PROBLEMS ON CYCLIC QUADRILATERAL AND INTERCEPTED ARC

Find the measure of angle and indicated arc.

Problem 1 :

Solution :

In a cyclic quadrilateral, the opposite angles are supplementary.

∠FGH = 180 - 119 ==> 61

∠FGH = (1/2) Measure of arc (VH + FV)  ---(1)

∠GFV = 180 - 65 ==> 115

∠GFV = (1/2) Measure of arc (GH + VH) ---(2)

From (2)

115 = (1/2) (168 + measure of arc VH)

230 = 168 + measure of arc VH

Measure of arc VH = 230 - 168

Measure of arc VH = 62

By applying this value in (1)

61= (1/2) (62 + measure of arc FV) 

122 - 62 = measure of arc FV

Measure of arc FV = 60

Problem 2 :

Solution :

In a cyclic quadrilateral, the opposite angles are supplementary.

∠KRQ = 180 - 110 ==> 70

∠KRQ = (1/2) (measure of arc PQ + KP)

70 = (1/2) (30x + 50)

140 = 30x + 50

140 - 50 = 30x 

30x = 90

x = 3

Problem 3 :

Find the values of x, y and z.

Solution :

∠DAB + ∠DCB = 180

∠DAB + z = 180 ---(1)

∠DAB = (1/2) measure of arc BD

∠DAB = (1/2) 136

∠DAB = 68

Applying the value of ∠DAB in (1), we get

z = 180 - 68

z = 112

∠ADC + ∠ABC = 180

x + y = 180

∠ADC = (1/2) Measure of arc AC

∠ADC = (1/2) 180

∠ADC = 90 = x

x = 90

Applying the value of x in x + y = 180, we get

90 + y = 180

y = 180 - 90

y = 90

Problem 4 :

Find the values of x, y and z.

Solution :

x = (1/2) (106 + 58)

x = (1/2) (164)

x = 82

y = 180 - 82 (Opposite angles are supplementary)

y = 98

z = 180 - 93

z = 87

Problem 5 :

Find measure of angles A, B, C and D.

Solution :

∠BAD = (1/2) Measure of BD

∠BAD = (1/2) (115 + 86)

∠BAD = 100.5

∠BCD = 180 - 100.5

∠BCD = 79.5

∠ADC = (1/2)(54 + 115)

∠ADC = (1/2)(169)

∠ADC = 84.5

∠ABC = 180 - 84.5

∠ABC = 95.5

Problem 6 :

Solution :

x = (1/2)(64 + 122)

x = (1/2)(186)

x = 93

y = 180 - 83

y = 97

Problem 7 :

Find ∠DBC, ∠ACD

Solution :

∠DBC = (1/2)ABC Measure of arc DC

∠DBC = (1/2)96

∠DBC = 48

∠ACD = (1/2)ABC Measure of arc AD

∠ACD = (1/2)90

∠ACD = 45

Problem 8 :

Find the value of each variable.

Solution :

Sum of opposite angles in a cyclic quadrilateral = 180

120 + y = 180

y = 180 - 120

y = 60

z + 80 = 180

z = 180 - 80

z = 100

So, the values of y and z are 60 and 100 respectively.

Problem 9 :

Solution :

x + 82 = 180

x = 180 - 82

= 98

y + 68 = 180

y = 180 - 68

= 112

So, the values of x and y are 98 and 112 degree respectively.

Problem 10 :

Solution :

c + 2c - 6 = 180

3c - 6= 180

3c = 180 + 6

3c = 186

c = 186/3

= 62

10x + 8x = 180

18x = 180

x = 180/18

x = 10

So, the values of x and c are 10 and 62 respectively.

Find the value of each variable.

Problem 11 :

Solution :

x + 80 = 180

x = 180 - 80

x = 100

y + 95 = 180

y = 180 - 95

y = 85

So, the values of x and y are 100 are 85 respectively.

Problem 12 :

In the figure O is the center of the circle and ∠AOC = 120, find ∠ABC.

Solution :

∠AOC = 120

Reflection of ∠AOC = 360 - 120

= 240

Angle created at the center of the circle, then it is equal to the angle created at circumference of the circle.

∠x = 2∠ABC

240 = 2 ∠ABC

∠ABC = 240/2

= 120

Problem 13 :

O is the center of the circle and AD bisects ∠BAC. Find ∠BCD

Solution :

Angles in the same segment will be equal.

∠BAD = ∠BCD

BC is the diameter of the circle, ∠BAC = 90

∠BAD = 90/2 ==> 45

∠BCD = 45

Problem 14 :

PQRS is a cyclic quadrilateral whose diagonals intersect at A. If ∠SQR = 80 and ∠QPR = 30. Find ∠SRQ.

Solution :

∠SQR = ∠SPR

∠SPR = ∠SQR = 80

Sum of opposite angles in cyclic quadrilateral = 180 degree

∠SPR + ∠RPQ = 80 + 30

= 110

∠SPQ + ∠SRQ = 180

110 + ∠SRQ = 180

∠SRQ = 180 - 110

∠SRQ = 70