A cyclic quadrilateral is a quadrilateral which has all its four vertices lying on a circle. It is also sometimes called inscribed quadrilateral.
In which opposite angles are supplementary.
Exterior angle is equal to opposite interior angle. That is, in the above cyclic quadrilateral
∠DCE = ∠BAD
Problem 1 :
Find the value of x.
Solution :
Since the opposite angles are supplementary,
∠G + ∠A = 180
21x - 2 + 38x + 5 = 180
59x + 3 = 180
59x = 180 - 3
59x = 177
x = 177/59
x = 3
Problem 2 :
Solution :
∠L + ∠N = 180
10x + 16y + 6 = 180
10x + 16y = 180 - 6
10x + 16y = 174
Dividing by 2, we get
5x + 8y = 87 ------(1)
∠X + ∠M = 180
4 + 18y + 10x - 6 = 180
10x + 18y - 2 = 180
10x + 18y = 182
Dividing by 2, we get
5x + 9y = 91 ------(2)
(1) - (2)
(5x + 8y) - (5x + 9y) = 87 - 91
-y = -4
y = 4
By applying the value of y in (1), we get
5x + 8(4) = 87
5x + 32 = 87
5x = 87 - 32
5x = 55
x = 11
Problem 3 :
Solution :
7x + 1 + 2y + 74 = 180
7x + 2y + 75 = 180
7x + 2y = 180 - 75
7x + 2y = 105 ----(1)
2y + 100 + 6x - 10 = 180
2y + 6x + 90 = 180
6x + 2y = 90 ----(2)
(1) - (2)
7x - 6x = 105 - 90
x = 15
By applying the value of x in (1), we get
7(15) + 2y = 105
105 + 2y = 105
2y = 0
y = 0
Problem 4 :
Solution :
7y - 16 + 4y - 2 = 180 11y - 18 = 180 11y = 180 + 18 11y = 198 y = 198/11 y = 18 |
4x + 5y = 180 9x = 180 x = 180 / 9 x = 20 |
Problem 5 :
Find ∠ABC, ∠AEC, ∠AGC, ∠ADC
Solution :
In triangle ABC, ∠ABC = 90 (angle in a semicircle)
∠AEC = 90 and ∠ADC = 90
In triangle AGC,
∠AGC + ∠ACG + ∠CAG = 180
∠AGC + 25 + 25 = 180
∠AGC + 50 = 180
∠AGC = 180 - 50
∠AGC = 130
Problem 6 :
PQRS is a cyclic quadrilateral, and the side PS is extended to the point A. If ∠PQR = 80°, find ∠ASR
Solution :
∠PQR = ∠ASR = 80 (Exterior angle)
Problem 7 :
ABCD is a cyclic quadrilateral whose diagonals intersect at O. If ∠ACB = 50° and ∠ABC = 110°, find ∠BDC
Solution :
∠ADC = 180 - ∠ABC
∠ADC = 180 - 110
∠ADC = 70
∠ADB = ∠ACB (Angle between the same arc AB)
∠ADB = ∠ACB (Angle between the same arc AB)
∠ADB = 50
∠ADC = ∠ADB + ∠BDC
70 = 50 + ∠BDC
∠BDC = 70 - 50
∠BDC = 20
Problem 8 :
In the given figure, ABCD is a cyclic quadrilateral whose diagonal intersect at P such that ∠DBC = 60^{o} and ∠BAC = 40^{o}. Find ∠BCD
Solution :
∠DBC = ∠DAC (Angle between same arc DC)
∠DAC = 60
∠DAB = ∠DAC + ∠CAB
∠DAB = 60 + 40
∠DAB = 100
∠DCB = 180 - 100
∠DCB = 80
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM