circles)
and MN = PQ (Given)
∴ By SSS criterion, we get
∆ MON = ∆ PO’Q
Hence, ∠ MON = ∠ PO’Q (By CPCT)

Question 2.
Prove that, if chords of congruent circles subtend equal angles at their centres, then the chords are equal.
Solution:
Given MN and PQ are two chords of congruent circles such that angles subtended by .
these chords at the centres O and O’ of the circles are equal.

To prove MN = PQ
Proof In ∆ MON and ∆ PO’Q, we get
MO = PO’ (Radii of congruent circles)
NO = QO’ (Radii of congruent circles)
and ∠MON = ∠PO’Q (Given)
∴ By SAS criteria, we get
∆ MON = ∆ PO’Q
Hence, MN = PQ (By CPCT)

We hope the NCERT Solutions for Class 9 Maths Chapter 11 Circles Ex 11.2, help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 11 Circles Ex 11.2, drop a comment below and we will get back to you at the earliest.