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##### 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. ( )

Solution:

• i) True
• ii) True
• iii) True
• iv) False
• v) False
• vi)True
• vii) True
##### Question 1. In the figure, if AB = CD and ∠AOB = 90° find ∠COD.

Solution:
‘O’ is the centre of the circle.
AB = CD (equal chords from the figure)
∴ ∠AOB = ∠COD
[ ∵ equal chords make equal angles at the centre]
∴ ∠COD = 90° [ ∵ ∠AOB = 90° given]

##### Question 2. In the figure, PQ = RS and ∠ORS = 48°. Find ∠OPQ and ∠ROS.

Solution:
‘O’ is the centre of the circle.
PQ = RS [given, equal chords]
∴∠POQ = ∠ROS [ ∵ equal chords make equal angles at the centre]
∴ In ΔROS
∠ORS + ∠OSR + ∠ROS = 180°
[angle sum property]
∴ 48° + 48° + ∠ROS = 180°
[ ∵ OR = OS(radii); ΔORS is isosceles]
∴ ∠ROS = 180° - 96° = 84°
Also ∠POQ = ∠ROS = 84°
∴ ∠OPQ = ∠OQP
=1/2 [180°-84°] = 48°

##### Question 3. In the figure, PR and QS are two diameters. Is PQ = RS ?

Solution:
‘O’ is the centre of the circle.
[ ∵ PR, QS are diameters]
OP = OR (∵ radii) .
∠POQ = ∠ROS [vertically opp. angles]
∴ ΔOPQ ≅ ΔORS [SAS congruence]
∴ PQ = RS [CPCT]

##### Question 1. Draw the following triangles and construct circumcircles for them. (OR) In ΔABC, AB = 6 cm, BC = 7 cm and ∠A = 60°. Construct a circumcircle to the triangle XYZ given XY = 6cm, YZ = 7cm and ∠Y = 60°. Also, write steps of construction.

Solution:

Steps of construction :

1. Draw the triangle with given mea-sures.
2. Draw perpendicular bisectors to the sides.
3. The point of concurrence of per-pendicular bisectors be S’.
4. With centre S; SA as radius, draw a circle which also passes through B and C.
5. This is the required circumcircle.

ii) In ΔPQR; PQ = 5 cm, QR = 6 cm and RP = 8.2 cm.

Steps of construction:

1. Draw ΔPQR with given measures.
2. Draw perpendicular to PQ, QR and RS; let they meet at ‘S’.
3. With S as centre and SP as radius draw a circle.
4. This is the required circumcircle.

iii) In ΔXYZ, XY = 4.8 cm, ∠X = 60° and ∠Y = 70°.

Steps of construction:

1. Draw ΔXYZ with given measures.
2. Draw perpendicular bisectors to the sides of ΔXYZ, let the point of con-currence be S’.
3. Draw the circle (S, SX).
4. This is the required circumcircle.

Solution:

##### Question 3. If two circles intersect at two points, then prove that their centres lie on the perpendicular bisector of the common chord.

Solution:

Let two circles with centre P and Q intersect at two distinct points say A and B.
Join A, B to form the common chord
AB Let ‘O’ be the midpoint of AB.
Join ‘O’ with P and Q.
Now in 8APO and ΔBPO
PO = PO (common)
AO = BO (∵ O is the midpoint)
∴ ΔAPO ≅ ΔBPO (S.S.S. congruence)
Also ∠AOP = ∠BOP (CPCT)
But these are linear pair of angles.
∴ ∠AOP = ∠BOP = 90°
Similarly in ΔAOQ and ΔBOQ
AO = BO (∵ O is the midpoint of AB)
OQ = OQ (common)
∴ AAOQ ≅ ABOQ
Also ∠AOQ = ∠BOQ (CPCT)
Also ∠AOQ + ∠BOQ = 180° (linear pair of angles)
∴ ∠AOQ = ∠BOQ =180°/2= 90°
Now ∠AOP + ∠AOQ = 180°
∴ PQ is a line.
Hence the proof.

##### Question 4. If two intersecting chords of a circle make equal angles with diameter pass¬ing through their point of intersection, prove that the chords are equal.

∠AEO = ∠DEO
Drop two perpendiculars OL and OM from ‘O’ to AB and CD;
Now in ΔLEO and ΔMEO
∠LEO = ∠MEO [given]
EO = EO [Common]
∠ELO = ∠EMO [construction 90°]
∴ ΔLEO ≅ ΔMEO
[ ∵ A.A.S. congruence]
∴ OL = OM [CPCT]
i.e., The two chords AB and CD are at equidistant from the centre ‘O’.
∴ AB = CD
[∵ Chords which are equi-distant from the centre are equal]
Hence proved.

##### Question 5. In the given figure, AB is a chord of circle with centre ‘O’. CD is the diam-eter perpendicular to AB. Show that AD = BD.

