Circles

i. Radius

ii. Diametre

iii. Minor arc

iv. Chord

v. Major arc

vi. Semi-circle

vii. Chord

viii. Minor segment

i. True

ii. True

iii. True

iv. False

v. False

vi. True

vii. True

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

∠COD = ∠AOB = 90⁰

In ORS,

∠ORS = ∠OSR (radius of circle)

∠OSR = ∠ORS = 48⁰

∠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 = 84⁰

In POQ,

∠POQ + ∠OPQ + ∠OQP = 180

84⁰ + x + x = 180

84⁰ + 2x = 180

2x = 180-84

2x = 96

x =

x = 48

∠OPQ = 48

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.

(i)

(ii)

(iii)

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 ΔOAO' and OBO' , we have
OA = OB (radii of same circle)
O'A = O'B (radii of same circle)
O'O = OO' (common side)
⇒ ΔOAO' ≅ ΔOBO' (SSS congruency)
⇒ ∠AOO' = ∠BOO'
⇒ ∠AOM = ∠BOM ......(i)
Now in ΔAOM and ΔBOM we have
OA = OB (radii of same circle)
∠AOM = ∠BOM (from (i))
OM = OM (common side)
⇒ ΔAOM ≅ ΔBOM (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.

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)
∴ ΔOLE ≅ ΔOME
⇒ OL = OM (CPCT)
Thus, AB = CD

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)

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

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°

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

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.

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,

OA² = OM² + AM² (Pythagoras Thm.)

⇒OA² = 3² + 4²

⇒OA² = 9 + 16

⇒OA² = 25

⇒OA =

⇒OA = 5 cm

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

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

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

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

70⁰ + 2x = 180

2x = 180-70

2x = 110

x =

x = 55

∠OCD = ∠ODC = 55

(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

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.

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.

(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