# Circles Solution of TS & AP Board Class 9 Mathematics

###### Exercise 12.1

**Question 1.**

Name the following parts from the adjacent figure where ‘O’ is the centre of the circle.

ii.

iii. BC

iv.

v. DCB

vi. ACB

vii.

viii. shaded region

**Answer:**

i. Radius

ii. Diametre

iii. Minor arc

iv. Chord

v. Major arc

vi. Semi-circle

vii. Chord

viii. Minor segment

**Question 2.**

State true or false.

i. A circle divides the plane on which it lies into three parts. ( )

ii. The area enclosed by a chord and the minor arc is minor segment. ( )

iii. The area enclosed by a chord and the major arc is major segment. ( )

iv. A diameter divides the circle into two unequal parts. ( )

v. A sector is the area enclosed by two radii and a chord ( )

vi. The longest of all chords of a circle is called a diameter. ( )

vii. The mid point of any diameter of a circle is the centre. ( )

**Answer:**

i. True

ii. True

iii. True

iv. False

v. False

vi. True

vii. True

###### Exercise 12.2

**Question 1.**

In the figure, if AB = CD and ∠AOB = 90° find ∠COD

**Answer:**

We know “Angles subtended by equal chords at the center of acircle are equal”.

∠COD = ∠AOB = 90^{0}

**Question 2.**

In the figure, PQ = RS and ∠ORS = 48°. Find ∠OPQ and ∠ROS.

**Answer:**

In ORS,

∠ORS = ∠OSR (radius of circle)

∠OSR = ∠ORS = 48^{0}

∠OSR + ∠ORS + ∠ROS = 180

⇒ 48 + 48 + ∠ROS = 180

⇒ 96 + ∠ROS = 180

⇒ ∠ROS = 180-96

⇒ ∠ROS = 180-96

⇒ ∠ROS = 84

We know “Angles subtended by equal chords at the center of a circle are equal”.

∠POQ = ∠ROS = 84^{0}

In POQ,

∠POQ + ∠OPQ + ∠OQP = 180

84^{0} + x + x = 180

84^{0} + 2x = 180

2x = 180-84

2x = 96

x =

x = 48

∠OPQ = 48

**Question 3.**

In the figure PR and QS are two diameters. Is PQ = RS?

**Answer:**

Considering a circle, where PR and QS being a diameter.

Let center of circle be O.

In triangles POQ and SOR

PO = OR (Radius of circle)

OS = OQ (Radius of circle)

Angle POQ = Angle SOR (Vertically opposite angles are always

equal)

Hence triangle POQ is congruent to triangle SOR (By Side Angle Side

Axiom)

PQ = RS (By C.P.C.T.C)

Hence, proved.

###### Exercise 12.3

**Question 1.**

Draw the following triangles and construct circumcircles for them.

(i) In ∆ ABC, AB = 6cm, BC = 7cm and ∠A = 60°

(iii) In ∆ XYZ, XY = 4.8cm, ∠X = 60° and ∠Y = 70°

(iii) In ∆ XYZ, XY = 4.8cm, ∠X = 60° and ∠Y = 70°

**Answer:**

(i)

(ii)

(iii)

**Question 2.**

Draw two circles passing through A, B where AB = 5.4cm

**Answer:**

**Question 3.**

If two circles intersect at two points, then prove that their centres lie on the perpendicular bisector of the common chord.

**Answer:**

Let two circles O and O' intersect at two points A and B so that AB is the common chord of two circles.

OO' is the line segment joining the centers.

Let OO' intersect AB at M

Now Draw line segments OA, OB , O'A and O'B

In Î”OAO' and OBO' , we have

OA = OB (radii of same circle)

O'A = O'B (radii of same circle)

O'O = OO' (common side)

⇒ Î”OAO' ≅ Î”OBO' (SSS congruency)

⇒ ∠AOO' = ∠BOO'

⇒ ∠AOM = ∠BOM ......(i)

Now in Î”AOM and Î”BOM we have

OA = OB (radii of same circle)

∠AOM = ∠BOM (from (i))

OM = OM (common side)

⇒ Î”AOM ≅ Î”BOM (SAS congruency)

⇒ AM = BM and ∠AMO = ∠BMO

But

∠AMO + ∠BMO = 180°

⇒ 2∠AMO = 180°

⇒ ∠AMO = 90°

Thus, AM = BM and ∠AMO = ∠BMO = 90°

Hence OO' is the perpendicular bisector of AB.

