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SCHEME OF WORK
WEEK TOPIC
THEME: ALGEBRAIC PROCESSES:
1. Revision of last term’s work.
2. Simple Equations and variations: (a) Change of subject of formulae – Formula involving brackets, roots and powers. (b) Subject of formula and substitution. (c) Types of variations; (i) Direct and inverse, (ii) joint and partial. (d) Application of variation.
3. Quadratic Equation: (a) Revision of factorization of quadratic expressions. (b) Solution of quadratic equation of the form: ab = 0 a = 0 or b = 0 (c) Formation of quadratic equation with given roots.
4. Quadratic Equation: (d) Drawing quadratic graph (e) Obtain roots from a quadratic graph. (f) Application of quadratic equation to real life situations.
5. Logical Reasoning: (a) Simple statements. (b) Meaning of simple statement – (i) True or false (ii) negation of simple statements. (c) Compound statements (i) Meaning (ii) Conjunction (iii) Disjunction (iv) Implication (v) bi-implication.
6. Logical Reasoning: (d) Logical operators and symbols. (i) list of logical operators and symbols (ii) True value of i (iii) A compound statement. (iv) Negation (NA) (v) Conjunction (vi) Disjunction (vii) Conditional statement (viii) bi-conditional statement.
THEME: GEOMETRY:
7. Constructions: (a) Revision of; (i) Construction of triangles with given sides. (ii) Bisection of an angle; 300, 450, 600 and 900.
8. Constructions:(b) Construction of; (i) An angle equal to a given angle. (ii) 4-sided plane figure given certain conditions. (iii) Locus of moving points equidistance from 2 lines, 2 points, and constant distance from a point, etc.
9. Proofs of some Basic Theorems: (a) Proofs of: Angle sum of a triangle is 1800 (b) The exterior angle of a triangle is equal to the sum of two interior opposite angles.
10. Proofs of some Basic Theorems:(c) Riders including – (i) angles of parallel lines (ii) angles in a polygon (iii) congruent triangles (iv) properties of parallelogram (v) intercept (vi) theorem.
11. Revision.

TOPIC: SIMPLE EQUATIONS AND VARIATIONS
CONTENT:
Change of subject of formulae
Subject of formula and substitution
Types of variations – direct and inverse
Joint and partial
Application of variation

Simple equation is an algebraic equation involving one variable and the power or index of the variable is one.
SUB-TOPIC 1: CHANGE OF SUBJECT OF FORMULAE
A literal equation is a simple equation that involves more than one variable (unknown). The procedure of solving such an equation is usually to find one of the variables (unknowns) in terms of the other(s).
Examples: (1) Given A=1/2 h(a+b), make a the subject of the formula
Solution: A=1/2 h(a+b)
Cross-multiply
2A=h(a+b)
2A=ah+bh
2A-bh=ah
Also, ah=2A-bh
a=(2A-bh)/h
a=2A/h-bh/h
∴a=2A/h-b
(2) Make v the subject of the formula;
H=(m(v^2-u^2))/2gx
Solution: H=(m(v^2-u^2))/2gx
Cross-multiply
2gxH= m(v^2-u^2)
2gxH= mv^2-mu^2
2gxH+mu^2=mv^2
(2gxH+mu^2)/m=v^2
Take square root of both sides;
v=±√((2gxH+mu^2)/m) or ±√(2gxH/m+u^2 )
(3) The period of a compound pendulum is given by T=2π√(((h^2+k^2)/gh) ) express k in terms of T,h and g, taking π^2 as 10
Solution: T=2π√(((h^2+k^2)/gh) )
Take the square of both side
T^2=(2π)^2 ((h^2+k^2)/gh)
T^2=〖4π〗^2 ((h^2+k^2)/gh)
Cross-multiply
T^2 gh=〖4π〗^2 (h^2+k^2 )
Divided both sides by 〖4π〗^2
(T^2 gh)/〖4π〗^2 =h^2+k^2
k^2=(T^2 gh)/〖4π〗^2 -h^2
Take square root of both sides
k=√((T^2 gh)/〖4π〗^2 -h^2 )
Using the value of 10 for π^2,
∴k=√((T^2 gh)/40-h^2 )
SUB-TOPIC 2: SUBJECT OF FORMULA AND SUBSTITUTION
A formula is an equation in which letters represent quantities. The value of one variable in a formula or algebraic equation may be found by substituting (i.e replacing) known values in the same formula
Examples: (1) The sum of the squares of the first n integers is given by s_n=(n(n+1)(2n+1))/6
Calculate (a) s_20 (b) the sum of the squares from 21 to 40 inclusive.
Solution: (a) s_20 means the values of s_n when n=20
s_20=(20(20+1)(2×20+1))/6
s_20=(20(21)(41))/6
=2870
(b) Sum of squares from 21 to 40 means = s_40-s_20
But, s_20 is already known to be 2870,
s_40=(40(40+1)(2×40+1))/6
=(40(41)(81))/6
=22140
∴s_40-s_20=22140-2870
=19270
(2) Find the value of 2π√(l/g) , when π=3 1/7 ,l=98 and g=32
Solution: 2π√(l/g) = 2×22/7 √(98/32)
= 44/7 √(49/16)
= 44/7×7/4
= 11
EVALUATION:
(1) Make the letters that appear as the subject of the formula
I=nE/(R+nr) , (n,R)
T=E/√(R^2-W^2 L^2 ) , (R,L)
V=πh^2 (r+h/3) , (r)
(2) (a) If v=kl(D^2-d^2), find d when v=1250,D=1000,k=5 and l=4
(b) Simplify p(q-2p)-(p+q)(q-2p) find the value of the expression when p=-2 and q=1
(c) Evaluate √((s^2-(b+c)s+bc)/bc) when b=25,c=7 and s=28
Assignment: NGM, page 82, Ex 6f nos 15,31 and 35; page 83, Ex 6g nos 10,13 and 16




TOPIC : VARIATIONS Date:………………………
Variation is a connection of sets of numerical values by an equation which indicates some kind of proportionality. We have four major types of variations which are direct, inverse, joint and partial variations.
SUB-TOPIC 3: DIRECT VARIATIONS:
Considering two quantities x and y. If when y increases, x also increases and when y decreases, x also decreases in a constant proportion, then x and y are said to be in direct variation. i.e. x varies directly as y or x is directly proportional to y: Written as xy  x = ky where k is the constant of proportionality or variation.
Other examples are as follows:
(i) A varies directly as the square of B
i.e. A=B2
* A = KB2 (where K is the constant of variation)
(ii) P varies directly as the square root of q.
i.e. P-q
* P = K-q (where k is the constant of variation).
Examples: (A) (i) if P varies directly with q and q = 2, p = 10. Find p when q is 5.
Solution: p∝q
p=kq
10=2k
k=5
Formula connecting p and q is p = 5q,
Then, p = 5 x 5 = 25
(ii) if x-3 is directly proportional to the square of y and x=5 when y=2,find x when y=6
Solution; x-3∝y^2
x-3=ky^2
5-3=k×2^2
2=4k
k=1/2
Formula connecting x-3 and y is x-3=1/2 y^2.
Then, x-3=1/2×6×6
x-3=18
∴x=21
Example 2:
The area of a box varies directly as the square of its length. When the area is 175cm2, the length is 5cm. Find the;
Equation connecting the quantities.
Area when the length is 8cm
Length when the Area is 100cm2
Solution:
Let A represent Area
L represent Length
A = L2
A * KL2 –- (1) (Where k is Constant)
If A = 175 and L = 5 then
175 * k(5)2
k = 175
25
k = 7
Substitute for k in equation (1) to get the law of variation.
A = 7L2
A = 7L2 is the equation connecting the quantities.
(ii) To find A when L = 8
put values in the law of variation A = 7L2 to have
A = 7(8)2
A =7 * 64
A = 448cm2.
(iii) To find L when A = 100cm2
put values in the law of variation A = 7L2 to have
100 = 7L2
100 = L2
7
L = 100
7
L = 10 or 3.78cm
7
EVALUATION
x is directly proportional to y. If x=5 when y=3,find y when x=2/7
The wages of a labourer varies directly as the number of hours worked by the labourer. The labourer earned N500 when he worked for 2 hours. Find
(i) The amount he would earn if he works for 7 hours.
(ii) The number of hours he would work if he is paid N800.
y varies directly as the square of x. If y = 98 when x = 7, calculate y when x = 5.
[WAEC]
If D∝S∝and D=140 when S=35,find
The relationship between D and S
The value of S when D = 176
SUB-TOPIC 4: INVERSE VARIATIONS.
Considering two quantities x and y. If when y increases, x decreases and when x increases y decreases in a constant ratio, then we say x and y varies inversely. It is normally written as x*1/y or y*1/x.
Note that if xµ1/y, then x = k/y where K is the constant of variation or proportionality.
Other examples are as follows:
(i) A varies inversely as the cube root of B
i.e. A µ ¬1
3*B
A = K x 1
3*B (where k is constant)
(ii) P varies inversely as the square of q
I.e. P µ 1
q2
P = K 1
q2 where k is constant
Example
(i) t∝1/d and t=0.15 when d=120
(a) find t when d = 45
(b) find d when t = 0.12
Solution: t∝1/d
t=k/d
0.15×120=k
k=18
Formula connecting t and d, t=18/d
t=18/45
=0.4
t=18/d

0.12d=18
d=18/0.12
∴d=150
Example
If P is inversely proportional to the square root of q, when P = 3, q = 25. Find
(a) P when q = 49
(b) q when p = 13

Solution:
P µ 1
q
P = k
q -------------------(1) (where k is constant)
When P = 3, q = 25
3 = k
25
3 x 5 = k
k = 15
Substitute in (1) to get the law of variation
P = 15
q
(a) To find p when q = 49
Substitute in the law of variation to have
P = 15
49
p = 15
7
 p = 21/7

(b) To find q when p = 13, substitute in the law of variation to have
13 = 15
p
13*p = 15

*p = 15
13
p = 15 2
13
p = 225
169
p = 156/169

EVALUATION:

if x varies inversely as the square of y and x=8 when y=4,find y when x=32
(2) If y varies inversely as x and y = 6 when x = 2. Find
(i) The law of variation
(ii) x when y = 10
(iii) y when x = 7

(3)If y varies inversely as the square of x and y = 8 when x = 3. Find
(i) The relationship between x and y.
(ii) x when y = 5
(iii) y when x = 9

(4) If P varies inversely as the square root of q and p = 12 when q = 4. Find
(i) p when q = 25
(ii) q when p = 8.

