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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)

∠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)

### 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

=> ∠ 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,

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