SCHEME OF WORK
WEEKS TOPICS
1. Deductive proofs: (i) The angle an arc subtends at the centre is twice the angle it subtends at the circumference of the circle. Solution to related problems (ii) Angles in the same segment are equal. Solution to related problems (iii) Riders based on angles subtended by chords at the centre perpendicular bisectors of the chords (iv) Angles in alternate segment are equal.
2. Solving problems on circle theorems.
3. Review of angles of elevation and depression, surface area, Volumes of solid shapes. (ii) Relationship between sector and cone (iii) Solving problems on surface area of solid shapes and volume.
4. Review of area and perimeter of plane shapes: Square, rectangle, trapezium, parallelogram, kite, circle, rhombus and triangle.
5. Probability: Experimental events (i) Throwing dice, tossing coins, playing cards, drawing balls of different colours etc (Teacher must practicalise e.g playing of ludo). (ii) Theoretical Probability: passing examination, playing football, giving birth to baby, rainfall, winning elections.
6. Addition and subtraction of probability for mutually exclusive events interpretation of ‘and ‘ ‘or’ in probability.
7. Review of first half term’s lesson (ii) periodic test
8. Matrices: (i) 2 x 2 and 3 x 3 matrices (ii) Addition, subtraction and multiplication of matrices.
9. (iii) Transpose and inverse of matrices (iv) Determinants of matrices (v) Application of determinants.
10. Problem solving on number bases. Expansion, Conversion and Relationship. Determination of missing number from a problem is to be emphasised.
11. Indicial Equations. Concept and relationship with quadratic equation. (ii) Revision
3RD TERM
WEEK 1
MAIN TOPIC: Deductive proofs
SPECIFIC TOPIC: The angle an arc subtends at the centre is twice the angle it subtends at the circumference of the circle.
REFERENCE BOOK : New General Mathematics For S.S.2
OBJECTIVE: At the end of the lesson, the students should be able to:
Recall, use and apply the properties of angles subtended by arcs and chords at the centre and circumference of a circle.
CONTENT: ANGLES IN A CIRCLE
THEOREM
The angle that an arc of a circle subtends at the centre is twice that which it subtends at any point on the remaining part of the circumference.
GIVEN : a circle APB with centre O.
To Prove : AOB = 2 X APB
CONSTRUCTION: Join PO and produce it to any point Q
PROOF:
NOTE: there are three possible diagrams, as in figure below
P
X2 y2
X1 y1
A Q B
With the lettering in figure above
/OA/ = /OP/ (radii)
Therefore x1 = x2 (base angles of isos. )
Therefore AOQ = X1 + X2 (ext. Angle of AOP)
Therefore AOQ = 2x2 (x1 = x2)
Similarly, BOQ = 2y2
In figure above AOB = AOQ + BOQ
= 2x2 + 2y2
= 2(x2 + y2)
= 2 x APB
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
P
X2 y2 B
Y1
A X1 0
Q
From the above diagram proof that the angle that an arc of a circle subtends at the centre is twice that which it subtends at any point on the remaining part of the circumference.
SPECIFIC TOPIC: Calculation of angles at the centre of a circle
OBJECTIVE: At the end of the lesson, the students should be able to:
Find the angle at centre of a given circle
CONTENT: DEDUCTIVE PROOF
Example 1: In figure below, O is the centre of circle WXYZ. If XWZ = 330, find XOZ and XYZ.
W O
330
X Z
Y
SOLUTION
XOZ = 2 x 330 (angle at centre = 2 x angle at circumference)
= 660
Reflex XOZ = 3600 - 660 (Angles at a point)
= 2940
XYZ = ½ of 2940 (angle at centre = 2 x angle at circumference)
= 1470
EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
Find the lettered angle in figure below (0 is the centre of the circle)
410
O
q
SPECIFIC TOPIC: The angle which an arc of a circle subtend at the centre is twice that which it subtends at a point on the remaining part of the circumference
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problem on angle at circle
CONTENT: DEDUCTIVE PROOFEDUCTIVE PROOF
Example 1: The angle which an arc (or chord) of a circle subtend at the centre is twice that which it subtends at a point on the remaining part of the circumference.
C
a b
O b
a
A P B
GIVEN : Arc AB subtends angle AOB at the centre O and subtends angle ACB at the circumference, as shown in the figure above
Required to prove that angle AOB = 2 angle ACB
Construction : Join CO and produce to a point P.