(5)Exercise 18c, page 217 of New General Mathematics for senior Secondary School, Book 1.
SUB-TOPIC 5: JOINT VARIATION
This type of variation involves three or more quantities joined together with a combination of two direct variations or one direct and one inverse or two inverse variations.
Examples:
(i) x varies directly as y and jointly as z
I.e. x*yz
x = kyz (where k is constant)
(ii) x varies directly as y and inversely as the square of z.
i.e. x*y
z2
* x = ky (where k is constant)
z2
(iii) P varies inversely as q and inversely as the square root of r.
I.e. p * 1
q*r
* p = k (where k is constant)
q*r



Example :
If x varies directly as y and inversely as the square of z. When y = 5 and z = 3, x = 20. Find (a) z when x = 213/5 and y = 15
(b) x when y = 6 and z = 4.

Solution:
X * y
z2
x = ky -------- (1) (where k is constant)
z2
when y = 5 and z = 3, x = 20
20 = k5
32
20 x 9 = k5
k = 20 x 9
5
k = 4 x 9
* k = 36.
Substitute in (1) above to get the law of variation.
X = 36y
z2
(a) To find z when x = 213/5 and y = 15
108 = 36 x 15
5 z2
108z2 = 36 x 15 x 5
z2 = 36 x 15 x 5
108
z = *25
* z = 5.
(b) To find x when y = 6 and z = 4.
x = 36 x 6
42
x = 36 x 6
16
*x = 13.5
Examples; (1) x varies directly as the product of u and v and inversely as their sum,if x=3 when u=3 and v=1,what is the value of x if u=3 and v=3?
Solution: x∝uv/(u+v)
x=kuv/(u+v)
3= (k×3×1)/(3+1)
3×4=3k
k=4
Law connecting x,u and v is x=4uv/(u+v)
Then, x=(4×3×3)/(3+3)
∴x=6
(2) A∝BC,when B=4 and C=9,A=6
(a) find the formula that connects A,B & C
(b) find A when B = 3 and C = 10
(c) find C if A = 20 and B = 15
Solution: A∝BC
A=kBC
6=k×4×9
k=1/6
A=1/6 BC
A=1/6 BC
A=1/6×3×10
A = 5
A=1/6 BC
20=1/6×15×C ∴C=8
EVALUATION
(2) A varies directly as B and inversely as C. When B = 3, C = 5, A = 40. Find
(i) The law of variation
(ii) A when B = 8 and C = 12.

(3) xµy. when y = 36 and z = 16, x = 81.
Öz
Find (i) The law of variation
(ii) y when x = 56 and z = 25
(iii) x when y = 27 and z = 9
(4) U varies directly as V and inversely as the square of W. When V = 3 and W = 4, U = 24. Find (i) U when V = 5 and W = 8
(ii) W when U = 30 and V = 8
(5) The electrical resistance of a copper wire varies directly as its length and inversely as the square of its radius. If a copper wire 500 meters long and radius 0.2cm has a resistance of 30 ohms, calculate the resistance of the same type of copper wire 750 meters long and radius 0.25cm.
(WAEC).
(6) If PµQÖR. When R = 16 and Q = 3, P = 48. Find (i) The law of variation
(ii) P when R = 25 and Q = 7
(iii) R when P = 36 and Q = 9

SUB-TOPIC 6:
PARTIAL VARIATION
This consist of two or more parts or quantities added together. Both parts may be made of variables or one part may be constant, while the other can either vary directly or inversely. The values of the constants of variation are usually found out by solving simultaneous equations.

Examples:
(i) x is partly constant, and partly varies as y
x = a+by (where a and b are constant)
(ii) p varies partly as q and partly as r.
p = aq + br (where a and b are constants)
(iii) v varies partly as u and partly as the reciprocal of w2.
Examples: (1) P is partly constant and partly varies as Q, when Q is 5, P is 20 and when Q is 8, P is 26. Find P when Q is 4.
Solution: P = c + kQ
Note: c & k are constants which must be obtained through simultaneous equation.
20=c+5k ………(i)
-26=c+8k……….(ii)
-6 = -3k
K = 2
substituting the value of k in equation (i)
20=c+5(2)
20-10=c
c=10
Thus, P = 10 + 2Q (formula connecting P & Q)
Then, P = 10 + 2(4)
P = 10 + 8
∴ P = 18
Example :
X is partly constant and partly varies as y. when y = 7, x = 15; and when y = 5, x = 7. Find
(a) The law of variation
(b) x when y = 2
(c) y when x = 11
Solution
(a) x = a + by --- (1) where a and b are constants of variation.
When y = 7, x = 15 and when y = 5, x = 7
15 = a + 7y ----------------- (2)
7 = a + 5y ----------------- (3)
Solve equation (2) and (3) simultaneously
Eqn (2): 15 = a + 7b _
Eqn (3): 7 = a + 5b
8 = 2b
8 = b
2
\ b = 4
Put in eqn (2) to have
15 = a + 7 x 4
15 = a + 28
a = 15 – 28
a = -13
Substitute a = -13 and b = 4 in eqn (1) to get the law of variation.
x = -13 + 4y -------------(law of variation)
(b) To find x when y = 2, put in the law of variation
x = -13 + 4 x 2
x = -13 + 8
\ x = -5
(c) To find y when x = 11, put in the law of variation
11 = -13 + 4y
11 + 13 = 4y
24 = 4y
y = 24
4
\ y = 6.

EVALUATION:
M is partly constant and partly varies with N, when N = 40, M = 150 and when N = 54, M = 192
(a) Find the formula connecting M and N, (b) Hence find M when N = 73
Two quantities P and Q are connected by a linear relation of the form P = aQ + b, where a & b are constants. If Q = 80 when P = 12 and Q = 300 when P = 50, find the equation connecting P and Q
The cost of producing a wooden frame varies directly as the width of the frame and partly as the square root of its length. When the width is 10cm and the length is 25cm, the cost is N115.00 and when the width is 18cm and the length is 36cm, the cost is N240. Find the
(a) Law of variation
(b) Cost of a frame of width 12cm and the length 49cm.
(2) The cost of producing a textbook is partly constant and partly varies as the number of books produced. It cost N4000 to produce 20 books and N6000 to produce 70 books. Find the
(a) Cost of producing 120 books
(b) Number of books produced at N10, 000.

(3) The cost of sinking a well varies partly as the depth of the well and partly as the number of laborers used for the job. If it cost N1500 to sink a well of 6 metres deep with 2 laborers and N2500 to sink a well of 9 metres deep with 5 laborers. Find the
(a) Law of variation
(b) Cost of sinking a well 15 metres deep with 6 laborers
(c) Number of laborers that would sink a well of 12 metres deep at N3000.


SUB-TOPIC 7: APPLICATION OF VARIATION
Examples; (1) A particle moves in such a way that its displacement S metres, at time t seconds is given by the relation S = at^2, where ‘a’ is constant. Calculate ‘a’ if S = 32 when t = 4
Solution; S = at^2
32 = a x 4^2
32/16=a
a = 2
(2) The cost of feeding a number of students of Deeper Life High School in a regional excursion is partly constant and partly varies directly as the number of students. The cost of feeding 75 students during the excursion is $875 and the cost of feeding 100 students during the same period of time is $1000. Find the cost of feeding 220 students over the same period of time.
Solution: The equation is given as; C = a + kN where C is the cost and N is the number of students
875 = a + 75k ……….(i)
1000 = a + 100k ………. (ii)
Solving both equations simultaneously, we obtain k = 5 & a = 500.
Formula connecting C and N implies; C = 500 + 5N
Then, C = 500 + 5(220)
C = 500 + 1100
∴ C = $1600
GENERAL EVALUATION:
The electrical resistance R of a wire varies inversely as the square of the radius r. use a constant k to show the law between R and r
Given that the energy E, varies directly as the resistance R and inversely as the square if the distance d, (i) obtain an equation connecting E, R and d given that E = 32/25 when R = 8 and d = 5. (ii) calculate (a) the value of R when E = 16 and d = 3 (b) the value of d when R = 5 and d = 5/6 (c) find the percentage increase in the value of R when each of E and d increases by 3%

READING ASSIGNMENT:
NGM for SSS book 1, pages 80-84, 213-219.
Mathematical Association of Nigeria (MAN) pages 63- 77

WEEKEND ASSIGNMENT:
Make W the subject of the formula; R-d=√(R^2-W^2 ) . Given that R = 1.25 and d = 0.25. calculate W
Given that T=2π√((l^2+k^2)/2gl)
Make k the subject of the formula
Find the value of k when l=1 and T^2=〖4π〗^2 g
The resistance R to the motion of a car is partly constant and partly proportional to the square of the speed v. when the speed is 30km/h, the resistance is 190O and when the speed is 50km/h, the resistance is 350O. find for what speed the resistance is 302.5O

