PRACTICE PROBLEMS ON CYCLIC QUADRILATERAL

Question 1 :

Find the value of x in the given figure. 

Solution :

In triangle ACB,

<ACB  =  90 (Angle in a semicircle)

Sum of opposite angles in a quadrilateral  =  180

<ADC + <ABC  =  180

120 + <ABC  =  180

<ABC  =  60

<ACB + <CAB + <ABC  =  180

x + 90 + 60  =  180

x + 150  =  180

x  =  180 - 150

x  =  30

Question 2 :

In the given figure, AC is the diameter of the circle with centre O. If <ADE = 30° ; <DAC = 35° and <CAB = 40°. Find (i) <ACD   (ii)  <ACB  (iii)  <DAE

Solution :

In triangle ADC,

<ADC  =  90 (angle in a semicircle)

<ADC + <DAC + <ACD  =  180

90 + 35 + <ACD  =  180

<ACD  =  180 - 125

<ACD  =  55

In triangle ACB,

<ACB + <ABC + <BAC  =  180

<ACB + 90 + 40  =  180

<ACB  =  180 - 130

<ACB  =  50

(iii)  <EDA + <ADC  + <EAD + <DAC  =  180

(Sum of opposite angles in a quadrilateral)

30 + 90 + <EAD + 35  =  180

<EDA  =  180 - 155

<EDA  =  25

Question 3 :

Find all the angles of the given cyclic quadrilateral ABCD in the figure.

Solution :

<A + <C  =  180

2y + 4 + 4y - 4  =  180

6y  =  180

y  =  180/6

y  =  30

<B + <D  =  180

6x - 4 + 7x + 2  =  180

13x - 2  =  180

13x  =  182

x  =  182/13

x  =  14

<A  =  2y + 4  =  2(30) + 4  =  60 + 4  =  64

<B  =  6x - 4  =  6(14) - 4  =  84 - 4  =  80

<C  =  4y - 4  =  4(30) - 4 =  120 - 4  =  116

<D  =  7x + 2  =  7(14) + 2  =  100

Question 4 :

In the given figure, ABCD is a cyclic quadrilateral where diagonals intersect at P such that <DBC = 40° and <BAC = 60° find

(i)  <CAD   (ii)  BCD

Solution :

Angles in a same segment will be equal.

<DAC  =  <DBC  =  40

In ΔBCD, we have:

∠BCD + ∠DBC + ∠BDC = 180° 

(Angle sum property of a triangle)

<BCD + 60° + 40° = 180°

∠BCD = (180° - 100°) = 80°

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