RS Aggarwal Solutions for Class 9 Maths Chapter 11–Circle

Class 9: Maths Chapter 11 solutions. Complete Class 9 Maths Chapter 11 Notes.

RS Aggarwal Solutions for Class 9 Maths Chapter 11–Circle

RS Aggarwal 9th Maths Chapter 11, Class 9 Maths Chapter 11 solutions

Ex 11A Solutions

Question 1.
Solution:
Let AB be a chord of a circle with centre O. OC⊥AB and OA be the radius of the circle, then
AB = 16cm, OA = 10cm .

OC ⊥ AB.
OC bisects AB at C
AC = 12 AB = 12 x 16 = 8cm

Question 2.
Solution:
Let AB be the chord of the circle with centre O and OC ⊥ AB, OA be the radius of the circle,
then OC = 3cm, OA = 5cm

Now in right ∆ OAC,
OA² = AC² = OC² (Pythagoras Theorem)

Question 3.
Solution:
Let AB be the chord, OA be the radius of
the circle OC ⊥ AB
then AB = 30 cm, OC = 8cm

Question 4.
Solution:
AB and CD are parallel chords of a circle with centre O.

Question 5.
Solution:
Let AB and CD be two chords of a circle with centre O.
OA and OC are the radii of the circle. OL⊥AB and OM⊥CD.

Question 6.
Solution:
In the figure, a circle with centre O, CD is its diameter AB is a chord such that CD⊥AB.
AB = 12cm, CE = 3cm.
Join OA.
∵ COD⊥AB which intersects AB at E

Question 7.
Solution:
A circle with centre O, AB is diameter which bisects chord CD at E
i.e. CE = ED = 8cm and EB = 4cm
Join OC.
Let radius of the circle = r

Question 8.
Solution:
Given : O is the centre of a circle AB is a chord and BOC is the diameter. OD⊥AB
To prove : AC || OD and AC = 20D
Proof : OD⊥AB
∵ D is midpoint of AB

Question 9.
Solution:
Given : O is the centre of the circle two
chords AB and CD intersect each other at P inside the circle. PO bisects ∠BPD.
To prove : AB = CD.

Question 10.
Solution:
Given : PQ is the diameter of the circle with centre O which is perpendicular to one chord AB and chord AB || CD.
PQ intersects AB and CD at E and F respectively
To prove : PQ⊥CD and PQ bisects CD.

Question 11.
Solution:
Two circles with centre O and O’ intersect each other.

To prove : The two circles cannot intersect each other at more than two points.
Proof : Let if opposite, the two circles intersect each other at three points P, Q and R.
Then these three points are non-collinear. But, we know that through three non- collinear points, one and only one circle can be drawn.
∵ Our supposition is wrong
Hence two circle can not intersect each other at not more than two points.
Hence proved

Question 12.
Solution:
Given : Two circles with centres O and O’ intersect each other at A and B. AB is a common chord. OO’ is joined.
AO and AO is joined.

Question 13.
Solution:
Given : Two equal circles intersect each other at P and Q.
A straight line drawn through
P, is drawn which meets the circles at A and B respectively
To prove : QA = QB

Question 14.
Solution:
Given : A circle with centre 0. AB and CD are two chords and diameter PQ bisects AB and CD at L and M
To Prove : AB || CD.

Question 15.
Solution:
Given : Two circles with centres A and B touch each other at C internally. A, B arc joined. PQ is the perpendicular bisector of AB intersecting it at L and meeting the bigger circle at P and Q respectively and radii of the circles are 5cm and 3cm. i.e. AC = 5cm,BC = 3cm
Required : To find the lenght of PQ

Question 16.
Solution:
Given : AB is a chord of a circle with centre O. AB is produced to C such that BC = OB, CO is joined and produced to meet the circle at D.
∠ ACD = y°, ∠ AOD = x°
To prove : x = 3y

Question 17.
Solution:
Given : O is the centre of a circle AB and AC are two chords such that AB = AC
OP⊥AB and OQ⊥AC.
which intersect AB and AC at M and N
respectively. PB and QC are joined.
To prove : PB = QC.

Question 18.
Solution:
Given : In a circle with centre O, BC is its diameter. AB and CD are two chords such that AB || CD.
To prove : AB = CD
Const. Draw OL⊥AB
OM⊥CD.
Proof : In ∆ OLB and ∆ OCM,
OB = OC (radii of the same circle)
∠ OLB = ∠ OMC (each 90°)
∠ OBL = ∠ OCM (alternate angles)

Question 19.
Solution:
Equilateral ∆ ABC in inscribed in a circle in which
AB = BC = CA = 9cm.