TOPIC: QUADRATIC EQUATIONS
CONTENT:
Revision of linear and quadratic expressions
Solution of quadratic expression of the form ab=0, a=0 or b=0
Formation of quadratic equation with given roots
SUBTOPIC 1: Revision of linear and quadratic expressions
Any expression in which highest power of the unknown is 1 is called a linear expression. Some examples of linear expressions are; (a) x + 1 (b) 2y + 3 (c) p -1/2
In general, linear expressions are expressions of the form ax+b, where a & b are constants and x is a variable.
A quadratic expression is that whose highest power of the unknown is 2. Examples are (a) x^2+3x (b) 〖2x〗^2-6x+10
Factorization of quadratic expressions;
Examples; (i) factorize the following quadratic expressions
x^2+3x (b) 〖2x〗^2-6x
Solution:
x^2+3x , x is a common factor of the terms x^2 & 3x. Hence x^2+3x can be written as x.x+3x isolating common factors, we have x(x+3)
〖2x〗^2-6x , the common factor of the terms 〖2x〗^2 & 6x is 2x
∴ 〖2x〗^2-6x can be written as 2x.x-2x.3, hence we obtain 2x(x-3)
(ii) Factorize the following; (a) x^2+8x-20 (b) 〖6a〗^2+15a+9 (c) 7-22x+〖3x〗^2
Solution: (a) x^2+8x-20 , find the product of the first and last terms x^2×(-20)=-20x^2. Find two terms such that their product is -20x^2 and their sum is +8x
Factors of -20x^2 sum of factors
-20x and+x -19x
+20x and-x +19x
-10x and+2x -8x
+10x and-2x +8x
-5x and+4x -x
+5x and-4x +x
Of these, only (d) gives the required result. Replace +8x with +10x and-2x in the given expression. Then factorize by grouping the terms.
x^2+8x-20=x^2+10x-2x-20
=x(x+10)-2(x+10)
=(x+10)(x-2)
(b) 〖6a〗^2+15a+9 , 3 is common factor , first take out the common factor.
3(〖2a〗^2+5a+3)
〖2a〗^2×3=〖6a〗^2
Factors of 〖+6a〗^2 Sum of factors
+6a and +a +2a
+3a and +2a +5a

〖6a〗^2+15a+9 = 3(〖2a〗^2+5a+3)
= 3(〖2a〗^2+3a+2a+3)
= 3[a(2a+3)+1(2a+3)]
= 3(2a+3)(a+1)
7-22x+〖3x〗^2 , find the product of the first and last terms i.e 7×(+〖3x〗^2 )=+21x^2
Find two terms such that their sum is -22x and their product is +21x^2. Since the middle term is negative, consider negative factors only. The terms are -21x and-x, replace -22x with -21x-x in the given expression.
7-22x+〖3x〗^2=7-21x-x+〖3x〗^2
=7(1-3x)-x(1-3x)
=(1-3x)(7-x)
EVALUATION:
Factorize the following quadratic expressions
〖6a〗^2-18a
〖3x〗^2 y-xy^2
x^2-4x+3
〖1-3x+2x〗^2
〖3m〗^2+5mn-2n^2
p^2-p+1/4
x^2-16
〖25x〗^2-1
SUB-TOPIC 2: Solution of quadratic expression of the form ab=0, a=0 or b=0
If the product of two numbers is 0, then one of the numbers (or possibly both of them) must be zero. For example, 3×0=0, 0×5=0 and 0×0=0.
In general, if a×b=0,then either a=0 or b=0 or both a & b are zero
Examples 1. Solve the equation (x-2)(x+7)=0
Solution: (x-2)(x+7)=0
If (x-2)(x+7)=0, then either (x-2)=0 or (x+7)=0
⇒ x=2 or x=-7
2. Solve the equation a(a+3)=0
If a(a+3)=0, then either a=0 or a+3=0
⇒ a=0 or a=-3
3. Solve the equations (i) 〖(2m-5)〗^2=0 (ii) d(d-4) (d+6)^2=0
Solution: (i) if 〖(2m-5)〗^2=0
Then, (2m-5)(2m-5)=0
(2m-5)=0 twice
⇒ m=5/2 twice
(ii) if d(d-4) (d+6)^2=0, then any one of the four factors of LHS may be 0
i.e d=0,d-4=0,(d+6)^2=0
⇒ d=0,d=4 or d=-6 twice
EVALUATION:
Solve the following equations;
(a-3)(a+5)=0
2y(y-1/3)=0
(m-2/3)^2 (m-1)=0
x^2 (x+5)(x-5)=0
SUB-TOPIC 3: Formation of quadratic equation with given roots
The roots of a quadratic equation are the solutions of that equation. Suppose the roots of a quadratic equation in x are a & b, then we can write; x=a & x=b
Examples; (1) Find the quadratic equation whose roots are -2 and+2
Solution: let x=-2 or x=2, then
x+2=0 or x-2=0
(x+2)(x-2)=0
On careful expansion, we obtain x^2-4=0
2. Find the quadratic equation whose roots are 2 1/2 and-1
Solution: If the roots are 2 1/2 and-1
Let x=2 1/2 and x=-1
x=5/2 and x=-1
2x=5 and x=-1
2x-5=0 or x+1=0
⇒ (2x-5)(x+1)=0
〖2x〗^2+2x-5x-5=0
〖2x〗^2-3x-5=0
EVALUATION:
Form the quadratic equations whose roots are;
3 and - 4
1/9 and 2/3
-5 and 1/10
(-1)/5 and (-1)/6
(-3)/8 and 0
4/7 and 7
READ ASSIGNMENT:
NGM for SSS book 1 pages 115 – 127
WEEKEND ASSIGNMENT:
NGM for SSS book 1 page 117, Ex. 10b, Q13,14,15 16. Page 119, Ex. 10d Q11-20 and page122, Ex. 10f, Q5-10.

TOPIC: QUADRATIC EQUATION
CONTENT:
Revision of linear graph and drawing quadratic graph
Obtaining roots from a quadratic graph
Finding an equation from a given graph
Application of quadratic equation to real life situations
SUB-TOPIC 1
Linear Graphs
Recall that any equation whose highest power of the unknown is 1 is a linear equation. To draw the graph of a linear equation, we need to
Make a table of value for the equation
Plot the graph of the linear equation
Example: Draw the graph of y=x-1
Solution; y = x - 1
x -2 0 2
-1 -1 -1 -1
y -3 -1 1
Scale: 2cm to 1unit on both axes
y-axis

1

Y=x-1 graph of y=x-1
x-axis

-1

-2

DRAWING QUADRATIC GRAPH: To draw a quadratic graph, we need to also follow the same process of drawing linear graph
Example: Draw the graph of y=x^2+2x+1
Solution; since y=x^2+2x+1, we shall now make a table for the values of x & y.
x -3 -2 -1 0 2 3
x^2 9 4 1 0 4 9
2x -6 -4 -2 0 4 6
1 1 1 1 1 1 1
y 4 1 0 1 9 16
Note: when plotting the graph,
we choose a scale such that our graph is as large as possible and also occupies the centre of the graph sheet. This will enable us to obtain the point where the graph cuts the x-axis more easily.
We join the points in the graph by a smooth curve

EVALUATION:
Draw the graph of y=〖2x〗^2-5x+3
SUB-TOPIC 2
Obtaining roots from a quadratic graph
To obtain the roots of a quadratic equation form a quadratic graph, we need to first plot the graph of the expression and then obtain the roots by reading the two values of x where the graph cuts or touches the x-axis, i.e where y=0
Example: Draw a graph to find the roots of the equation y=〖4x〗^2-20x+25
Solution; 〖y=4x〗^2-20x+25
x 0 1 2 3 4 5
〖4x〗^2 0 4 16 36 64 100
-20x 0 -20 -40 -60 -80 -100
25 25 25 25 25 25 25
y 25 9 1 1 9 25

From the graph it is clear that the curve does not cut the x-axis. It appears to touch the x-axis where x=2.5. this result can be checked by factorisation.
i.e 〖4x〗^2-20x+25=0
(2x-5)(2x-5)=0
(2x-5)^2=0
.∴x=2.5 twice
Note: when the curve touches the x-axis, the roots are said to be coincident
EVALUATION:
Use the table of values below to solve the equation y=x^2+x-8 graphically for -4≤x≤3
x -4 -3 -2 -1 0 1 2 3
y -8 4
SUB-TOPIC 3
Finding an equation from a given graph
It is possible to find the equation of a curve from its graph. The graph of y=x^2-2x-3 cuts the x-axis (i.e the line y=0) at the points x=-1 and x=3. This implies that -1 and 3 are the roots of the equation x^2-2x-3=0. Therefore in general if a graph cuts the x-axis at points a & b, it satisfies the equation (x-a)(x-b)=0
Example 1: Obtain the equation of the graph below
y-axis
2 P


0 1 2 x-axis


Solution; From the graph when y=0, x=1 and x=2,
Then (x-1)(x-2)=0
.x^2-3x+2=0, at point P, y=2 when x=0
.∴the equation of the curve is y=x^2-3x+2
Example 2: obtain the equation of the graph below (WAEC)
y-axis


-1 -0.5 0 1 2 x-axis


-2
Solution: In the graph above, where y=0, x=(-1)/2 and x=2
(x--1/2)(x-2)=0
(x+1/2)(x-2)=0
x^2-1 1/2 x-1=0 ……..(i)
Second, in the curve above, at point P, y=-2 when x=0. However the constant term in equation (i) is only -1. So we multiply both sides of the equation (i) by 2
〖2x〗^2-3x-2=0 ….(ii)
Equation (ii) satisfies(x--1/2)(x-2)=0 and the requirement that the constant term should be -2
∴ The equation of the curve isy= 〖2x〗^2-3x-2





EVALUATION:
Find the equation of the graphs below
Y-axis 2. Y-axis




-2.5 -1 0 1 2.5 3.5 -2 -1.5 -1 0 1 1.5 2 x-axis
-1

SUB-TOPIC 4
Application of quadratic equation to real life situations
Example 1: The area of a rectangle is 〖60cm〗^2. The length is 11cm more than the width. Find the width.
Solution:
Let the width be xcm, length = area/width
.∴ length will be 60/x cm.
The length is 11cm more than the width gives 60/x=x+11
Simplifying; we have 60=x^2+11x
i.e x^2+11x-60=0
factorizing completely we have, x=-15 or x=4
.∴ the width is 4cm since it cannot be negative.
Check: length = 60/4 = 15 = 4 + 11 and 15 x 4 = 60
EVALUATION:
When 11 times a certain integer is subtracted from twice the square of the integer, the result is 21. What is the integer?
A rectangular lawn is 4cm longer than its width. If its area is 165〖cm〗^2, calculate its width
Musa is 60years old and Joy is 25years old. How many years from now will the product of their ages be 2244years?
GENERAL EVALUATION:
Solve the following equations graphically and obtain the least value of y
〖y=x〗^2-4x+3=0
〖y=2x〗^2-3x+1
y=〖3x〗^2-4x+1
The sum of the ages of a mother and her child is 63. If the product of their ages four years ago was 484. What are their ages now?
Find the equations of the graphs below
Y-axis (b) y-axis
5