Proof: Triangles AOC and BOC are isosceles triangles
Since AO = OC = OB = radius of circle.
Put
Angle OAC = Angle OCA = a0
Angle OCB = Angle OBC = b0
So that angle ACB = angle OCA + angle OCB = a0 + b0
Using the fact that exterior angle of a triangle is equal to the sum of the two interior opposite angles , we have
Angle POA = angle OAC + angle OCA = 2a0 .................................... (1)
Angle POB = angle OCB + angle OBC = 2b0 .................................... (2)
Add equations 1 and 2
Angle POA + POB = 2(a0 + b0)
i.e angle AOB = 2 angle ACB
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
p
O
P Q
P,Q,R are points on a circle, centre O as shown in figure above. If angle OPQ = 420, find angle PRQ
ASSIGNMENT:
Find the lettered angles in figure below (O is the centre of the circle)
1260
O
s
SPECIFIC TOPIC: The angle an arc subtends at the centre is twice the angle it subtends at the circumference of the circle.
REFERENCE BOOK : New General Mathematics For S.S.2
OBJECTIVE: At the end of the lesson, the students should be able to:
Recall, use and apply the properties of angles subtended by arcs and chords at the centre and circumference of a circle.
CONTENT: ANGLES IN A CIRCLE
THEOREM
The angle that an arc of a circle subtends at the centre is twice that which it subtends at any point on the remaining part of the circumference.
GIVEN : a circle APB with centre O.
To Prove : AOB = 2 X APB
CONSTRUCTION: Join PO and produce it to any point Q
PROOF:
NOTE: there are three possible diagrams, as in figure below
P
X2 y2
X1 y1
A Q B
With the lettering in figure above
/OA/ = /OP/ (radii)
Therefore x1 = x2 (base angles of isos. )
Therefore AOQ = X1 + X2 (ext. Angle of AOP)
Therefore AOQ = 2x2 (x1 = x2)
Similarly, BOQ = 2y2
In figure above AOB = AOQ + BOQ
= 2x2 + 2y2
= 2(x2 + y2)
= 2 x APB
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
P
X2 y2 B
Y1
A X1 0
Q
From the above diagram proof that the angle that an arc of a circle subtends at the centre is twice that which it subtends at any point on the remaining part of the circumference.
SPECIFIC TOPIC: Calculation of angles at the centre of a circle
OBJECTIVE: At the end of the lesson, the students should be able to:
Find the angle at centre of a given circle
CONTENT: DEDUCTIVE PROOF
Example 1: In figure below, O is the centre of circle WXYZ. If XWZ = 330, find XOZ and XYZ.
W O
330
X Z
Y
SOLUTION
XOZ = 2 x 330 (angle at centre = 2 x angle at circumference)
= 660
Reflex XOZ = 3600 - 660 (Angles at a point)
= 2940
XYZ = ½ of 2940 (angle at centre = 2 x angle at circumference)
= 1470
EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
Find the lettered angle in figure below (0 is the centre of the circle)
410
O
q
SPECIFIC TOPIC: The angle which an arc of a circle subtend at the centre is twice that which it subtends at a point on the remaining part of the circumference
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problem on angle at circle
CONTENT: DEDUCTIVE PROOFEDUCTIVE PROOF
Example 1: The angle which an arc (or chord) of a circle subtend at the centre is twice that which it subtends at a point on the remaining part of the circumference.
C
a b
O b
a
A P B
GIVEN : Arc AB subtends angle AOB at the centre O and subtends angle ACB at the circumference, as shown in the figure above
Required to prove that angle AOB = 2 angle ACB
Construction : Join CO and produce to a point P.
Proof: Triangles AOC and BOC are isosceles triangles
Since AO = OC = OB = radius of circle.
Put
Angle OAC = Angle OCA = a0
Angle OCB = Angle OBC = b0
So that angle ACB = angle OCA + angle OCB = a0 + b0
Using the fact that exterior angle of a triangle is equal to the sum of the two interior opposite angles , we have
Angle POA = angle OAC + angle OCA = 2a0 .................................... (1)
Angle POB = angle OCB + angle OBC = 2b0 .................................... (2)
Add equations 1 and 2
Angle POA + POB = 2(a0 + b0)
i.e angle AOB = 2 angle ACB
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
p
O
P Q
P,Q,R are points on a circle, centre O as shown in figure above. If angle OPQ = 420, find angle PRQ
ASSIGNMENT:
Find the lettered angles in figure below (O is the centre of the circle)
1260
O
s
WEEK 2
SPECIFIC TOPIC: Solving problems on circle theorems.
