## Important Examples

**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

⇒10 units.

**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 \(BAC = 30^o\)

Therefore,

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