Example1: ABCD is a cyclic quadrilateral. AB and CD are produced to meet P. If angle ADC = \(70^o \)and angle\( DAB = 60^o\), then what will be angle PBC + angle PCB?
Solution: First step: Make an appropriate figure using statements provided in questions.
- Cyclic quadrilateral has all its vertices on circle and sum of all angles is 360^o.
- Sum of opposite angles = 180^o.
- External angle = Opposite internal angle.
Therefore, angle \(PBC = 70^o\)
angle\( PCB = 60^o\)
So, \(PBC + PCB = 130^0\)
See its very simple question, if you know the properties of geometry. Let's try more examples
Example2: ABCD is a parallelogram and P is any point within it. If area of parallelogram is 20 units, then what will be the sum of areas of triangle PAB and PCD?
Solution: According to the question, figure will be as follows:
Properties for this question:
- Area of parallelogram = Base times Height.
- Area of triangle = 1/2 Base times altitude.
Given that: Area of parallelogram = 20
i.e. Base times Height = 20
⇒ b times h = 20
To find = Area of ( PAB +PCD)
⇒ (1/2) b times (PY) + (1/2) b times (PX) ( See Figure)
⇒1/2 b ( PX +PY)
⇒1/2 b h
⇒(1/2) times 20
Example3: ABCD is a cyclic trapezium with AB parallel DC and AB diameter of circle. If angle\( CAB = 30^o\), then angle ADC will be?
Solution: According to ques, figure will be as follows:
- Angle subtended by diameter is always 90^o.
- Sum of angles of triangle = \(180^o.\)
- Sum of opposite angles = \(180^o\)
angle \(ACB = 90^o\)
angle \(BAC = 30^o\)
angle \(ABC = 60^o\)
angle ABC + angle \(ADC = 180^o ⇒ 60^o + ADC = 180^o \)⇒ angle \(ADC = 120^o\)
Example4: Two side of plot measures 30m and 22m and angle between them is \(90^o\). The other two sides measures 24m and the three remaining angles are not right angles. Find the area of plot.
Solution: Figure becomes: Center point of BD = O
ABD is a right angle triangle, therefore, Pythagoras theorem followed.
\((BD)^2 = (AB)^2 + (AD)^2 ⇒ BD = 40\)
Given that, BC = CD Therefore, line drawn from C to BD given right angles Also, this will lead to DO = BO
So, DO = BO = 20
Similarly, using Pythagoras theorem, OC = 15 m
Now, Area of triangles (ABD + BOC + COD) ⇒ 1/2 (24 times 32) + 2 ((1/2) 20 times 15) ⇒ 684 m^2