REFERENCE BOOK : New General Mathematics For S.S.2
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on circle
CONTENT: SOLVING PROBLEMS ON CIRCLE THEOREMS.
Example 1: In figure below, PQ is a diameter of circle PMQN, centre O. If PQM = 630, find QNM.
P
0 M
N 630
Q
In QPM,
PMQ = 900 (angle in semicircle)
Therefore, QPM = 1800 - 900 - 630 (angle sum of triangle)
= 270
Therefore, QNM = 270 (in same segment as QPM)
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
Find the lettered angles in the circle below, where a point O is the centre of the circle
400
a b
320 c
ASSIGNMENT:
Find the lettered angles in the circle below, where a point O is given is the centre of the circle
f
g 560
h 0
SPECIFIC TOPIC: SOLVING PROBLEMS ON CIRCLE THEOREMS.
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on circle theorem
CONTENT: SOLVING PROBLEMS ON CIRCLE THEOREMS.
Theorem : The opposite angles of a cyclic quadrilateral are supplementary.
Or
Angles in opposite segments are supplementary.
Note: the sum of supplementary angles is 1800.
GIVEN: a cyclic quadrilateral ABCD.
To prove : BAD + BCD =1800
A
x
2y B
D 2x
y
C
EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.
A
X1 B
y x2
D X
SPECIFIC TOPIC: SOLVING PROBLEMS ON CIRCLE THEOREMS.
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on circle
CONTENT: SOLVING PROBLEMS ON CIRCLE THEOREMS.
In figure below CE is a diameter of circle ABCDE. If ABC = 1270, find ACE
ADC = 1800 - 1270 (opp. Angles of cycl. Quad.)
= 530
EDC = 900 (angle in semicircle)
EDA = 900 - 530
= 370
ACE = 370 (In same segment as EDA)
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
In figure below P,Q,R,S are points on a circle centre O. QP is produced to X. If XPS = 770 and PSO = 680, find PQO.
P X
Q 770
680 S
0
R
EXAMPLE 1: The diagram shows a circle ABCD with centre O and radius 7cm. The reflex angle AOC = 1900 and angle DAO = 350. Find (i) angle ABC (ii) angle ADC.
B
A
350 7cm C
1900
D
Angle ABC = ? Reflex angle
Angle AOC = 2 x angle ABC (angle at centre = 2 x angle at circumference)
190 = 2 x angle ABC
Angle ABC = 190/2
Angle ABC = 950
(ii) Angle ADC = ?
Angle ADC + Angle ABC = 1800 (Opposite interior angles of a cyclic quadrilateral)
Angle ADC + 950 = 1800
Angle ADC = 180 – 95
= 850
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
In the diagram, angle 2000 is subtended at the centre of the circle. Find the value of x
2000
300 x
ASSIGNMENT:
New General Mathematics for senior secondary school II page28 number 4 a,b,c,d. Exercise 2c number 8
CONTENT: TEST
(i) In the diagram below, O is the centre of the circle. If angle PQR = 1140, calculate angle PQR.
0
1140 R
P
Q
(II) In the diagram below , 0 is the centre of the circle . If angle POQ = 800 and angle PRQ = 5x, find the value of x. R
5x
0
P 800 Q
ASSIGNMENT:
New General Mathematics For senior secondary school book two page 32 exercise 2d number 3 b,d,f
REFERENCE BOOK : New General Mathematics For S.S.2
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on circle
CONTENT: SOLVING PROBLEMS ON CIRCLE THEOREMS.
Example 1: In figure below, PQ is a diameter of circle PMQN, centre O. If PQM = 630, find QNM.
P
0 M
N 630
Q
In QPM,
PMQ = 900 (angle in semicircle)
Therefore, QPM = 1800 - 900 - 630 (angle sum of triangle)
= 270
Therefore, QNM = 270 (in same segment as QPM)
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
Find the lettered angles in the circle below, where a point O is the centre of the circle
400
a b
320 c
ASSIGNMENT:
Find the lettered angles in the circle below, where a point O is given is the centre of the circle
f
g 560
h 0
SPECIFIC TOPIC: SOLVING PROBLEMS ON CIRCLE THEOREMS.