-5 0 1 x-axis -1 0 3 x-axis


-3

WEEKEND ASSIGNMENT
Solve the following questions from NGM for SS1 page 117 Ex. 10b nos 30-35.
Page 120 Ex. 10e nos 45-50, page 124 Ex. 10g nos 1, 2, 3, 5, 8, 9, 10

TOPIC: LOGICAL REASONING
CONTENT:
Simple statements
Meaning of simple statement (i) True or False (ii) Negation of simple statements
Compound statements (i) meaning (ii) conjunction (iii) disjunction (iv) implication (v) bi-implication

SUB-TOPIC 1: SIMPLE STATEMENTS
Logic is the science of thinking about or explaining the reason for something. It is a particular method or system of reasoning which arrive at conclusions by way of valid evidence.
Mathematical logic can be defined as the study of the relationship between certain objects such as numbers, functions, geometric figures etc. .
Example: The following are logical statements;
Nigeria is in Africa
The river Niger is in Enugu
2 + 5 = 3
3 ≤ 7
(N.B The educator should ask the students to give their examples)
Example: The following are not logical statements because they are neither true nor false.
What is your name?
Oh what a lovely day
Take her away
Who is he?
Mathematics is a simple subject (note that this statements is true or false depending on each individual, so it is not logical).
EVALUATION
Educator to ask the students to give their own examples.


SUB – TOPIC 2. MEANING OF SIMPLE STATEMENT.

Statements are verbal or written declarations or assertions. The fundamental (i.e logical) property of a statement is that it is either true or false but not both. So logical statements are statements that are either reasonably true or false but not both
True and False statements: To determine the truth or falsity of a simple statement, one requires pre-knowledge and/or definition of the main concepts related to the statements. For example, the simple statement ‘it is hot’ is true if ‘it’ refers to a hot object or weather. Otherwise the statement is false. A true statement is said to have a truth value T while a false statement is said to have a truth value F.
Example: indicate T or F for the truth value of each statement.
〖10〗_two is equal to 10
Green is one of the colours on the Nigerian flag
How far is Abuja from here?
3 ∈ {2, 4, 6, 8, …}
The perimeter of a room 2.5m by 3.5m is 6m
Solution:
F
T
Not applicable
F
F
(note: educator to explain closed statements and open statements as in question 3)

NEGATION OF SIMPLE STATEMENTS The opposite of a statement is called the negation of the statement. Given any logical statement P, the negation (or the contradiction or the denial) of P is written symbolically as ~P
Examples: write the negation of each of the following statements.
I am a Mathematician
2 > 4
1/5<1/2
x+1≥4
The sky is the limit
Solution:
I am not a mathematician
2≯ 4
1/5≮1/2
x+1≤4 or x+1≱4
The sky is not the limit


Truth table for ~P:

P ~P
T F
F T


EVALUATION: Write the negation of the following statements:
All polygons are quadrilaterals
It is a sunny day
XYZ is an isosceles right angled triangle
The figure is a cube
X is not a prime number
SUB-TOPIC 3: Compound statements—When two or more simple statements are combined, we have a compound statement. To do this, we use the words: ‘and’, ‘or’, ‘if … then’, ‘if and only if’, ‘but’. Such words are called connectives.
Conjunction (or ˄) of logical reasoning: Any two simple statements p,q can be combined by the word ‘and’ to form a compound (or composite) statement ‘p and q’ called the conjunction of p,q denoted symbolically as p˄q.
Example: 1. Let p be “The weather is cold” and q be “it is raining”, then the conjunction of p,q written as p˄q is the statement “the weather is cold and it is raining”.
2. The symbol ‘˄’ can be used to define the intersection of two sets A and B as follows;
A∩B={x:x∈A ˄ x∈B}
The truth table for p˄q is given below;
EVALUATION:
Form compound statements using ‘and’, and express the following compound statements in symbol form.
P: It is cold.
Q: It is wet.
P: x+3
Q: x= -3
P: f(x) = 5x2 + 2
Q: f(1) = 7
P: (x+2)2 is a perfect square.
Q: when x = 1, (x+2)2 = 9
SUB-TOPIC 4:
Disjunction (or ˅) of logical statements. Any compound statement formed by using the word ‘or’ to combine simple statements is called a disjunction. The symbol ‘˅’ stands for ‘or’.
Examples
Let `p ‘ be “Bola studied Mathematics”, and `q’ be “Ngozi studied French”. Then the disjunction of p, q (p˅q) is the statement “Bola studied Mathematics or Ngozi studied French”.
p: You will read your notes . q: You will fail p˅q: You will read your notes or fail
P: x+1=2
Q: x=1.
P ˅ Q : x+1=2 or x=1.
P: The solution of x^(2 )-2x -15 =0 is 5.
Q: The solution of x^2-2x-15=0 is -3.
P˅Q: The solution of x^2-2x-15=0 or-3

The truth table for p˅q is illustrated below
P Q p˅q
T T T
F F F
T F T
F T T

EVALUATION Express the compound statements in symbolic form.
P: √2 is a rational number , q: √2 is an even number.
P: the trade union is stubborn, q: the workers strike will soon be ended.
P: 5 is a prime number, q: 7 is an even number.
P: a person who has taken physics can go to geophysics, q: a person who has taken geology can go for geophysics.

IMPLICATIONS (CONDITIONAL STATEMENTS) when the connective ‘if…then’ is used to combine simple statements, the result is called an implicative or conditional proposition. We denote implication symbolically by ⇒ i.e p⇒q means if p is true, then q is true. (or p implies q or p only if q, etc.)
Examples: form compound statements using ‘if … then’
P: The triangle is an equilateral triangle
Q: The angles are equal
P ⇒ Q: if the triangle is equilateral then the angles are equal.
P: -∞<x<10
Q: 100<x^2<∞
P ⇒ Q: if -∞<x<10 then 100<x^2<∞
P: Isa is a mathematician .
Q: He is intelligent.
P ⇒ Q: if Isa is a mathematician then he is intelligent.
N.B : Educator to explain antecedent and consequent. Example should be given.
The truth or falsity of the implication P ⇒ Q: is shown below;
P Q P ⇒ Q
T T T
T F F
F T T
F F T

EVALUATION: form compound statements using ‘if … then ’
P: y = 2
Q: y^2=4
P: A student reads Mathematics
Q: the student reads science
P: Damilola is a youth corper
Q: she has a degree
Identify the antecedent and consequent in the statement below;
If mathematics teachers work very hard then they will be compensated.
SUB-TOPIC 5
BI-IMPLICATION OR BI-CONDITIONAL STATEMENT (EQUIVALENCE). Another common statement in Mathematics is of the form “p if and only if q”. This statement is actually the combination of two conditional statements and so it is called bi-conditional or equivalence and is denoted by p↔q or sometimes p iff q (if and only if) i.e implies and is implied by.
Examples: 1. Let p be “he is a handsome man” and q be “10 > 6” then p↔q is the statement “he is a handsome man if and only if 10 > 6”, then p ↔q is the statement "He is a handsome man if and only if 10 >6" ”.
2. P: A number is divisible by 3
Q: the sum of the digits of the number is divisible by 3
P↔Q: A number is divisible by 3 iff the sum of its digits is divisible by 3
The truth table for p↔q is shown below;
P Q P↔Q
T T T
T F F
F T F
F F T

EVALUATION:
Copy and complete the table
P Q ~Q P ⇒ Q P ˅~Q (P⇒Q)↔(P˅~Q)
T T T
T F
F T F
F F T

GENERAL EVALUATION:
Determine the truth value of the compound statement ~ (P ˄ ~Q)

READING ASSIGNMENT:
Study the following;
Antecedent
Consequent
Converse, inverse and contrapositive statements
WEEKEND ASSIGNMENT:
If P and Q are two logical statements, copy and complete the following truth table
P Q P ˅ Q ~(P ˅ Q) (P ˅ Q) ˄ ~P ~(P ˅ Q) ⇒ ~P






Use a truth table to prove that; ~(p⇒q)↔(p˄~q)

TOPIC: LOGICAL REASONING
CONTENT: Logical operators and symbols
List of logical operators and symbols
A compound statement
Negation
Conjunction
Disjunction
Conditional statement
Bi-conditional statement
SUB-TOPIC 1
List of logical operators and symbols
The word ‘not’ and the four connectives ‘and’, ‘or’, ‘if … then’, ‘if and only if’ are called logic operators. They are also referred to as logical constants. The symbols adopted for the logic operators are given below.
Logic Operators Symbols
‘not’ - or~
‘and’ ˄
‘or’ ˅
‘if … then’ →
‘if and only if’ ↔
When the symbols above are applied to propositions p and q, we obtain the representations in the table below:
Logic operation Representation
‘not p’ ~p or p ̅
‘P and q’ p˄q
‘p or q’ p˅q
‘if p then q’ p→q
‘p if and only if q’ p↔q

SUB-TOPIC 2
Compound statements—When two or more simple statements are combined, we have a compound statement. To do this, we use the words: ‘and’, ‘or’, ‘if … then’, ‘if and only if’, ‘but’. Such words are called connectives.
Conjunction (or ˄) of logical reasoning: Any two simple statements p,q can be combined by the word ‘and’ to form a compound (or composite) statement ‘p and q’ called the conjunction of p,q denoted symbolically as p˄q.
Example: 1. Let p be “The weather is cold” and q be “it is raining”, then the conjunction of p,q written as p˄q is the statement “the weather is cold and it is raining”.
2. The symbol ‘˄’ can be used to define the intersection of two sets A and B as follows;
A∩B={x:x∈A ˄ x∈B}
The truth table for p˄q is given below;
P q p˄q
T T T
T F F
F F F

SUB-TOPIC 3
Negation of simple statements: The opposite of a statement is called the negation of the statement. Given any logical statement P, the negation (or the contradiction or the denial) of P is written symbolically as ~P
Examples: write the negation of each of the following statements.
I am a Mathematician
2 > 4
1/5<1/2
x+1≥4
The sky is the limit
Solution:
I am not a mathematician
2≯ 4
1/5≮1/2
x+1≤4 or x+1≱4
The sky is not the limit