Question 20.

Solution:
Given : AB and AC are two equal chords of a circle with centre O
To Prove : O lies on the bisector of ∠ BAC

Question 21.
Solution:
Given : OPQR is a square with centre O, a circle is drawn which intersects the square at X and Y.
To Prove : Q = QY
Const. Join OX and OY

Ex 11B

Question 1.
Solution:
(i) O is the centre of the circle
∠OAB = 40°, ∠OCB = 30°
Join OB.

Question 2.
Solution:
O is the centre of the cirlce and ∠AOB = 70°
∵ Arc AB subtends ∠AOB at the centre and ∠ACB at the remaining part of the circle.
∵ ∠ACB = 12 ∠AOB = 12 x 70°
=> ∠ACB = 35°
or ∠OCA = 35°
In ∆OAC,
OA = OC (radii of the same circle)
∴ ∠OAC = ∠OCA = 35° Ans.

Question 3.
Solution:
In the figure, O is the centre of the circle. ∠PBC = 25°, ∠APB =110°
∠ APB + ∠ BPC = 180° (Linear pair)
=> 110° + ∠ BPC = 180°

Question 4.
Solution:
O is the centre of the circle
∠ABD = 35° and ∠B AC = 70°
BOD is the diameter of the circle
∠BAD = 90° (Angle in a semi circle)
But ∠ADB + ∠ABD + ∠BAD = 180° (Angles of a triangle)
=> ∠ADB + 35° + 90° = 180°
=> ∠ADB + 125° = 180°
=> ∠ADB = 180° – 125° = 55°
But ∠ACB = ∠ADB (Angles in the same segment of the circle)
∠ACB = 55° Ans.

Question 5.
Solution:
O is the centre of a circle and ∠ACB = 50°
∴ arc AB subtends ∠ AOB at the centre and ∠ ACB at the remaining part of the circle.
∴ ∠ AOB = 2 ∠ ACB
= 2 x 50° = 100
∴ OA = OB (radii of the same circle)
∴ ∠ OAB = ∠ OBA (Angles opposite to equal sides)
Now in ∆ OAB,
∠ OAB + ∠ OBA + ∠ AOB = 180°
=> ∠ OAB + ∠ OAB + ∠ AOB = 180° (∠OAB = ∠OBA)
=> 2 ∠ OAB + 100°= 180°
=> 2 ∠ OAB = 180° – 100° = 80°
=> ∠OAB = 80o2 = 40°
Hence, OAB = 40° Ans.

Question 6.
Solution:
(i) In the figure,
∠ABD = 54° and ∠BCD = 43°
∠BAD = ∠BCD (Angles in the same segment of a circle)
∠BAD = 43°

Question 7.
Solution:
Chord DE || diameter AC of the circle with centre O.
∠CBD = 60°
∠CBD = ∠ CAD
(Angles in the same segment of a circle)
∠CAD = 60°
Now in ∆ ADC,

Question 8.
Solution:
In the figure,
chord CD || diameter AB of the circle with centre O.
∠ ABC = 25°
Join CD and DO.
AB || CD
∠ ABC = ∠ BCD (alternate angles)

Question 9.
Solution:
AB and CD are two straight lines passing through O, the centre of the circle and ∠AOC = 80°, ∠CDE = 40°
∠ CED = 90° (Angle in a semi circle)

Question 10.
Solution:
O is the centre of the circle and ∠AOB = 40°, ∠BDC = 100°
Arc AB subtends ∠AOB at the centre and ∠ ACB at the remaining part of the circle
∠ AOB = 2 ∠ ACB

Question 11.
Solution:
Chords AC and BD of a circle with centre O, intersect each other at E at right angles.
∠ OAB = 25°. Join OB.
In ∆ OAB,
OA = OB (radii of the same circle)

Question 12.
Solution:
In the figure, O is the centre of a circle ∠ OAB = 20° and ∠ OCB = 55° .
In ∆ OAB,

Question 13.
Solution:
Given : A ∆ ABC is inscribed in a circle with centre O and ∠ BAC = 30°
To Prove : BC = radius of the circle
Const. Join OB and OC