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on circle theorem
CONTENT: SOLVING PROBLEMS ON CIRCLE THEOREMS.
Theorem : The opposite angles of a cyclic quadrilateral are supplementary.
Or
Angles in opposite segments are supplementary.
Note: the sum of supplementary angles is 1800.
GIVEN: a cyclic quadrilateral ABCD.
To prove : BAD + BCD =1800
A
x
2y B
D 2x
y
C
EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.
A
X1 B
y x2
D X
SPECIFIC TOPIC: SOLVING PROBLEMS ON CIRCLE THEOREMS.
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on circle
CONTENT: SOLVING PROBLEMS ON CIRCLE THEOREMS.
In figure below CE is a diameter of circle ABCDE. If ABC = 1270, find ACE
ADC = 1800 - 1270 (opp. Angles of cycl. Quad.)
= 530
EDC = 900 (angle in semicircle)
EDA = 900 - 530
= 370
ACE = 370 (In same segment as EDA)
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
In figure below P,Q,R,S are points on a circle centre O. QP is produced to X. If XPS = 770 and PSO = 680, find PQO.
P X
Q 770
680 S
0
R
EXAMPLE 1: The diagram shows a circle ABCD with centre O and radius 7cm. The reflex angle AOC = 1900 and angle DAO = 350. Find (i) angle ABC (ii) angle ADC.
B
A
350 7cm C
1900
D
Angle ABC = ? Reflex angle
Angle AOC = 2 x angle ABC (angle at centre = 2 x angle at circumference)
190 = 2 x angle ABC
Angle ABC = 190/2
Angle ABC = 950
(ii) Angle ADC = ?
Angle ADC + Angle ABC = 1800 (Opposite interior angles of a cyclic quadrilateral)
Angle ADC + 950 = 1800
Angle ADC = 180 – 95
= 850
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
In the diagram, angle 2000 is subtended at the centre of the circle. Find the value of x
2000
300 x
ASSIGNMENT:
New General Mathematics for senior secondary school II page28 number 4 a,b,c,d. Exercise 2c number 8
CONTENT: TEST
(i) In the diagram below, O is the centre of the circle. If angle PQR = 1140, calculate angle PQR.
0
1140 R
P
Q
(II) In the diagram below , 0 is the centre of the circle . If angle POQ = 800 and angle PRQ = 5x, find the value of x. R
5x
0
P 800 Q
ASSIGNMENT:
New General Mathematics For senior secondary school book two page 32 exercise 2d number 3 b,d,f
WEEK 3
MAIN TOPIC: Review of angles of elevation and depression, surface area, Volumes of solid shapes. (ii) Relationship between sector and cone (iii) Solving problems on surface area of solid shapes and volume.
SPECIFIC TOPIC: Review of angles of elevation and depression
REFERENCE BOOK : New General Mathematics For S.S.2
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on elevation and depression
CONTENT: REVIEW OF ANGLES OF ELEVATION AND DEPRESSION
Example 1: A man is at a point 18m away from the foot of a tree. From that point the angle of elevation of the top of the tree is 290. Calculate the height of the tree
Solution T
h
P 290 Q
18m
From the diagram, TQ is the tree of height h. The point is P. The angle of elevation which is 290 is as shown.
To calculate the height h
Tan 290 = h/18m ( note: Tan = opp/Adj)
h= 18m x tan 290
= 18m x 0.5543
= 9.977m
h= 9.98m
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
The angle of depression from the top of a building of height 30m of a stationary car is 410. Find the distance between the car and the top of the building
ASSIGNMENT:
A man 1.83m tall, stands at a distance of 14.8m away from the base of a tower. He discovers that the angle of elevation of the top of the tower is 630. Calculate the height of the tower.
MAIN TOPIC: Review of angles of elevation and depression, surface area, Volumes of solid shapes. (ii) Relationship between sector and cone (iii) Solving problems on surface area of solid shapes and volume.
SPECIFIC TOPIC: surface area, Volumes of solid shapes.
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on surface area of solid shapes
CONTENT: SURFACE AREA, VOLUMES OF SOLID SHAPES.
Example 1: A cylindrical container has a base radius of 6m and its height is 17m.