Truth table for ~P:

P ~P
T F
F T


EVALUATION: Write the negation of the following statements:
All polygons are quadrilaterals
It is a sunny day
XYZ is an isosceles right angled triangle
The figure is a cube
X is not a prime number
SUB-TOPIC 4
Conjunction (or ˄) of logical reasoning: Any two simple statements p,q can be combined by the word ‘and’ to form a compound (or composite) statement ‘p and q’ called the conjunction of p,q denoted symbolically as p˄q.
Example: 1. Let p be “The weather is cold” and q be “it is raining”, then the conjunction of p,q written as p˄q is the statement “the weather is cold and it is raining”.
2. The symbol ‘˄’ can be used to define the intersection of two sets A and B as follows;
A∩B={x:x∈A ˄ x∈B}
The truth table for p˄q is; similar to that of compound statement truth table given above.
SUB-TOPIC 5
Disjunction (or ˅) of logical statements: Any compound statement formed by using the word ‘or’ to combine simple statements is called a disjunction. The symbol ‘˅’ stands for ‘or’.
Examples; 1. Let p be “Bola studied Mathematics”, and q be “Ngozi studied French”. Then the disjunction of p,q is the statement “Bola studied Mathematics or Ngozi studied French”
2. p: You will read your notes
q: You will fail
p˅q: You will read your notes or fail
3. p: x+1=2
q: x=1
p˅q: x+1=2 or x=1
The truth table for p˅q is illustrated below
P Q p˅q
T T T
F F F
T F T
F T T

EVALUATION: Express the compound statements in symbolic form.
P: √2 is a rational number , q: √2 is an even number
P: the trade union is stubborn , q: the workers strike will soon be ended
P: 5 is a prime number , q: 7 is an even number
P: a person who has taken physics can go to geophysics , q: a person who has taken geology can go for geophysics
Form compound statements using ‘or’
SUB-TOPIC 6
Implications (conditional statements): when the connective ‘if…then’ is used to combine simple statements, the result is called an implicative or conditional proposition. We denote implication symbolically by ⇒ i.e p⇒q means if p is true, then q is true. (or p implies q or p only if q)
Examples: form compound statements using ‘if … then’
P: the triangle is an equilateral triangle
Q: the angles are equal
P ⇒ Q: if the triangle is equilateral then the angles are equal.
P: -∞<x<10
Q: 100<x^2<∞
P ⇒ Q: if -∞<x<10 then 100<x^2<∞
P: Isa is a mathematician
Q: he is intelligent
P ⇒ Q: if Isa is a mathematician then he is intelligent
N.B : Educator to explain antecedent and consequent. Example should be given
The truth or falsity of the implication P ⇒ Q: is shown below;
P Q P ⇒ Q
T T T
T F F
F T T
F F T

EVALUATION: form compound statements using ‘if … then ’
P: y = 2
Q: y^2=4
P: A student reads Mathematics
Q: the student reads science
P: Damilola is a youth corper
Q: she has a degree
Identify the antecedent and consequent in the statement below;
If mathematics teachers work very hard then they will be compensated.

SUB-TOPIC 7
Bi-implication or Bi-conditional statement (equivalence): Another common statement in Mathematics is of the form “p if and only if q”. This statement is actually the combination of two conditional statements and so it is called bi-conditional or equivalence and is denoted by p↔q or sometimes p iff q (if and only if) i.e implies and is implied by.
Examples: 1. Let p be “he is a handsome man” and q be “10 > 6” then p↔q is the statement “he is a handsome man if and only if 10 > 6 ”
2. P: A number is divisible by 3
Q: the sum of the digits of the number is divisible by 3
P↔Q: A number is divisible by 3 iff the sum of its digits is divisible by 3


The truth table for p↔q is shown below;
P Q P↔Q
T T T
T F F
F T F
F F T

EVALUATION:
Copy and complete the table
P Q ~Q P ⇒ Q P ˅~Q (P⇒Q)↔(P˅~Q)
T T T
T F
F T F
F F T

GENERAL EVALUATION:
Determine the truth value of the compound statement ~ (P ˄ ~Q)
READING ASSIGNMENT:
Study the following;
Antecedent
Consequent
Converse, inverse and contrapositive statements
Tautology and contradiction
Laws of algebra of logical statements
WEEKEND ASSIGNMENT:
If P and Q are two logical statements, copy and complete the following truth table
P q P ˅ q ~(P ˅ q) (P ˅ q) ˄ ~p ~(P ˅ q) ⇒ ~p






Use a truth table to prove that; ~(p⇒q)↔(p˄~q)

TOPIC: LOGICAL REASONING
CONTENT: Logical operators and symbols
List of logical operators and symbols
A compound statement
Negation
Conjunction
Disjunction
Conditional statement
Bi-conditional statement
SUB-TOPIC 1
List of logical operators and symbols
The word ‘not’ and the four connectives ‘and’, ‘or’, ‘if … then’, ‘if and only if’ are called logic operators. They are also referred to as logical constants. The symbols adopted for the logic operators are given below.
Logic Operators Symbols
‘not’ - or~
‘and’ ˄
‘or’ ˅
‘if … then’ →
‘if and only if’ ↔
When the symbols above are applied to propositions p and q, we obtain the representations in the table below:
Logic operation Representation
‘not p’ ~p or p ̅
‘P and q’ p˄q
‘p or q’ p˅q
‘if p then q’ p→q
‘p if and only if q’ p↔q

SUB-TOPIC 2
Compound statements—When two or more simple statements are combined, we have a compound statement. To do this, we use the words: ‘and’, ‘or’, ‘if … then’, ‘if and only if’, ‘but’. Such words are called connectives.
Conjunction (or ˄) of logical reasoning: Any two simple statements p,q can be combined by the word ‘and’ to form a compound (or composite) statement ‘p and q’ called the conjunction of p,q denoted symbolically as p˄q.
Example: 1. Let p be “The weather is cold” and q be “it is raining”, then the conjunction of p,q written as p˄q is the statement “the weather is cold and it is raining”.
2. The symbol ‘˄’ can be used to define the intersection of two sets A and B as follows;
A∩B={x:x∈A ˄ x∈B}
The truth table for p˄q is given below;
P q p˄q
T T T
T F F
F F F

SUB-TOPIC 3
Negation of simple statements: The opposite of a statement is called the negation of the statement. Given any logical statement P, the negation (or the contradiction or the denial) of P is written symbolically as ~P
Examples: write the negation of each of the following statements.
I am a Mathematician
2 > 4
1/5<1/2
x+1≥4
The sky is the limit
Solution:
I am not a mathematician
2≯ 4
1/5≮1/2
x+1≤4 or x+1≱4
The sky is not the limit

Truth table for ~P:

P ~P
T F
F T


EVALUATION: Write the negation of the following statements:
All polygons are quadrilaterals
It is a sunny day
XYZ is an isosceles right angled triangle
The figure is a cube
X is not a prime number
SUB-TOPIC 4
Conjunction (or ˄) of logical reasoning: Any two simple statements p,q can be combined by the word ‘and’ to form a compound (or composite) statement ‘p and q’ called the conjunction of p,q denoted symbolically as p˄q.
Example: 1. Let p be “The weather is cold” and q be “it is raining”, then the conjunction of p,q written as p˄q is the statement “the weather is cold and it is raining”.
2. The symbol ‘˄’ can be used to define the intersection of two sets A and B as follows;
A∩B={x:x∈A ˄ x∈B}
The truth table for p˄q is; similar to that of compound statement truth table given above.
SUB-TOPIC 5
Disjunction (or ˅) of logical statements: Any compound statement formed by using the word ‘or’ to combine simple statements is called a disjunction. The symbol ‘˅’ stands for ‘or’.
Examples; 1. Let p be “Bola studied Mathematics”, and q be “Ngozi studied French”. Then the disjunction of p,q is the statement “Bola studied Mathematics or Ngozi studied French”
2. p: You will read your notes
q: You will fail
p˅q: You will read your notes or fail
3. p: x+1=2
q: x=1
p˅q: x+1=2 or x=1
The truth table for p˅q is illustrated below
P Q p˅q
T T T
F F F
T F T
F T T

EVALUATION: Express the compound statements in symbolic form.
P: √2 is a rational number , q: √2 is an even number
P: the trade union is stubborn , q: the workers strike will soon be ended
P: 5 is a prime number , q: 7 is an even number
P: a person who has taken physics can go to geophysics , q: a person who has taken geology can go for geophysics
Form compound statements using ‘or’
SUB-TOPIC 6
Implications (conditional statements): when the connective ‘if…then’ is used to combine simple statements, the result is called an implicative or conditional proposition. We denote implication symbolically by ⇒ i.e p⇒q means if p is true, then q is true. (or p implies q or p only if q)
Examples: form compound statements using ‘if … then’
P: the triangle is an equilateral triangle
Q: the angles are equal
P ⇒ Q: if the triangle is equilateral then the angles are equal.
P: -∞<x<10
Q: 100<x^2<∞
P ⇒ Q: if -∞<x<10 then 100<x^2<∞
P: Isa is a mathematician
Q: he is intelligent
P ⇒ Q: if Isa is a mathematician then he is intelligent
N.B : Educator to explain antecedent and consequent. Example should be given
The truth or falsity of the implication P ⇒ Q: is shown below;
P Q P ⇒ Q
T T T
T F F
F T T
F F T

EVALUATION: form compound statements using ‘if … then ’
P: y = 2
Q: y^2=4
P: A student reads Mathematics
Q: the student reads science
P: Damilola is a youth corper
Q: she has a degree
Identify the antecedent and consequent in the statement below;
If mathematics teachers work very hard then they will be compensated.