Question 14.
Solution:
In a circle with centre O and PQ is its diameter. ∠PQR = 65°, ∠SPR = 40° and ∠PQM = 50°
(i) ∠PRQ = 90° (Angle in a semicircle) and ∠PQR + ∠RPQ + ∠PQR = 180° (Angles of a triangle)

Ex 11C

Question 1.
Solution:
In cyclic quad. ABCD, ∠ DBC = 60° and ∠BAC = 40°
∴∠ CAD and ∠ CBD are in the same segment of the circle.
∴∠ CAD = ∠ CBD or ∠ DBC
=> ∠ CAD = 60°
∴∠BAD = ∠BAC + ∠CAD
= 40° + 60° = 100°
But in cyclic quad. ABCD,
∠BAD + ∠BCD = 180°
(Sum of opposite angles)
=> 100° + ∠BCD = 180°
=> ∠BCD = 180° – 100°
∴ ∠ BCD = 80°
Hence (i) ∠BCD = 80° and
(ii) ∠CAD = 60° Ans.

Question 2.
Solution:
In the figure, POQ is diameter, PQRS is a cyclic quad, and ∠ PSR =150° In cyclic quad. PQRS.
∠ PSR + ∠PQR = 180°
(Sum of opposite angles)
=> 150° + ∠PQR = 180°
=> ∠PQR = 180°- 150° = 30°
=> ∠PQR =180° – 150° = 30°
Now in ∆ PQR,
∴∠ PRQ = 90° (Angle in a semicircle)
∴∠ RPQ + ∠PQR = 90°
=> ∠RPQ + 30° = 90°
=> ∠RPQ = 90° – 30° = 60° Ans.

Question 3.
Solution:
In cyclic quad. ABCD,
AB || DC and ∠BAD = 100°
∠ ADC = ∠BAD =180°
(co-interior angles)

=> ∠ ADC + 100° = 180°
=> ∠ADC = 180° – 100° = 80°
∴ ABCD is a cyclic quadrilateral.
∴ ∠BAD + ∠BCD = 180°
=> 100° + ∠ BCD = 180°
=> ∠BCD = 180° – 100°
=> ∠BCD = 80°
Similarly ∠ABC + ∠ADC = 180°
=> ∠ABC + 80° = 180°
=> ∠ABC = 180° – 80° = 100°
Hence (i) ∠BCD = 80° (ii) ∠ADC = 80° and (iii) ∠ABC = 100° Ans.

Question 4.
Solution:
O is the centre of the circle and arc ABC subtends an angle of 130° at the centre i.e. ∠AOC = 130°. AB is produced to P
Reflex ∠AOC = 360° – 130° = 230°
Now, arc AC subtends reflex ∠ AOC at the centre and ∠ ABC at the remaining out of the circle.

Question 5.
Solution:
In the figure, ABCD is a cyclic quadrilateral in which BA is produced to F and AE is drawn parallel to CD.
∠ABC = 92° and ∠FAE = 20°
ABCD is a cyclic quadrilateral.

Question 6.
Solution:
In the figure, BD = DC and ∠CBD = 30°
In ∆ BCD,
BD = DC (given)
∠ BCD = ∠ CBD
(Angles opposite to equal sides)
= 30°
But ∠BCD + ∠CBD + ∠BDC = 180° (Angles of a triangle)
=> 30°+ 30°+ ∠BDC = 180°
=> 60°+ ∠BDC = 180°
=> ∠ BDC =180° – 60° = 120°
But ABDC is a cyclic quadrilateral
∠BAC + ∠BDC = 180°
=> ∠BAC + 120°= 180°
=> ∠ BAC = 180° – 120° = 60°
Hence ∠ BAC = 60° Ans.

Question 7.
Solution:
(i) Arc ABC subtends ∠ AOC at the centre , and ∠ ADC at the remaining part of the circle.
∠ AOC = 2 ∠ ADC

Question 8.
Solution:
In the figure, ABC is an equilateral triangle inscribed is a circle
Each angle is of 60°.
∠ BAC = ∠ BDC
(Angles in the same segment)
∠BDC = 60°
BECD is a cyclic quadrilateral.
∠BDC + ∠BEC = 180°
(opposite angles of cyclic quad.)
=> 60°+ ∠BEC = 180°
=> ∠BEC = 180° – 60°= 120°
Hence ∠BDC = 60° and ∠BEC = 120° Ans.

Question 9.
Solution:
ABCD is a cyclic quadrilateral.
∠BCD + ∠BAD = 180°
(opposite angles of a cyclic quad.)
=> 100°+ ∠BAD = 180°
so ∠BAD = 180° – 100° = 80°
Now in ∆ ABD,
∠BAD + ∠ABD + ∠ADB = 180° (Angles of a triangle)

=> 80° + 50° + ∠ADB = 180°
=> 130°+ ∠ADB = 180°
=> ∠ADB = 180° – 130° = 50°
Hence, ∠ADB = 50° Ans.

Question 10.
Solution:
Arc BAD subtends ∠ BOD at the centre and ∠BCD at the remaining part of the circle.

Question 11.
Solution:
In ∆ OAB,
OA = OB (radii of the same circle)
∠OAB = ∠OBA = 50°
and Ext ∠BOD = ∠OAB + ∠OBA
=>x° = 50° + 50° – 100°
ABCD is a cyclic quadrilateral
∠BAD + ∠BCD = 180°
(opposite angles of a cyclic quad.)
=> 50°+ y° = 180°
=> y° = 180° – 50° = 130°
Hence x = 100° and y = 130° Ans.

Question 12.
Solution:
Sides AD and AB of cyclic quadrilateral ABCD are produced to E and F respectively.
∠CBF = 130°, ∠CDE = x.
∠CBF + ∠CBA = 180° (Linear pair)
=> 130°+ ∠CBA = 180°
=> ∠CBA = 180° – 130° = 50°
But Ext. ∠ CDE = Interior opp. ∠ CBA (In cyclic quad. ABCD)
=> x = 50° Ans.

Question 13.
Solution:
In a circle with centre O AB is its diameter and DO || CB is drawn. ∠BCD = 120°
To Find : (i) ∠BAD (ii) ABD
(iii) ∠CBD (iv) ∠ADC
(v) Show that ∆ AOD is an equilateral triangle.
(i) ABCD is a cyclic quadrilateral.
∠BCD + ∠BAD = 180°
120° + ∠BAD = 180°

Question 14.
Solution:
AB = 6cm, BP = 2cm, DP = 2.5cm
Let CD = xcm
Two chords AB and CD

Question 15.
Solution:
O is the centre of the circle
∠ AOD = 140° and ∠CAB = 50°
BD is joined.
(i) ABDC is a cyclic quadrilateral.

Question 16.
Solution:
Given : ABCD is a cyclic quadrilateral whose sides AB and DC are produced to meet each other at E.
To Prove : ∆ EBC ~ ∆ EDA
Proof : In ∆ EBC and ∆ EDA
∠ E = ∠ E (common)
∠ECB = ∠EAD
{Exterior angle of a cyclic quad, is equal to its interior opposite angle}
and ∠ EBC = ∠EDA
∆ EBC ~ ∆ EDA (AAS axiom)
Hence proved

Question 17.
Solution:
Solution Given : In an isosceles ∆ ABC, AB = AC
A circle is drawn x in such a way that it passes through B and C and intersects AB and AC at D and E respectively.
DE is joined.
To Prove : DE || BC
Proof : In ∆ ABC,
AB = AC (given)
∠ B = ∠ C (angles opposite to equal sides)
But ∠ ADE = ∠ C (Ext. angle of a cyclic quad, is equal E to its interior opposite angle)
∠ADE = ∠B
But, these are corresponding angles
DE || BC.
Hence proved.

Question 18.
Solution:
Given : ∆ ABC is an isosceles triangle in which AB = AC.
D and E are midpoints of AB and AC respectively.
DE is joined.
To Prove : D, B, C, E are concyclic.
Proof: D and E are midpoints of sides AB and AC respectively.
DE || BC
In ∆ ABC, AB = AC
∠B = ∠C
But ∠ ADE = ∠ B (alternate angles)
∠ADE =∠C
But ∠ADE is exterior angle of quad. DBCE which is equal to its interior opposite angle C.
DBCE is a cyclic quadrilateral.
Hence D, B, C, E are con cyclic.
Hence proved.

Question 19.
Solution:
Given : ABCD is a cyclic quadrilateral whose perpendicular bisectors l, m, n, p of the side are drawn

To prove : l, m, n and p are concurrent.
Proof : The sides AB, BC, CD and DA are the chords of the circle passing through the vertices’s of quad. A, B, C and D. and perpendicular bisectors of a chord always passes through the centre of the circle.
l,m, n and p which are the perpendicular bisectors of the sides of cyclic quadrilateral will pass through O, the same point Hence, l, m, n and p are concurrent.
Hence proved.