Calculate its:
(a) Total surface area
(b) Curved surface area
(c) Volume
Solution
We shall consider both when the two tops are closed and only when only one top is closed
17m
6m
(a) Total surface area when both tops are closed
Total surface area = 2∏(r + h)
= 2 x 22/7 x 6(6 + 17)
= 264/7 (23)
= 264 x 23/7
= 6072/7
= 867.43
= 867.4cm2
(b) Total surface area when one top is closed
Total surface area = ∏r(r + 2h)
= 22/7 x 6(6 + 2 x 17)
= 22/7 x 6/7(6 + 34)
= 132 x (40)/7
= 5280/7
= 754.3cm2
(c) Volume = ∏r2h
= 22/7 x 6 x 6 x17
= 22 x 36 x 17/7
= 13464/7
= 1923.4cm3
EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
A cylindrical tank has a base radius of 6.5m and height 7.2m.Calculate
(a) Total surface area if one end is open
(b) Volume
ASSIGNMENT:
A cylindrical tank has a base radius of 4.7m and height 18m.Calculate
(a) Total surface area if one end is open
(b) Total surface area if both ends are closed
(c) Volume
MAIN TOPIC: Review of angles of elevation and depression, surface area, Volumes of solid shapes. (ii) Relationship between sector and cone (iii) Solving problems on surface area of solid shapes and volume.
SPECIFIC TOPIC: Relationship between sector and cone
OBJECTIVE: At the end of the lesson, the students should be able to:
Determine the relationship between sector and cone
CONTENT: RELATIONSHIP BETWEEN SECTOR AND CONE
Example 1: A cone has a base radius of 6cm and height 9cm. Calculate its
(i) Curved surface area
(ii) Total surface area
Solution
9cm L
6cm
L2 = 62 + 92
= 36 + 81
= 117
L = 117
L = 10.82
L = 10.8cm
(i) Curved surface area = ∏rl
= 22/7 x 6 x 10.8
= 132 x 10.82
7
= 1428.24
7
= 204.03
= 204cm2
(ii) Total surface area = ∏r2 + ∏rl
= ∏r(r + l)
= 22/7 x 6 ( 6 + 10.82)
= 132 x 16.82
7
= 2220.24
7
= 317.18
= 317.2cm2
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
A sector of a circle of radius 7cm subtending an angle 2100 at the centre of the circle, is used to form a cone. Calculate to the nearest whole number the:
(i) Base radius of the cone
(ii) Height of the cone
(iii) Total surface area of the cone
ASSIGNMENT:
A sector of a circle of radius 7cm subtending an angle 2700 at the centre of a circle is used to form a cone.
(i) Find the base radius of the cone.
(ii) Calculate the total surface area of the cone
MAIN TOPIC: Review of angles of elevation and depression, surface area, Volumes of solid shapes. (ii) Relationship between sector and cone (iii) Solving problems on surface area of solid shapes and volume.
SPECIFIC TOPIC: Solving problems on surface area of solid shapes and volume.
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve volume of solid shapes
CONTENT: VOLUME OF SOLID SHAPES
EXAMPLE 1: A cube has a length of 5cm, calculate its volume
Solution 5cm
5cm
Volume of the cube = x3
= 53
= 125cm3
Example 2: A cylindrical container has a base radius of 6m and its height is 17m. Calculate its volume
Solution
6m
17m
Volume = ∏r2h
= 22/7 x 6 x 6 17
= 22 x 36 x 17
7
= 13464
7
= 1923.4cm3
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
A cylindrical tank has a base radius of 6.5m and height 7.2m. Calculate its volume
ASSIGNMENT:
A cylindrical tank has a base radius of 15m and height 18m. Calculate its volume
MAIN TOPIC: Review of angles of elevation and depression, surface area, Volumes of solid shapes. (ii) Relationship between sector and cone (iii) Solving problems on surface area of solid shapes and volume.
SPECIFIC TOPIC: Solving problems on surface area of solid shapes and volume.
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on surface area and volume of plane shapes
CONTENT: AREA AND VOLUME OF SURFACE AREA OF PLANE SHAPES
TEST
(I) A cylindrical container has a base radius of 6m and its height is 17m.
Calculate its:
(a) Total surface area
(b) Curved surface area
(c) Volume
(II) A cone has a base radius of 6cm and height 9cm. Calculate its
(a) Curved surface area
(b) Total surface area
(c) volume
ASSIGNMENT:
A right pyramid consists of a square base of side 13cm and four isosceles triangles whose equal sides are each are each 27cm.