SUB-TOPIC 7
Bi-implication or Bi-conditional statement (equivalence): Another common statement in Mathematics is of the form “p if and only if q”. This statement is actually the combination of two conditional statements and so it is called bi-conditional or equivalence and is denoted by p↔q or sometimes p iff q (if and only if) i.e implies and is implied by.
Examples: 1. Let p be “he is a handsome man” and q be “10 > 6” then p↔q is the statement “he is a handsome man if and only if 10 > 6 ”
2. P: A number is divisible by 3
Q: the sum of the digits of the number is divisible by 3
P↔Q: A number is divisible by 3 iff the sum of its digits is divisible by 3


The truth table for p↔q is shown below;
P Q P↔Q
T T T
T F F
F T F
F F T

EVALUATION:
Copy and complete the table
P Q ~Q P ⇒ Q P ˅~Q (P⇒Q)↔(P˅~Q)
T T T
T F
F T F
F F T

GENERAL EVALUATION:
Determine the truth value of the compound statement ~ (P ˄ ~Q)
READING ASSIGNMENT:
Study the following;
Antecedent
Consequent
Converse, inverse and contrapositive statements
Tautology and contradiction
Laws of algebra of logical statements
WEEKEND ASSIGNMENT:
If P and Q are two logical statements, copy and complete the following truth table
P q P ˅ q ~(P ˅ q) (P ˅ q) ˄ ~p ~(P ˅ q) ⇒ ~p






Use a truth table to prove that; ~(p⇒q)↔(p˄~q)

TOPIC: Constructions
Content:
 Construction Of Perpendiculars
 How To Construct Perpendiculars
 Circumscribed And Inscribed Circles Of A Triangle
 Construction Of Quadrilaterals:
 Locus
Construction Of Perpendiculars
How To Construct A Perpendicular To A Given Straight Line AB From A Point P Outside The Line. P
Given: A line AB with a point P outside the line AB

A B
Step 1: Fix Compasses Pin at point P. Extend
the Compasses wide enough to draw an arc to
cut the given line AB at two points x1 and x2 .
Step 2: Fix Compasses Pin at x1 and x2, with
equal radii, draw arcs to cut each other at Q.

Step 3: Join QP cutting AB at C.
Then ACP = BCP = 900 P


C
A x1 x2 B



Q
Examples 1:
Using ruler and a pair of Compasses only,
(a) Constant
(i) triangle XYZ with XY = 8cm, YXZ = 600
and XŶ Z = 300
(ii) the perpendicular ZT to meet XY in T;
(iii) the locus l1 of points equidistant from ZY and XY.
(b) If l1 and ZT intersect at S, Measure ST
WASSCE, JUNE 2002, № 10.
Z
Solution: l1

S
600 T 300

X 8cm Y




How To Construct A Perpendicular To A Given Straight Line AB From A Point P, On The Line .
Given: A line AB with a point P on the line AB.

A P B

Step 1: Fix Compasses Pin at P and draw an Q
arc to cut AB at X1 and X2.
Step 2: Fix compasses Pin at X1 and X2., with
equal radii, draw arcs to cut each other at Q.
Step 3: Join QP A x1 P x2 B

EVALUATION:
New General Mathematics for Senior Secondary School , Book 1, pages 196 to 197, Exercise 16b, Nos. 3, 4, 5 and 7
SUB-TOPIC 2:
Circumscribed And Inscribed Circles Of A Triangle
How To Draw A Circle Passing Through The Three Vertices Of A Triangle ( Circumscribed Circle Of A Triangle)
Step 1: – Construct the perpendicular bisectors of each of the three sides of the triangle. The lines of bisection will all meet at a point.
Step 2: – Fix pin at the meeting point of the three lines and extend compass to one of the vertex and draw the circle.
Example 2:
Construct a triangle PQR such that PQ = 8cm, QR = 7cm and PR = 6cm. Construct a circle passing through the points P, Q and R. What is the radius of the circle?
R






P Q





Radius = 4.1cm

How To Construct The Inscribed Circle Of A Given Triangle ABC
Step 1: – Construct an internal bisector of each of the angles A, B and C. The bisectors of the
three angles will meet at a point O.
Step 2: – Construct a perpendicular from O to AB to meet AB at P
Step 3: – Join OP ,(OP is the radius of the circle)
Step 4: – Draw the inscribed circle with center O and radius OP.
Example 3:
Construct a triangle ABC such that
AB = 8cm, BC = 7cm and AC = 6cm.
Construct an inscribed circle of the triangle .





7cm

A P 8cm B

EVALUATION:
New General Mathematics for Senior Secondary School , Book 1, pages 196 to 197 , Exercise 16b, Nos. 1, 2, and 6
SUB-TOPIC 3:
Construction Of Quadrilaterals:

Parallelograms:

Example 4:
Using a ruler and a pair of compasses
only construct a parallelogram RXYZ
such that XY = 7cm, XYZ = 1200 and
the diagonal XZ = 9.5cm.
Solution:
Step 1: – Draw the line XY = 7cm with
your ruler.
Step 2: – Construct the angle 1200 at point Y. R Z

Step 3: – Measure 10.5cm with your 9.5cm
pair of compasses. Fix pin at X and
draw an arc to cut the line YZ at Z. 1200
Step 4: – Measure YZ with your X 7cm Y
pair of compasses. Fix pin at the
point X and draw an arc at the suspected position of R (Since opposite sides of a parallelogram are equal XR = YZ).
Step 5: – Measure 7cm with your pair of compasses. Fix pin at Z and cut an arc at the suspected position of R to cut the first arc in step 4. (Opposite sides of a parallelogram are equal RZ = XY). The meeting point of the arcs is R.
Step 6: – Join RZ and XR.
* Put into considerations the properties of a parallelogram
Trapezium:

Example 5:
Using a ruler and a pair of compasses only, construct a trapezium ABCD, in which AB//DC, AB = 8cm, ABC = 600, BC = 5.5cm and BD = 8.3cm.





Solution:
Step 1: – Draw the line
AB = 8cm using a ruler. X D C
Step 2: – Construct
angle 600 at point B.
Step 3: – Measure 5.5cm
with your Compasses,
fix pin at B and cut C to get BC. 600
Step 4: – Using the measurement A 8cm B
of BC = 5.5cm and properties of a parallelogram. Draw AX
parallel to BC. Measure BA = 8cm,
fix pin at C, draw CX parallel to BA (Use broken lines).
Step 5: – Measure 8.3cm with your pair of compasses. Fix pin at B and draw an arc to cut the line CX at D.
Step 6: – Join CD with a thick line, Join AD with a thick line.

EVALUATION

(1) Using a ruler and a pair of compasses only, construct the following parallelograms.
(a) //gm ABCD, such that AB = 8cm, ABC = 1350 and BC = 4.5cm. Measure AC.
(b) //gm PQSR, such that PQ = 7cm, SPQ = 1200 and QR = 5cm. Measure SQ.
(c) //gm ABCD, such that AB = 7.5cm, ABC = 1050 and AD = 4cm. Measure BD.
(d) //gm ABCD, such that AB = 8cm, AD = 5cm and BD = 6cm. Measure BCD


(2) Using a ruler and a pair of compasses only, construct the following trapeziums.
(a) ABCD, such that AB = 8cm, ABC = 750, DAB = 600, AD = 4.5cm and AB//DC. Measure BC.
(b) PQRS, such that PQ = 7.6cm, SPQ = 900, PS = 4cm, SR = 5.7cm and PQ//SR. Measure QR.
(c) ABCD, such that AB = 7cm, BC = 5cm, ABC = 600, CD = 4cm and DC//AB. Measure AD.
(d) ABCD, such that AB = 6cm, BC = 4.3cm, ABC = 1200, CD = 8.5cm and AB//DC. Measure DAB.
LOCUS:
DEFINITION
The Locus of a point is the set of all possible positions occupied by an object, which varies its position according to some given law. The plural of locus is loci. Below are some examples of common loci.

***************************************************************************
1. The locus l1 of points equidistant
from two given fixed point A and B is
the perpendicular bisector of the line l1
joining the two points A and B.
A B


***************************************************************************
2. The locus of point equidistant from
two intersecting lines is the pair of bisectors of
the angles between the lines e.g. locus l2 of
point equidistant from AB and CD.

l2
Step 1: Fix pin at the point of intersection O , and draw
an arc to cut the two lines A C
Step 2: Fix pin at points where the arc cut the two lines
and draw arcs to intersect
Step 3: Join the point of intersection of the l2
arc to the point of intersection
of the line , O and produce through.
STEP 4 : carry out the same steps above for D B
the other angle not bisected on the intersecting
line . (the two part of the locus
will intersect at right angles )

Note Also That: A
The locus of points equidistant
from AB and BC is given below. B l2
(Note that the two lines AB and BC
meets at a point B hence bisect the C
angle B)







3. The locus l3 of points at a fixed
distance d from a fixed point A is a circle
drawn from a point A with a radius of l3
length d units. e.g. the locus l3 of points 3cm A
from A. dcm



***************************************************************************
4. The locus of points equidistant from parallel A B
lines AB and CD is a line parallel to AB and CD ----------------------------------  locus
at equal distance from each C D

***************************************************************************
5. Locus of points at a given distance x cm l
from a straight line AB

Step 1: Draw the line AB x cm x cm
Step 2: Measure x cm ,fix pin at A and draw A B
an arc above and below the line x cm x cm
l
Step 3: With the same x cm, fix pin at B and draw an arc above and below the line.
Step 4: Join the arcs on top with a straight line parallel to AB .
Step 5: Join the arc below AB,
with a straight line parallel to AB.
***************************************************************************
EVALUATION
(1) Using a ruler and a pair of compasses only, construct
(i) A triangle XYZ in which YZ = 8cm, XYZ = 600 and XZY = 750. Measure XY.
(ii) The locus l1 of points equidistant from Y and Z.
(iii) The locus l2 of points equidistant from YX and YZ.
(iv) The locus l3 ,4cm from X.
(b) Measure QY where Q is the point of intersection of l1 and l2.
SSCE, June 1994, No 8 (WAEC).
(2) Using a ruler and a pair of compasses only
(a) Construct a triangle ABC such that AB = 6cm, AC = 8.8cm and BAC = 1200.
(b) Construct a locus l1 of points equidistant from point A and B.
(c) Construct the locus l2 of points equidistant from AB and AC.
(d) Find the points of intersection P1 and P2 of l1 and l2 and measure P1 P2.
G.C.E, Nov 1990, No 10 (WAEC).
Example 7:
Using a ruler and a pair of compasses only construct
a triangle ABC such that AB = 7.5cm
ABC = 750, BC = 6.5cm.
(a) Find the locus l1 of points equidistant from A and B.
(b) The locus l2 of points 4cm from C
(c) Locate the points of intersection x1 and x2 of l1 and l2.
Measure |x1 x2|



Solution:
x1


l2
C 4cm



x2 6.5cm

750

A 7.5cm B
l1

/x1x2/= 6.9cm

Example 8:
(a) Using a ruler and a pair of compasses
only, construct a parallelogram RXYZ
such that XY = 6.4cm, YZ = 4.7cm and XYZ = 1200.
(b) Construct in the plane of the parallelogram, RXYZ
(i) The locus l1 of points equidistant from XR and XY,
which lie inside the parallelogram.
(ii) The locus l2 of points at a distance 4cm from Y.
(iii) The locus l3 of points equidistant from R and Z.
(c) (i) Find the points of intersection A and B of l1 and l2.
(ii) Find the points of intersection C and D of l2 and l3.
(iii) Measure AC. G.C.E Nov. 1982 (W.A.E.C)
Solution:

.
L3


R Z
cCCc

A

X Y l2




D
/AC/= 4cm .

