THEOREMS AND PROOF RELATING TO CYCLIC QUADRILATERAL

**SUBJECT: MATHEMATICS**

**CLASS: SS 2**

**DATE:**

**TERM: 2nd TERM**

**REFERENCE BOOKS**

- New General Mathematics SSS2 by M.F. Macraeetal.
- Essential Mathematics SSS2 by A.J.S. Oluwasanmi.

WEEK NINE

**TOPIC: THEOREMS AND PROOF RELATING TO CYCLIC QUADRILATERAL **

**CONTENT**

-Definition of Cyclic Quadrilateral

-Theorems and proof relating to cyclic quadrilateral

-Corrolary from Cyclic Quadrilateral

-Solving problems on Cyclic Quadrilateral

**CYCLIC QUADRILATERAL**

**Definition**: A cyclic quadrilateral is described as any quadrilateral having its vertices lying on certain parts of the circumferences of a circle. i..e its four vertices.

Note: that opposite angles of a cyclic quadrilateral lies in opposite segment of a circle.

**Theorem**:

**The opposite angle of a cyclic quadrilateral are supplementary “or angle in opposite segment are supplementary i.e. They sum up to 180****0****.**

**Proof**:

**Given**: A cyclic quadrilateral ABCD.

**To prove**:< BAD +< BCD = 1800

**Construction**: join B and D to O the centre

**Proof**:< BOD = 2y (angle of centre = 2 x angle at circumference)

Reflex< BOD = 2x (angle at centre = 2x angle at circumference)

2x + 2y = 3600 (angle at a point)

2(x + y) = 3600

x + y = 3600

2

x + y = 1800

**Example:**Find the value of x

x + 720=1800 (opp. Angle of a cyclic quadrilateral)

x = 1800-720 = 1080

**Evaluation**

Find x and y

- 2.

**Corollary from Cyclic Quadrilateral**

**Theorem**:

**The exterior angle of a cyclic quadrilateral to the interior opposite angle.**

**Proof**:

**Given**: A cyclic quadrilateral ABCD

**To Prove**: x1 = x2 or x2 = x1

**Construction**: Extend DC to x

Proof: x1 + y = 1800 (opp. Angle in a cyclic quad)

x2 + y = 1800 (angle in a straight line)

x1 = x2 = (180-y)

< BCX =< BAD

**Example**:

In the fig. below PQRS are points on a circle centre O. QP is produced to x if< XPS = 770 and

< QPS = 1800 – 770 = 1030 (angle on a straight line)

PQO = 3600 – 3250

PQO = 350

**Example **

BEC is a triangle

BCE = 1800 – 850 (angle on a straight line)

CBE = 620 (exterior angle of cyclic quadrilateral)

x =< BEC = 1800 – (620 + 950) [sum of angles in a â ]

1800 – 1570 = 230

**Evaluation**

In the figure below AB is a diameter of semi circle ABCD. If

2.

In the fig, A,B,C,D are points on a circle such that

**Application of Cyclic Quadrilateral [Circle Geometry]**

Solution

< NOM = 180 – (20 + 20)[ sum of angle in a triangle 1800 – 400 = 1400]

2

**Evaluation**

Find the marked angle.

**GENERAL EVALUATION/REVISION QUESTIONS**

Find the marked angle in each of the following.Where a point O is the centre of the circle.

- 2.

- A right pyramid on a base 8cm square has a slant edge of 6cm, calculate the volume of the pyramid.
- Calculate the volume and total surface area of a cylinder which has a radius of 12cm and height 6cm

**READING ASSIGNMENT**

Essential Mathematics SSS2, pages 143-144, Exercise10.5,numbers 6-10.

**WEEKEND ASSIGNMENT**

**Objective**

1.In the diagram below, O is the centre of the circle,

Calculate

(a) 1000 (b) 860 (c) 940 (d) 1440

2.In the diagram |PS| is a diameter of circle PQRS. |PQ| =|QR|and

(a) 320 (b) 370 (c) 480 (d) 530

3.In the diagram below, O is the centre of the Circle PQRS and

(a) 360 (b) 1440 (c) 720 (d) 1080

4.In the diagram below: PQRS is a cyclic quadrilateral,

(a) 430 (b) 480 (c) 530 (d)580

5.In the diagram below; 0 is the centre of the circle. If

(a) 1050 (b) 750 (c) 150 (d) 1500

**Theory**

1.In the fig.Calculate the value of x giving a reason for each step in your answer.

- I n the diagram below,
0.Find

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