Class 10 SELINA Solutions Maths Chapter 17 - Circles
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 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 25
Solution 26
Solution 27
Solution 28
Solution 29
Solution 30
Solution 32
Solution 33
Solution 34
Solution 35
Solution 36
Solution 38
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
Solution 16
Solution 24
Solution 31
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 39
Circles Exercise Ex. 17(B)
Solution 1
Solution 2
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 3
Solution 10
Circles Exercise Ex. 17(C)
Solution 1
Solution 2
Solution 4
Solution 5
Solution 6
Solution 7
Solution 9
Solution 11
Solution 13
Solution 14
Solution 15
Solution 18
Solution 19
Solution 20
Solution 22
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°
Solution 3
Solution 8
Solution 10