(a) Total surface area
(b) Volume
SPECIFIC TOPIC: Review of angles of elevation and depression
REFERENCE BOOK : New General Mathematics For S.S.2
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on elevation and depression
CONTENT: REVIEW OF ANGLES OF ELEVATION AND DEPRESSION
Example 1: A man is at a point 18m away from the foot of a tree. From that point the angle of elevation of the top of the tree is 290. Calculate the height of the tree
Solution T
h
P 290 Q
18m
From the diagram, TQ is the tree of height h. The point is P. The angle of elevation which is 290 is as shown.
To calculate the height h
Tan 290 = h/18m ( note: Tan = opp/Adj)
h= 18m x tan 290
= 18m x 0.5543
= 9.977m
h= 9.98m
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
The angle of depression from the top of a building of height 30m of a stationary car is 410. Find the distance between the car and the top of the building
ASSIGNMENT:
A man 1.83m tall, stands at a distance of 14.8m away from the base of a tower. He discovers that the angle of elevation of the top of the tower is 630. Calculate the height of the tower.
MAIN TOPIC: Review of angles of elevation and depression, surface area, Volumes of solid shapes. (ii) Relationship between sector and cone (iii) Solving problems on surface area of solid shapes and volume.
SPECIFIC TOPIC: surface area, Volumes of solid shapes.
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on surface area of solid shapes
CONTENT: SURFACE AREA, VOLUMES OF SOLID SHAPES.
Example 1: A cylindrical container has a base radius of 6m and its height is 17m.
Calculate its:
(a) Total surface area
(b) Curved surface area
(c) Volume
Solution
We shall consider both when the two tops are closed and only when only one top is closed
17m
6m
(a) Total surface area when both tops are closed
Total surface area = 2∏(r + h)
= 2 x 22/7 x 6(6 + 17)
= 264/7 (23)
= 264 x 23/7
= 6072/7
= 867.43
= 867.4cm2
(b) Total surface area when one top is closed
Total surface area = ∏r(r + 2h)
= 22/7 x 6(6 + 2 x 17)
= 22/7 x 6/7(6 + 34)
= 132 x (40)/7
= 5280/7
= 754.3cm2
(c) Volume = ∏r2h
= 22/7 x 6 x 6 x17
= 22 x 36 x 17/7
= 13464/7
= 1923.4cm3
EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
A cylindrical tank has a base radius of 6.5m and height 7.2m.Calculate
(a) Total surface area if one end is open
(b) Volume
ASSIGNMENT:
A cylindrical tank has a base radius of 4.7m and height 18m.Calculate
(a) Total surface area if one end is open
(b) Total surface area if both ends are closed
(c) Volume
MAIN TOPIC: Review of angles of elevation and depression, surface area, Volumes of solid shapes. (ii) Relationship between sector and cone (iii) Solving problems on surface area of solid shapes and volume.
SPECIFIC TOPIC: Relationship between sector and cone
OBJECTIVE: At the end of the lesson, the students should be able to:
Determine the relationship between sector and cone
CONTENT: RELATIONSHIP BETWEEN SECTOR AND CONE
Example 1: A cone has a base radius of 6cm and height 9cm. Calculate its
(i) Curved surface area
(ii) Total surface area
Solution
9cm L
6cm
L2 = 62 + 92
= 36 + 81
= 117
L = 117
L = 10.82
L = 10.8cm
(i) Curved surface area = ∏rl
= 22/7 x 6 x 10.8
= 132 x 10.82
7
= 1428.24
7
= 204.03
= 204cm2
(ii) Total surface area = ∏r2 + ∏rl
= ∏r(r + l)
= 22/7 x 6 ( 6 + 10.82)
= 132 x 16.82
7
= 2220.24
7
= 317.18
= 317.2cm2
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
A sector of a circle of radius 7cm subtending an angle 2100 at the centre of the circle, is used to form a cone. Calculate to the nearest whole number the:
(i) Base radius of the cone
(ii) Height of the cone
(iii) Total surface area of the cone
ASSIGNMENT:
A sector of a circle of radius 7cm subtending an angle 2700 at the centre of a circle is used to form a cone.
(i) Find the base radius of the cone.
(ii) Calculate the total surface area of the cone
MAIN TOPIC: Review of angles of elevation and depression, surface area, Volumes of solid shapes. (ii) Relationship between sector and cone (iii) Solving problems on surface area of solid shapes and volume.
SPECIFIC TOPIC: Solving problems on surface area of solid shapes and volume.