EVALUATION:
(1) Using a ruler and a pair of compasses only(a) Construct
(i) a  ABC such that AB = 5cm, AC = 7.5cm and CAB = 1200.
(ii) The locus l1 of points equidistant from A and B.
(iii) The locus l2 of points equidistant from AB and AC which passes through triangle ABC.
(b) Label the point P where l1 and l2 intersects.
(c) Measure CP
SSCE, June 1988, No 11 and SSCE, June1992, No 7 (WAEC).

(2) (a) Using a ruler and a pair of compasses only, construct a triangle ABC such that ôABô = 9cm, ôBCô = 7cm and ôACô = 6cm.
(b) Construct the locus l1 equidistant from AB and BC.
(c) Construct the locus l2, 4cm from A.
(d) Locate the points of intersection P1 and P2 of l1 and l2. Measure ôP1 P2ô.

(3) Using a ruler and a pair of compasses only, construct the following
(a) A trapezium ABCD such that AB = 7cm, DAB = 600, AD = 5cm and DC = 4cm. DC//AB.
(b) Locus l1 equidistant from A and B
(c) Locus l2, 4cm from B.
(d) Locate the points of intersection X1 and X2 of l1 and l2. Measure ôX1 X2ô.
SUB-TOPIC
Dividing A Line Segment Into N Equal Parts
Example 13:
Divide the line AB = 10cm in the ratio 5:2
Solution:
The line would be divided into 5 + 2 = 7 parts.
Step 1: – Draw the line AB to be divided in the ratio 5:2
Step 2: – Draw any other line AP through A.
Step 3: – Set your compasses at any convenient radius, divide the line drawn in step 2 into 7
equal parts AC, CD, DE, EF, FG, GH and HI.
Step 4: – Join BI.
Step 5: – Construct lines parallel to BI at the points H, G, F, E, D, C using your setsquare and a ruler. (The assistance of a teacher is needed for detailed explanations).
Step 6: – The line is in the ratio 5:2 at the point q on AB.

A 10cm q B

C D
E F
G
H I P








Example 14:
Divide the line AB = 8cm in the ratio 3:2

Solution: A 8cm B










Example 15:
(a) Using a ruler and a pair of compasses only,
construct a triangle ABC with AB = 7.5cm,
BC = 8.1cm and ABC = 1050.
(b) Locate the point D on BC such that
BD: DC is 3:2
(c) Through D construct a line, l perpendicular to BC.
(d) If the line l meets AC at P measure BP.
SSCE, June 1995, No 9 (WAEC).











Solution:


C
l

P

D

8.1cm

1050

A 7.5cm B /BP/ = 5.3cm


GENERAL EVALUATION:

(1) (a) Using a ruler and a pair of compasses only, construct
(i) A triangle QRT with QR = 8cm, RT = 6cm and QT = 3cm.
(ii) A trapezium PQRS, which has a common side QR with  QRT, given that PQ is parallel to SR, PQ = 7cm, QR = 8cm, RS = 4cm and PTQ is a straight line.
(iii) The locus l1 of points equidistant from PQ and PS.
(iv) The locus l2 of points equidistant from T and R.
(b) Measure TX, where X is the point of intersection of l1 and l2.
G.C.E, Nov 1985, No 8 (WAEC).

(2) (a) Using a ruler and compasses only, construct
(i) A triangle PQR such that PQ = 6cm, QR = 7cm and PQR = 1350.
(ii) The locus l1 of points equidistant from P and Q.
(iii) The locus l2 of points equidistant from PQ and QR.
(iv) The locus l3 of point at which QR subtends an angle of 900.
(b) Locate;
(i) The point of intersection X of l1 and l2.
(ii) The point of intersection Y of l2 and l3.
(c) Measure XY.
G.C.E, Nov 1985, No 21 (WAEC).

(3) Using a ruler and a pair of compasses only construct a triangle ABC in which AB = 8cm, AC = 5cm and BAC = 450. Measure BC. Construct a circle with center P on BC such that AB and AC produce are tangent of circle. Measure the radius of the circle.
SSCE, Nov 1992, No 11 (WAEC).




References
1. New General Mathematic for Senior Secondary School, Book 1, By Channon , Smith Et al.
2. MAN Mathematics for Senior Secondary School, Book 1
3. Fundamental General Mathematics for Senior Secondary School, By Idode G. O

TOPIC: PROOFS OF SOME BASIC THEOREMS
CONTENT:
Proof: The sum of the angles in a triangle
The exterior angle of a triangle is equal to the sum of the opposite interior angles
Geometry is the study of the properties of shapes. In theoretical or formal geometry the facts are proved for general cases by a method of argument or reasoning rather than by measurement. Geometrical basic facts are called theorems. Theorems are the foundations upon which geometry is built.
Interior and exterior angles of triangles and polygons.
Recall: Angles on a straight line is 180°. Thus,

50° 60° θ
θ+50+60=180
θ=180-110
= 70°


Using the diagram below;
P T

Q θ θ S
Use the angle properties related to parallel lines to explain why;
Angle TRS = angle PQR corresponding or ‘F’ angles
Angle TRP = angle QPR. Alternate or ‘z’ angles
Explain why the sum of the three angles at R is 180°. Angles on a straight line
Theorem 1: The sum of the angles of a triangle is 180°
Given: any ∆ABC
To prove: A ̂+B ̂+C ̂=180°
Construction: produce (BC) ̅ to a point X.
Draw (CP) ̅ parallel to (BA) ̅
A P
〖 a〗_2 a_1
b_2 c b_1
B C X
Proof:
With the lettering of the diagram
a_1=a_(2 ) (alternate angles)
b_1=b_(2 ) (corresponding angles)
〖C+a〗_1+b_1=180° (angles on straight line)
〖C+a〗_2+b_2=180°
AC ̂B+A ̂+B ̂ = 180°
∴ A ̂+B ̂+C ̂ = 180°


EVALUATION:
Prove that a=90°-b in the figure below

a
b



Find the value of angle ‘a’ below and state clearly any theorem applied

20°
a° 50°
45°

In the figure below, <WXP=115° and <ZYP=158°,calculate <XPY
P



W 158° 115°
X Y Z
What is the value of the angle marked m in the figure below?
X

33°

m°

22°





Theorem 2: The exterior angle of a triangle is equal to the sum of the opposite interior angles.
A
P°
q° θ m°
B C X
Given: any ∆ABC with (BC) ̅ produce to X
Proof: With the lettering of the diagram,
AC ̂X+AC ̂B=180° (angles on a straight line)
∴ AC ̂X=A ̂+B ̂(=180°-AC ̂B) or
The diagram shows that m = p + q as stated above,
Since p + q + θ = 180° (sum of angles in a ∆)
m + θ = 180° (angles on a straight line)
m + θ = p + q + θ
i.e m = p + q
EVALUATION:
Find the sizes of the lettered angles in the diagrams below. State clearly the reason for each statement
i. k


115° h 31°



ii.
m°

64° P° 2P°

iii.
k°
h°

128° g°

The angles of a triangle are 5a/2, 3a/4 and 7a/4 . Find the value of the largest angle

Theorem 3: The sum of the interior angles of any n-sided convex polygon is (2n-4)right angles
Recall: in basic 8 the sum of interior angles of any n-sided polygon is (n-2)180°
Note: A convex polygon does not contain any reflex angles

O



Given: any convex polygon ABCDE… with n-sides
To prove: A ̂+B ̂+C ̂+⋯=(2n-4)right angles
Construction: Join the vertices A, B, C, … to any point O inside the polygon.
Proof:
By construction there are n triangles (polygon ABCDE… has n sides)
Sum of angles of 1∆=180° or 2 right angles
Sum of angles of n ∆s=2n right angles
Sum of angles at O =360° or 4 right angles
Sum of angles of polygon ABCDE… = sum of angles of n ∆s- sum of angles at O
∴A ̂+B ̂+C ̂+⋯=(2n-4)right angles
EVALUATION:
The angles of a quadrilateral taken in order are a, 5a, 4a & 2a. find these angles, Draw a rough sketch of the quadrilateral. What kind of quadrilateral is it?
PQRST is a regular pentagon. Find the angles of ∆PSR.
Find the interior angles of a regular polygon which has (a) 12 sides (b) 20 sides
Four angles of a pentagon are equal and the fifth is 60°, find the equal angles and show that two sides of the pentagon are parallel
THEOREM 4: The sum of the exterior angles of any convex polygon is 4 right angles.
Note: while convex polygon has all the interior angles less than 180°, a reflex or re-entrant polygon has some interior angles more than 180°.