**Question 4.**

If two intersecting chords of a circle make equal angles with diameter passing through their point of intersection, prove that the chords are equal.

**Answer:**

Given that AB and CD are two chords of a circle, with center O intersecting at a point E.

PQ is a diameter passing through E, such that ∠ AEQ = ∠ DEQ

Draw OL ⊥ AB and OM ⊥ CD.

In right angled OLE

∠LOE + 90° + ∠ LEO = 180° (Angle sum property of a triangle)

∴∠LOE = 90° – ∠LEO

= 90° – ∠AEQ = 90° – ∠DEQ

= 90° – ∠MEO = ∠MOE

In triangles OLE and OME,

∠LEO = ∠MEO

∠LOE = ∠MOE (Proved)

OE = OE (Common side)

∴ Î”OLE ≅ Î”OME

⇒ OL = OM (CPCT)

Thus, AB = CD

**Question 5.**

In the adjacent figure, AB is a chord of circle with centre O. CD is the diameter perpendicualr to AB. Show that AD = BD.

**Answer:**

Given, AB is a chord of circle with centre O. CD is the diameter

perpendicular to AB.

We know that line drawn from the center of a circle to the chord

Perpendicular to it bisects the chord.

AP = BP

In ADP and BDP,

AP = BP

APD = BPD = 90

PD = PD (Common)

ADP BDP

AD = BD (CPCT)

###### Exercise 12.4

**Question 1.**

In the figure, ‘O’ is the centre of the circle. ∠AOB = 100° find ∠ADB.

**Answer:**

Given, ∠AOB = 100°

Join AB, which is a chord.

∠ ACB = (Angle subtended at the center is twice the angle subtended at circumference by the same chord)

⇒∠ ACB = 50

Now, ACBD is a quadrilateral and angles on the opposite sides in the quadrilateral are supplementary.

∠ ADB = 180-∠ ACB

⇒∠ ADB = 180-50

⇒∠ ADB = 130^{o}

**Question 2.**

In the figure, ∠BAD = 40° then find ∠BCD.

**Answer:**

Given, ∠BAD = 40°

Join BD, which is a chord.

We know “Two or more angles subtended by the chord at the circumference are same”.

∠BCD = ∠BAD = 40°

**Question 3.**

In the figure, O is the centre of the circle and ∠POR = 120°. Find ∠PQR and ∠PSR

**Answer:**

Given, ∠POR = 120°

We know that “Angle subtended at the center is twice the angle subtended at circumference by the same chord”.

∠PQR =

⇒∠PQR =

⇒∠PQR = 60

Now, PQRS is a quadrilateral and angles on the opposite sides in the quadrilateral are supplementary.

∠ PSR = 180-∠ PQR

⇒∠ PSR = 180-60

⇒∠ PSR = 120^{o}

**Question 4.**

If a parallelogram is cyclic, then it is a rectangle. Justify.

**Answer:**

Let ABCD be a cyclic parallelogram.

A rectangle is a parallelogram with one angle 90. So, we have to prove angle 90.

Since ABCD is a parallelogram,

∠A = ∠C

In cyclic parallelogram ABCD,

∠A + ∠C = 180

∠A + ∠A = 180

2∠A = 180

∠A =

∠A = 90

Hence, proved.

**Question 5.**

In the figure, ‘O’ is the centre of the circle. OM = 3cm and AB = 8cm. Find the radius of the circle

**Answer:**

Joining OA we see that OA is radius of the circle.

Let radius (OA) be r cm.

Given, AB = 8cm

OM = 3cm

We know that “perpendicular from center divides the chord into equal parts”.

AM =

⇒ AM = 4 cm

In right angled triangle OAM,

OA^{2} = OM^{2} + AM^{2} (Pythagoras Thm.)

⇒OA^{2} = 3^{2} + 4^{2}

⇒OA^{2} = 9 + 16

⇒OA^{2} = 25

⇒OA =

⇒OA = 5 cm

**Question 6.**

In the figure, ‘O’ is the centre of the circle and OM, ON are the perpendiculars from the centre to the chords PQ and RS. If OM = ON and PQ = 6cm. Find RS.