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve volume of solid shapes
CONTENT: VOLUME OF SOLID SHAPES
EXAMPLE 1: A cube has a length of 5cm, calculate its volume
Solution 5cm
5cm
Volume of the cube = x3
= 53
= 125cm3
Example 2: A cylindrical container has a base radius of 6m and its height is 17m. Calculate its volume
Solution
6m
17m
Volume = ∏r2h
= 22/7 x 6 x 6 17
= 22 x 36 x 17
7
= 13464
7
= 1923.4cm3
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
A cylindrical tank has a base radius of 6.5m and height 7.2m. Calculate its volume
ASSIGNMENT:
A cylindrical tank has a base radius of 15m and height 18m. Calculate its volume
MAIN TOPIC: Review of angles of elevation and depression, surface area, Volumes of solid shapes. (ii) Relationship between sector and cone (iii) Solving problems on surface area of solid shapes and volume.
SPECIFIC TOPIC: Solving problems on surface area of solid shapes and volume.
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on surface area and volume of plane shapes
CONTENT: AREA AND VOLUME OF SURFACE AREA OF PLANE SHAPES
TEST
(I) A cylindrical container has a base radius of 6m and its height is 17m.
Calculate its:
(a) Total surface area
(b) Curved surface area
(c) Volume
(II) A cone has a base radius of 6cm and height 9cm. Calculate its
(a) Curved surface area
(b) Total surface area
(c) volume
ASSIGNMENT:
A right pyramid consists of a square base of side 13cm and four isosceles triangles whose equal sides are each are each 27cm.
(a) Total surface area
(b) Volume
WEEK 4
MAIN TOPIC: AREA OF PLANE FIGURES
SPECIFIC TOPIC: AREA OF PARALLELOGRAM
REFERENCE BOOK : New General Mathematics For S.S.2
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on area of parallelogram
CONTENT: AREA OF PARALLELOGRAM
The area of parallelogram can be got in two ways, as a result of the dimensions given
a h
θ
b
fig: 1
area = b x h i.e base x perpendicular height
or
area = a x b x hsinθ i.e product of two sides and the sine of the included angle.
Example 1: Calculate the area of each of the following figures below
(a) 11cm 7cm (b) 1060
21cm 18cm
Solution
(a) Base = 21cm
Height = 7cm
Area = base x height
Area = 21 x 7
Area = 147cm2
(b) Base = 18cm
Other side 7cm
Included angle = 1060
Area = ABsinθ
Area = 18 x 7 x sin 1060
= 18 x 7 x 0.9613
= 121.1cm2
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
Calculate the area of each of the following figures below
(a) 22cm 5.5cm (b) 1220
23cm 25cm
ASSIGNMENT:
Calculate the area of each of the following figures below
(a) 10cm 8cm (b) 1090
18cm 17cm
SPECIFIC TOPIC: AREA OF TRAPEZIUM
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on area of trapezium
CONTENT: AREA OF TRAPEZIUM
The area of a trapezium is half the product of the sum of the parallel sides and the perpendicular distance between them.
a
d h c
b
Area = ½ (a + b)h
The area can also be found if an included angle is given. The included angle and the hypotenuse side to it are used in expressing the height
a
d h
b θ
Area = ½ (a + b) x c sinθ
Example 1: ABCD is a trapezium in which AB//DC. /AB = 12cm, /BC/ = 9cm, /CD/ = 6cm and ABC = 460. Calculate the area of ABCD
D 6cm C
460
A 12cm B
Solution
Area = ½ (a + b) c sinθ
= ½ (6 + 12) 9 sin 460
= ½ (18) 9 x 0.7193
= 9 x 9 x 0.7193
= 58.3cm2
EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
Example 1: ABCD is a trapezium in which AB//DC. /AB/ = 16cm, /BC/ = 6.2cm, /CD/ = 8cm and ABC = 560. Calculate the area of ABCD
D 8cm C
560
A 16cm B
ASSIGNMENT:
Example 1: ABCD is a trapezium in which AB//DC. /AB/ = 22cm, /BC/ = 19cm, /CD/ = 9cm and ABC = 660. Calculate the area of ABCD
D 9cm C
660
A 22cm B
SPECIFIC TOPIC: AREA OF A KITE
