Practice Problems on Cyclic Quadrilateral :
Here we are going to see some example problems on cylic 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°
After having gone through the stuff given above, we hope that the students would have understood, "Practice Problems on Cyclic Quadrilateral"
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