**Answer:**

Given, OM = ON

We know that “If two chords are equidistant from the center then the two chords are equal in length”.

PQ = RS = 6 cm

**Question 7.**

A is the centre of the circle and ABCD is a square. If BD = 4cm then find the radius of the circle.

**Answer:**

Given, BD = 4cm

We know that “a square has equal diagonals” and BD is a diagonal.

AC, which is the radius of the circle, is also a diagonal of the square.

AC = BD = 4 cm

Thus, Radius of the circle = 4 cm

**Question 8.**

Draw a circle with any radius and then draw two chords equidistant from the centre.

**Answer:**

Circle with center O and two chords CD and EF equidistant from center .

**Question 9.**

In the given figure ‘O’ is the centre of the circle and AB, CD are equal chords. If ∠AOB = 70°. Find the angles of the ∆OCD.

**Answer:**

We know “Angles subtended by equal chords at the center of a circle are equal”.

∠DOC = ∠AOB = 70

In OCD,

∠OCD + ∠ODC + ∠DOC = 180

x + x + 70 = 180

70^{0} + 2x = 180

2x = 180-70

2x = 110

x =

x = 55

∠OCD = ∠ODC = 55

###### Exercise 12.5

**Question 1.**

Find the values of x and y in the figures given below.

i.

ii.

iii.

**Answer:**

(i) We know “Sum of all angles of a triangle is 180”.

x + y + 30 = 180

Since it is an isosceles triangle, x = y.

x + x + 30 = 180

⇒ 2x + 30 = 180

⇒ 2x = 180-30

⇒ 2x = 150

⇒ x =

⇒ x = 75

x = y = 75

(ii) We know that “Angles on the opposite sides in the cyclic quadrilateral are supplementary”.

x + 110 = 180

⇒ x = 180-110

⇒ x = 70

y + 85 = 180

⇒ y = 180-85

⇒ y = 95

(iii) Given, x = 90°

x + y + 50 = 180

⇒ 90 + y + 50 = 180

⇒ y + 140 = 180

⇒ y = 180-140

⇒ y = 40

**Question 2.**

Given that the vertices A, B, C of a quadrilateral ABCD lie on a circle. Also ∠A + ∠C = 180°, then prove that the vertex D also lie on the same circle.

**Answer:**

In a quadrilateral, if the sum of opposite angles is 180°, then it is a

cyclic quadrilateral.

In quad. ABCD, A + C = 180°,

Therefore, ABCD is a cyclic quad.

In a cyclic quad., the vertices lie on the same circle. Thus, D also

lies on the same circle.

Hence, proved.

**Question 3.**

Prove that a cyclic rhombus is a square.

**Answer:**

To prove rhombus inscribed in a circle is a square, we need to prove that either any one of its interior angles is equal to 90° or its diagonals are equal.

∠ABD = ∠DBC = b

∠ADB = ∠BDC = a

In the figure, diagonal BD is angular bisector of angle B and angle D.

In triangle ABD and BCD,

AD = BC (sides of rhombus are equal)

AB = CD (sides of rhombus are equal)

BD = BD (common side)

△ABD ≅ △BCD. (SSS congruency)

In the figure,

2a + 2b = 180° (as, opposite angles of a cyclic quadrilateral are always supplementary)

2(a + b) = 180°

a + b = 90°

In △ABD,

Angle A = 180°-(a + b)

= 180°-90°

= 90°

Therefore, proved that one of its interior angle is 90°

Hence, rhombus inscribed in a circle is a square.

**Question 4.**

For each of the following, draw a circle and inscribe the figure given. If a polygon of the given type can’t be inscribed, write not possible.

(a) Rectangle

(b) Trapezium

(c) Obtuse triangle

(d) Non-rectangular parallelogram

(e) Accute issosceles triangle

(f) A quadrilateral PQRS with as diameter.

**Answer:**

(a) Rectangle = Possible

(b) Trapezium = Possible

(c) Obtuse triangle = Not Possible

(d) Non-rectangular parallelogram = Not Possible

(e) Acute isosceles triangle = Possible

(f) A quadrilateral PQRS with as diameter = Possible