Class 10 SELINA Solutions Maths Chapter 17 - Circles
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Circles Exercise Ex. 17(A)
Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Solution 12
Solution 13
Solution 14
Solution 15
Solution 16
Solution 17
Solution 18
Solution 19
Solution 19(b)
▭ABPQ is a cyclic quadrilateral.
⟹∠A = ∠P …..(Exterior angle property of cyclic quadrilateral) …(1)
▭ABCD is a parallelogram.
⟹ ∠A = ∠C …..(Opposite angles of a parallelogram) ….. (2)
From (1) and (2),
∠P = ∠C…..….(3)
But ∠C + ∠D = 180° …. (Sum of interior angles of a parallelogram is 180°)
From (3), we get
∠P + ∠D = 180°
⟹ PCDQ is a cyclic quadrilateral.
Solution 20
Solution 21
Solution 22
Solution 23
Solution 24
Solution 25
Solution 26
Solution 27
Solution 28
Solution 29
Solution 30
Solution 31
Solution 32
Solution 33
Solution 34
Solution 35
Solution 36
Solution 37
(i) 
AEB = 
(Angle in a semicircle is a right angle)
Therefore EBA = - EAB = - 
= 
(ii) AB 
ED
Therefore DEB = EBA = (Alternate angles)
Therefore BCDE is a cyclic quadrilateral
Therefore DEB + BCD = 
[Pair of opposite angles in a cyclic quadrilateral are supplementary]
Therefore BCD = - = 
Solution 38
Solution 39
Solution 40
Solution 41
Solution 42
Solution 43
Solution 44
Solution 45
Solution 46
Solution 47
Solution 48
Solution 49
- ABCD is a cyclic quadrilateral
m∠DAB = 180° - ∠DCB
= 180° - 130°
= 50°
- In ∆ADB,
m∠DAB + m∠ADB + m∠DBA = 180°
⇒50° + 90° + m∠DBA = 180°
⇒m∠DBA = 40°
Solution 50
Solution 51
Solution 52
Solution 53
Solution 54
Solution 55
Solution 56
Solution 57
Circles Exercise Ex. 17(B)
Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Circles Exercise Ex. 17(C)
Solution 1
Solution 2
Solution 3
Given - In Δ ABC, AB = AC and a circle with AB as diameter is drawn which intersects the side BC at D.
To prove - D is the midpoint of BC.
Construction - Join AD.
Proof:
∠1 = 90° [Angle in a semi circle]
But ∠1 + ∠2 = 180° [Linear pair]
∴ ∠2 = 90°
Now in right ΔABD and Δ ACD,
Hyp. AB = Hyp. AC [Given]
Side AD = Ad [Common]
∴ By the Right angle - Hypotenuse - Side criterion of congruence, we have
Δ ABD ≅ ΔACD [RHS criterion of congruence]
The corresponding parts of the congruent triangles are congruent.
∴ BD = DC [c.p.c.t]
Hence D is the mid point of BC.
Solution 4
Join OE.
Arc EC subtends ∠EOC at the centre and ∠EBC at the remaining part of the circle.
∠EOC = 2 ∠EBC = 2 × 65° = 130°.
Now in Δ OEC, OE = OC [Radii of the same circle]
∴ ∠OEC = ∠OCE
But, in Δ EOC,
∠OEC + ∠OCE + ∠EOC = 180° [Angles of a triangle]
⇒ ∠OCE + ∠OCE + ∠EOC = 180°
⇒ 2 ∠OCE + 130° = 180°
⇒ 2 ∠OCE = 180° - 130°
⇒ 2 ∠OCE + 50°
⇒ ∠OCE = 
= 25°
∴ AC || ED [given]
∴ ∠DEC = ∠OCE [Alternate angles]
⇒ ∠DEC = 25°
Solution 5
Given - ABCD is a cyclic quadrilateral and PRQS is a quadrilateral formed by the angle.
Bisectors of angle ∠A, ∠B, ∠C and ∠D.
To prove - PRQS is a cyclic quadrilateral.
Proof - In Δ APD,
∠PAD + ∠ADP + ∠APD = 180° ......(1)
Similarly, in Δ BQC,
∠QBC + ∠BCQ + ∠BQC = 180° ......(2)
Adding (1) and (2), we get
∠PAD + ∠ADP + ∠APD + ∠QBC + ∠BCQ + ∠BQC = 180° + 180°
⇒ ∠PAD + ∠ADP + ∠QBC + ∠BCQ + APD + ∠BQC = 360° ....(3)
But ∠PAD + ∠ADP + ∠QBC + ∠BCQ = 1/2 [∠A + ∠B + ∠C + ∠D]
= 1/2 × 360° = 180°
∴ ∠APD + ∠BQC = 360° - 180° = 180° [from (3)]
But these are the sum of opposite angles of quadrilateral PRQS.
∴ Quad. PRQS is a cyclic quadrilateral.
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Solution 12
Solution 13
Solution 14
Solution 15
Solution 16
Solution 17
Solution 18
Solution 19
Solution 20
Solution 21
Solution 22
Given - In the figure, CP is the bisector of ∠ABC
To prove - DP is the bisector of ∠ADB
Proof - Since CP is the bisector of ∠ACB
∴ ∠ACP = ∠BCP
But ∠ACP = ∠ADP [Angles in the same segment of the circle]
and ∠BCP = ∠BDP
But ∠ACP = ∠BCP
∴ ∠ADP = ∠BDP
∴ DP is the bisector of ∠ADB
Solution 23
Solution 24
i. AD is parallel to BC, i.e., OD is parallel to BC and BD is transversal.
Solution 25
∠DAE and ∠DAB are linear pair
So,
∠DAE + ∠DAB = 180°
∴∠DAB = 110°
Also,
∠BCD + ∠DAB = 180°……Opp. Angles of cyclic quadrilateral BADC
∴∠BCD = 70°
∠BCD = 
∠BOD…angles
subtended by an arc on the center and on the circle
∴∠BOD = 140°
In ΔBOD,
OB = OD……radii of same circle
So,
∠OBD =∠ODB……isosceles triangle theorem
∠OBD + ∠ODB + ∠BOD = 180°……sum of angles of triangle
2∠OBD = 40°
∠OBD = 20°