Question 20.
Solution:
Given : ABCD is a rhombus and four circles are drawn on the sides AB, BC, CD and DA as diameters. Diagonal AC and BD intersect each other at O.

Question 21.

Solution:
Given: ABCD is a rectangle whose diagonals AC and BD intersect each other at O.
To prove : O is the centre of the circle passing through A, B, C and D

Question 22.
Solution:
Construction.
(i) Let A, B and C are three points
(ii) With A as centre and BC as radius draw an arc
(iii) With centre C, and radius AB, draw another arc which intersects the first arc at D.
D is the required point.
Join BD and CD, AC and BA and CB

BC = BC (common)
AC = BD (const.)
AB = DC
∴ ∆ ABC ≅ ∆ DBC (SSS axiom)
∴ ∠BAC ≅ ∠BDC (c.p.c.t.)
But these are angles on the same sides of BC
Hence these are angles in the same segment of a circle
A, B, C, D are concyclic Hence D lies on the circle passing througtvA, B and C.
Hence proved.

Question 23.
Solution:
Given : ABCD is a cylic quadrilateral (∠B – ∠D) = 60°
To prove : The small angle of the quad, is 60°

Question 24.

Solution:
Given : ABCD is a quadrilateral in which AD = BC and ∠ ADC = ∠BCD
To prove : A, B, C and D lie on a circle

Question 25.
Solution:
Given : In the figure, two circles intersect each other at D and C
∠BAD = 75°, ∠DCF = x° and ∠DEF = y°

Question 26.
Solution:
Given : ABCD is a cyclic quadrilateral whose diagonals AC and BD intersect at O at right angle.

Question 27.
Solution:
In a circle, two chords AB and CD intersect each other at E when produced.
AD and BC are joined.

Question 28.
Solution:
Given : Two parallel chords AB and CD of a circle BD and AC are joined and produced to meet at E.

Question 29.
Solution:
Given : In a circle with centre O, AB is its diameter. ADE and CBE are lines meeting at E such that ∠BAD = 35° and ∠BED = 25°.
To Find : (i) ∠DBC (ii) ∠DCB (iii) ∠BDC
Solution. Join BD and AC,

RS Aggarwal Solutions for Class 9 Maths Chapter 11: Download PDF

RS Aggarwal Solutions for Class 9 Maths Chapter 11–Circle

Download PDF: RS Aggarwal Solutions for Class 9 Maths Chapter 11–Circle PDF

Chapterwise RS Aggarwal Solutions for Class 9 Maths :

• Chapter 1–Real Numbers
• Chapter 2–Polynomials
• Chapter 3–Introduction to Euclid’s Geometry
• Chapter 4–Lines and Triangles
• Chapter 5–Congruence of Triangles and Inequalities in a Triangle
• Chapter 6–Coordinate Geometry
• Chapter 7–Areas
• Chapter 8–Linear Equations in Two Variables
• Chapter 9–Quadrilaterals and Parallelograms
• Chapter 10–Area
• Chapter 11–Circle
• Chapter 12–Geometrical Constructions
• Chapter 13–Volume and Surface Area
• Chapter 14–Statistics
• Chapter 15–Probability

About RS Aggarwal Class 9 Book

Investing in an R.S. Aggarwal book will never be of waste since you can use the book to prepare for various competitive exams as well. RS Aggarwal is one of the most prominent books with an endless number of problems. R.S. Aggarwal’s book very neatly explains every derivation, formula, and question in a very consolidated manner. It has tonnes of examples, practice questions, and solutions even for the NCERT questions.

He was born on January 2, 1946 in a village of Delhi. He graduated from Kirori Mal College, University of Delhi. After completing his M.Sc. in Mathematics in 1969, he joined N.A.S. College, Meerut, as a lecturer. In 1976, he was awarded a fellowship for 3 years and joined the University of Delhi for his Ph.D. Thereafter, he was promoted as a reader in N.A.S. College, Meerut. In 1999, he joined M.M.H. College, Ghaziabad, as a reader and took voluntary retirement in 2003. He has authored more than 75 titles ranging from Nursery to M. Sc. He has also written books for competitive examinations right from the clerical grade to the I.A.S. level.

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