Solution:
CD is diameter, O is the centre.
CD ⊥ AB; Let M be the point of inter-section.
Now in ΔAMD and ΔBMD
AM = BM [ ∵ radius perpendicular to a chord bisects it]
∠AMD =∠BMD [given]
DM = DM (common)
∴ ΔAMD ≅ ΔBMD

##### Question 1. In the figure, ‘O’ is the centre of the circle. ∠AOB = 100°, find ∠ADB.

Solution:
’O’ is the centre
∠AOB = 100°

##### Question 2. In the figure, ∠BAD = 40° then find ∠BCD.

Solution:
‘O’ is the centre of the circle.
∴ In ΔOAB; OA = OB (radii)
∴ ∠OAB = ∠OBA = 40°
(∵ angles opp. to equal sides)
Now ∠AOB = 180° - (40° + 40°)
(∵ angle sum property of ΔOAB)
= 180°-80° = 100°
But ∠AOB = ∠COD = 100°
Also ∠OCD = ∠ODC [OC = OD]
= 40° as in ΔOAB
∴ ∠BCD = 40°
(OR)
In ΔOAB and ΔOCD
∠AOB = ∠COD (vertically opp. angles)
∴ ΔOAB ≅ ΔOCD
∴ ∠BCD = ∠OBA = 40°
[ ∵ OB = OA ⇒ ∠DAB = ∠DBA]

##### Question 4. If a parallelogram is cyclic, then it is a rectangle. Justify.

Solution:

Let □ABCD be a parallelogram such
that A, B, C and D lie on the circle.
∴∠A + ∠C = 180° and ∠B + ∠D = 180°
[Opp. angles of a cyclic quadri lateral are supplementary]
But ∠A = ∠C and ∠B = ∠D
[∵ Opp. angles of a ‖gm are equal]
∴∠A = ∠C =∠B =∠D =180/2= 90°
Hence □ABCD is a rectangle

##### Question 5. In the figure, ‘O’ is the centr of the circle. 0M = 3 cm and AB = 8 cm. Find the radius of the circle.

Solution:

‘O’ is the centre of the circle.
OM bisects AB.
∴ AM =AB/2=8/2= 4 cm
OA2= OM2+ AM2[ ∵ Pythagoras theorem]
OA =√32+42
=√9+16=√25
= 5cm

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

Solution:
‘O’ is the centre of the circle.
OM = ON and 0M ⊥ PQ; ON ⊥ RS
Thus the chords FQ and RS are equal.
[ ∵ chords which are equidistant from the centre are equal in length]
∴ RS = PQ = 6cm

##### Question 7. A is the centre of the circle and ABCD is a square. If BD = 4 cm then find the radius of the circle.

Solution:

A is the centre of the circle and ABCD is a square, then AC and BD are its diagonals. Also AC = BD = 4 cm But AC is the radius of the circle.

##### Question 8. Draw a circle with any radius and then draw two chords equidistant from the centre.

Solution:

1. Draw a circle with centre P.
3. Mark off two points M and N oh these radii. Such that PM = PN.
4. Draw perpendicular through M and N to these radii.
##### 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 ΔOCD.

Solution:
‘O’ is the centre of the circle.
AB, CD are equal chords
⇒ They subtend equal angles at the centre.
∴ ∠AOB =∠COD = 70°
Now in ΔOCD
∠OCD = ∠ODC [∵ OC = OD; radii angles opp. to equal sides]
∴ ∠OCD + ∠ODC + 70° = 180°
= ∠OCD +∠ODC = 180° – 70° = 110°
∴ ∠OCD + ∠ODC = 110° = 55°

##### Question 1. Find the values of x and y in the figures given below.

i)

Solution:
From the figure x = y [∵ angles opp. to opp. to equal sides]
But x + y + 30° = 180°
∴ x + y = 180° - 30° = 150°
⇒ x + y = 150°/2= 75°

ii)

Solution:
From the figure x° + 110° = 180°
[ ∵ Opp. angles of a cyclic quad, are supplementary]
y + 85°= 180°
∴ x= 180° - 110°; y = 180° – 85°
x = 70°; y = 95°

iii)

Solution:
From the figure x = 90° [angle in a semi-circle]
∴ y = 90° – 50° [∵ angle sum property]
= 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.

Solution:
Given : ∠A + ∠C = 180°
∴ ∠B + ∠D = 360° – 180°
[ ∵ sum of the four angles of a quad. is 360 ].
Now in □ABCD, sum of the pairs of opp. angles is 180°.
∴ □ABCD must be a cyclic quadrilateral, i.e., D also lie on the same circle on which the vertices A, B and C lie. Hence proved.

##### Question 3. Prove that a cyclic rhombus is a square.

Solution:

Let □ABCD be a cyclic rhombus,
i.e., AB = BC = CD = DA and
∠A + ∠C = ∠B + ∠D = 180°
But a rhombus is basically a parallelo-gram.

##### Question 4. For each of the following, draw a circle and inscribe the figure given. If a poly¬gon of the given type can’t be in-scribed, write not possible.
• a) Rectangle
• b) Trapezium .
• c) Obtuse triangle
• d) Non-rectangular parallelogram
• e) Acute isosceles triangle
• f) A quadrilateral PQRS with PR as diameter.

Solution: