# RS Aggarwal Solutions Class 9 Maths Chapter 12 Circles (Updated For 2021-22) RS Aggarwal Solutions Class 9 Maths Chapter 12 Circles: Maths can always become easy when you are studying the RS aggarwal Solutions Class 9 Maths. The solutions of RS Aggarwal Solutions Class 9 Maths Chapter 12 Circles are explained in a way for you to understand them better, are accurate and very creible, all thanks to the subject matter experts.

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RS Aggarwal Solutions Class 9 Maths Chapter 12 Circles

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### RS Aggarwal Solutions Class 9 Maths Chapter 12 Circles Ex 12.1

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 ### RS Aggarwal Solutions Class 9 Maths Chapter 12 Circles Ex 12.2

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)  Question 7.
Solution:
Chord DE || diameter AC of the circle with centre O.
∠CBD = 60°
(Angles in the same segment of a circle) 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)   ### RS Aggarwal Solutions Class 9 Maths Chapter 12 Circles Ex 12.3

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
= 40° + 60° = 100°
(Sum of opposite angles)
=> 100° + ∠BCD = 180°
=> ∠BCD = 180° – 100°
∴ ∠ BCD = 80°
Hence (i) ∠BCD = 80° and

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:
AB || DC and ∠BAD = 100°
(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°  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°
∠BDC + ∠BEC = 180°
=> 60°+ ∠BEC = 180°
=> ∠BEC = 180° – 60°= 120°
Hence ∠BDC = 60° and ∠BEC = 120° Ans.

Question 9.
Solution:
(opposite angles of a cyclic quad.)
so ∠BAD = 180° – 100° = 80°
Now in ∆ ABD, => 80° + 50° + ∠ADB = 180°
=> ∠ADB = 180° – 130° = 50°

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°
(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
(v) Show that ∆ AOD is an equilateral triangle.
(i) ABCD is a cyclic quadrilateral.  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)
{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)
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)
But ∠ADE is exterior angle of quad. DBCE which is equal to its interior opposite angle C.
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:
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. 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,  This is the complete blog on the RS Aggarwal Solutions Class 9 Maths Chapter 12 Circles. To know more about the CBSE Class 9 Maths exam, ask in the comments.

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