# SELINA Solutions for Class 10 Maths Chapter 17 - Circles

Page / Exercise

## Chapter 17 - 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)

⟹∠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
1. ABCD is a cyclic quadrilateral

mDAB = 180° - DCB

= 180° - 130°

= 50°

mDAB + mADB + mDBA = 180°

50° + 90° + mDBA = 180°

mDBA = 40°

Solution 50

Solution 51

Solution 52

Solution 53

Solution 54

Solution 55

Solution 56

Solution 57

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

## Chapter 17 - Circles Exercise Ex. 17(C)

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 20

Solution 21

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 = 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

2OBD = 40°

OBD = 20°

### STUDY RESOURCES

REGISTERED OFFICE : First Floor, Empire Complex, 414 Senapati Bapat Marg, Lower Parel, Mumbai - 400013, Maharashtra India.