Fig I
Given: any convex polygon ABCDE..….N with n sides, each side produced to give exterior angles x,y,z,…
To prove: x+y+z+⋯=4 right angles
Construction: From any point O, draw lines OP, OQ, OR, OS, … parallel to sides of ABCDE……N in turn
Proof: With the lettering of the diagram
a=x (OP//NA and OQ//AB)
Similarly, b=y,c=z,…
But a+b+c+⋯=4 right angles(angles at a point)
∴x+y+z+⋯=360°
Example: In the figure below, side (BC) ̅ of ∆ABC is produced to D. The bisector of AC ̂D meets (BA) ̅ produced at X.
Prove that X ̂=1/2(BA ̂C-B ̂) X
A

B b^° 〖 z〗^° D
Since (CX) ̅ is the bisector of AC ̂D, AC ̂X = DC ̂X = z
In ∆XAC,
x+y+z=180 (angle sum of a ∆)
x=180-y-z ……..(i)
At A, a+y=180 (angles on a straight line)
y=180-a
In ∆ABC,
2z=a+b (exterior angle of a trianle)
z=1/2 a+1/2 b
Substitute for y and z in (i)
x=180-(180-a)-(1/2 a+1/2 b)
=180-180+a-1/2 a-1/2 b
=1/2 a-1/2 b
=1/2(a-b)
∴ X ̂=1/2(BA ̂C-B ̂)
EVALUATION:
In ∆ABC, the bisectors of B ̂ and C ̂ meet at P. Prove that BP ̂C=90+1/2 A ̂
In ∆ABC, the side (BC) ̅ is produced to D. If the bisector of AC ̂D is parallel to (AB,) ̅ prove that two angles of ∆ABC are equal.
In a given regular polygon, the ratio of exterior angle to the interior angle is 1:3, how many sides has the polygon?
If the exterior angles of a pentagon are a°,〖(a+5)〗^°, 〖(a+10)〗^°,〖(a+15)〗^° and 〖(a+20)〗^°, find the value of the largest angle.

WEEKEND ASSIGNMENT:
Find the sum of the exterior angles of a polygon with 8 sides.
One angle of a pentagon is 120°. The other angles are regular, find one of the other angles.
New General Maths page 32, Ex. 2a nos 13 & 14

REFERENCES
M.F Macrae etal (2011), New General Mathematics for Senior Secondary Schools 1.
MAN Mathematics for senior Secondary Schools 1.
New school mathematics for senior secondary school et al; Africana publishers limited
Fundamental General Mathematics For Senior Secondary School by Idode G. O


WEEK 10 Date…………………………
TOPIC: Proofs of some Basic theorems.
Sub- Topics 1
Contents
Riders including angles of parallel lines
Angles in a polygon
Congruent triangles
Properties of Parallelogram
Intercept theorem
Period 1
Angles of Parallel line
Recall Basic geometrical facts are called theorems. The first is the sum of the angles of a triangle is 1800. Many other theorems depend on it. For this reason they are often called Riders. (Since they ride on theorem 1)
If two parallel lines are intersected by a Transversal.
The alternate angles are equal
The corresponding angles are equal
The interior angles on the same side of the transversal are supplementary viz:
(a)

a

a
Transversal
a = a alternate angle
(b)
b

b

b = b corresponding angle
(c) d

c

co-interior / allied
c + d = 1800. Supplementary angle
Other angles formed are: vertically opposite angles, they are equal.
Angle at a point (3600) and on
Straight line (1800) a
a
b
b
a = a vertically opposite angle
Examples:
In the diagram below YPA = (3X +10), APQ = (4x -20) and PQD = x + 700.
Prove that ll

Y
A P B


C Q D
X
Given: as above
Proof: ll
3x - 10 + 4x – 20 = 1800 (angle on straight line)
3x + 4x – 10 – 20 =1800.
7x – 300 =1800.
7x = 1800 + 300
7x = 2100
X = 300.
If 4x – 200 = x + 700 (alternate angle)
Substituting for x
4(30) – 20 = 30 + 70
120 -20 = 30 + 70
100 = 100
∴ ll where is transversal.
2. Calculate the size of the marked angle in the diagram below.

53° 47° ⇒ 53° a b
M 47°
a = 530 (alternate angle)
b = 470 (alternate angle)
M = 3600 – (a + b) (angle at a point)
M = 3600 – 1000
M = 2600.
Evaluation
In the diagram below, is parallel to . Find the value of x
A E

x°
62° C
312°
B D








In the diagram below: and are parallel. If PQR = 1450 and RST = 1550. Find the value of x
Q P
145°

R x°

155°
S T
In the diagram below: ll , PST = 520, QTS = 260. and lPQl = lQTl. Calculate PRQ.
P

Q R

S T
PERIOD 2
Congruent Triangles
When there exists equality relationship in terms of corresponding sides and angles of triangles.
We say that the triangles are congruent
Two triangles are congruent if proved that
Two sides and the included angle of one are equal to the corresponding two sides and the included angle of the other (SAS)
Two angles and a side of one are equal to the corresponding two angles and a side of the other. ASA or AAS
Three sides of one are respectively equal to three sides of the other (SSS)
In right – angled triangles, the hypotenuse and another side are equal to the hypotenuse and a side of the other (RHS)
EXAMPLES
Name the triangle which is congruent to ∆XYZ, giving the letters in the correct order. State the condition of Congruency.

F H X (b) Y
D

G Y Z

X Z
(c) Z (d) X
X


Y Y Z A


C
The congruency is as follows:
XYZ ≡ GHF (SAS)
XYZ ≡ ZDX (AAS) (ASA)
XYZ ≡ YXC (SSS)
XYZ ≡ XAZ (RHS)
Name the triangles that are congruent. Give reasons
X


L
O 120°


120°
Y M Z
∆YZL≡∆XZM (SAS)
∇XOL≡∆YOM (ASA)
In each of the following, the statements refers to ∆s XYZ and MNO. In each case, sketch the triangles and mark in what is given. If the triangles are congruent, state three other pairs of equal elements and give the condition of congruency.
|XY|=|MN|, |YZ|=|NO|, Y ̂=N ̂
Solution:
Y N

X M
Z O
X ̂=M ̂ Z ̂=O ̂ (SAS)
EVALUATION:
Page 35, Ex. 2b nos; 2e, 2g, 2i and 2j (NGM sss1)
Page 35, Ex. 2b nos; 3e, 3f, 3g and 3h (NGM sss1)
PERIOD 3
Properties of parallelogram
Definition: A parallelogram is a quadrilateral which has both pairs of opposite sides parallel.
The properties of parallelogram are:
The opposite sides are parallel
The opposite angles are equal
The opposite sides are equal
The diagonal bisects one another
The diagonals make pairs of alternate angles which are equal
The allied or co-interior forming adjacent vertices of a parallelogram are supplementary



Theorem 1: In a parallelogram, (a) the opposite sides are equal (b) the opposite angles are equal
A B
y_2 x_1
x_2 y_1
D C
Given: Parallelogram ABCD
To prove: (i) |AB|=|CD|, |BC|=|AD|
(ii) B ̂=D ̂, A ̂=C ̂
Construction: Draw the diagonal AC
Proof: In ∆s ABC and CDA, x_1=x_2 (alt.angles,AB||DC)
y_1=y_2 (alt.angles,AD||BC)
AC is common
∴∆ABC≡∆CDA (ASA)
∴(i) |AB|=|CD| ,|BC|=|DA| corresponding dises
(ii) B ̂=D ̂ corresponding angles
A ̂=x_1+y_2
=x_2+y_1
=C ̂
Theorem 2: The diagonals of a parallelogram bisect one another
A B

O
D C
Given: Parallelogram ABCD with diagonals AC and BD intersecting at O.
To prove: |AO|=|OC|, |BO|=|OD|
Proof: In ∆s AOB and COD,
x_1=x_2 (alt. Angles, AB||CD)
y_1=y_2 (alt. Angles, AB||CD)
|AB|=|C D| (opp sides of llgm)
∴ ∆AOB≡∆COD (ASA)
∴|AO|=|CO| (corresponding sides)
and |BO|=|DO| (corresponding sides)
EVALUATION:
Page 39 NGM book 1 EX. 2e, nos 1, 2, 3, 6
Period 4: Intercept Theorem
Intercept: This is to stop something that is going from one place to another. Mathematically from the diagram below; a transversal line is stopped by two lines there by forming an intercept between the lines.
P
A x B
intercept
C y
D
If a transversal cut two or more parallel lines, the parallel lines cut the transversal into line segments known as intercept



Examples:
X
P

Q

R

Y
(PQ) ̅ and (QR) ̅ are intercept of (XY) ̅

X M

A P U B

C Q V D

E R W F

Y N
(PQ) ̅and (QR) ̅ are the intercept of (XY) ̅ while (UV) ̅ and (VW) ̅are the intercept of (MN) ̅

EVALUATION:
In the diagram below, |LM|=|LN|, (MN) ̅//(OP) ̅ and <OPN = 45°. Find <PLM
L


M N


O P
New General Mathematics for SSS 1, page 41, Ex. 2f nos 1a, 1b, 1f and 4

Having known what and how to prove a given theorem, there is need to identify them in order as ‘riders’.
The sum of the angles of a triangle is 180°
The exterior angle of a triangle is equal to the sum of the opposite interior angles
The sum of the interior angles of any n-sided convex polygon is (2n-4)right angles
The sum of the exterior angles of any convex polygon is 4 right angles
The base angles of an isosceles triangle are equal
In a parallelogram, (a) the opposite sides are equal (b) the opposite angles are equal
The diagonals of a parallelogram bisect one another
If three or more parallel lines cut off equal intercepts on a transversal, then they cut off equal intercepts on any other transversal

EVALUATION: Try to memorize these theorems as riders
GENERAL ASSIGNMENT:
NGM for SS1, page 87, revision exercise 2, nos 1, 2, 3,4
Revision test 2, nos 1, 2, 4, 6 and 8
Page 37, Ex. 2c, nos 13, 16 and 17


REFERENCES
M.F Macrae etal (2011), New General Mathematics for Senior Secondary Schools 1.
MAN Mathematics for senior Secondary Schools 1.
New school mathematics for senior secondary school et al; Africana publishers limited
Fundamental General Mathematics For Senior Secondary School by Idode G. O