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 1800.

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

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

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

0 (ext angle of a cyclic quadrilateral)

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

0(the exterior angle of a cyclic quad=interior opp.angle)

0=1540(angle at the centre=2×angle at the circumference)

0 – (1540 + 1030 + 680) sum of angle in a quadrilateral

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 0,calculate

In the fig, A,B,C,D are points on a circle such that 0.CD is produced to E so that 0.Calculate

Application of Cyclic Quadrilateral [Circle Geometry]

Solution

0 (base angle of Isosceles triangle ONM)

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

1400= 700 (2x angle at circum = angle at centre)

0 (base angle of Isos triangle MNT)

0 (fe. 32 + 32) (extension of triangle MNT)

0 + 640) sum of angle in a triangle

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.

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|and0 find

(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, 0 and 0, Calculate

(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

Lesson Notes By Weeks and Term - Senior Secondary School 2