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on kite
CONTENT: AREA OF A KITE
A kite is four- sided polygon. It looks like two triangles joined together at their bases. The diagonals bisects at right-angle B
A y b
A h1 h2 D
A y b
C
Area = Area of triangle ABC + Area of triangle BCD
= ½BC x h1 + ½ BC x h2
= ½ (y + y) x h1 + ½ (y + y) + h2
= ½ (2y)h1 + ½ (2y)h2
= yh1 + yh2
= (h1 + h2)y
Example 1; Find the area of the figure below
7cm 5cm
4.8cm 5cm 12cm
13cm
Solution
Area = (h1 + h2)y
= (4.8 + 12)5
= 16.8 x 5
= 84cm2
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
Find the area of the figure below
6cm 4cm
2.8cm 7cm 11cm
10cm
ASSIGNMENT:
Find the area of the figure below
9cm 7cm
7.9cm 8cm 14cm
16cm
SPECIFIC TOPIC: AREA OF CYCLE
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on area of a cycle
CONTENT: AREA OF A CYCLE
A cycle is also a plane shape bounded by a curved surface
Circumfrence
radius
area = ∏r2
example 1: Calculate the area of a circle radius 10.5cm
Solution
Area = ∏r2
= 22/7 x 10.5 x 10.5
= 346.5cm2
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
Calculate the area of the circles with radius
(a) 13.5cm
(b) 11.2cm
(c) 22.4cm
(d) 14.2cm
ASSIGNMENT:
Calculate the area of the circles with radius
(a) 23.1cm
(b) 22.7cm
(c) 20.5cm
(d) 12.2cm
CONTENT: AREA OF PLANE SHAPES
TEST
Find the area of the following figures
(a) ABCD is a trapezium in which AB//DC. /AB = 15cm, /BC/ = cm, /CD/ = 9cm and ABC = 260. Calculate the area of ABCD
D 9cm C
260
A 15cm B
(b) Find the area of the figure below
8cm 5cm
2.3cm 5cm 12cm
16cm
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
ASSIGNMENT:
Calculate the area of each of the following figures below
(a) 16cm 8cm (b) 1260
15cm 16cm
SPECIFIC TOPIC: AREA OF PARALLELOGRAM
REFERENCE BOOK : New General Mathematics For S.S.2
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on area of parallelogram
CONTENT: AREA OF PARALLELOGRAM
The area of parallelogram can be got in two ways, as a result of the dimensions given
a h
θ
b
fig: 1
area = b x h i.e base x perpendicular height
or
area = a x b x hsinθ i.e product of two sides and the sine of the included angle.
Example 1: Calculate the area of each of the following figures below
(a) 11cm 7cm (b) 1060
21cm 18cm
Solution
(a) Base = 21cm
Height = 7cm
Area = base x height
Area = 21 x 7
Area = 147cm2
(b) Base = 18cm
Other side 7cm
Included angle = 1060
Area = ABsinθ
Area = 18 x 7 x sin 1060
= 18 x 7 x 0.9613
= 121.1cm2
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
Calculate the area of each of the following figures below
(a) 22cm 5.5cm (b) 1220
23cm 25cm
ASSIGNMENT:
Calculate the area of each of the following figures below
(a) 10cm 8cm (b) 1090
18cm 17cm
SPECIFIC TOPIC: AREA OF TRAPEZIUM
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on area of trapezium
CONTENT: AREA OF TRAPEZIUM
The area of a trapezium is half the product of the sum of the parallel sides and the perpendicular distance between them.
a
d h c
b
Area = ½ (a + b)h
The area can also be found if an included angle is given. The included angle and the hypotenuse side to it are used in expressing the height
a
d h
b θ
Area = ½ (a + b) x c sinθ
Example 1: ABCD is a trapezium in which AB//DC. /AB = 12cm, /BC/ = 9cm, /CD/ = 6cm and ABC = 460. Calculate the area of ABCD
D 6cm C
460
A 12cm B
Solution
Area = ½ (a + b) c sinθ
= ½ (6 + 12) 9 sin 460
= ½ (18) 9 x 0.7193
= 9 x 9 x 0.7193
= 58.3cm2
EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
Example 1: ABCD is a trapezium in which AB//DC. /AB/ = 16cm, /BC/ = 6.2cm, /CD/ = 8cm and ABC = 560. Calculate the area of ABCD
D 8cm C
560
A 16cm B
ASSIGNMENT:
Example 1: ABCD is a trapezium in which AB//DC. /AB/ = 22cm, /BC/ = 19cm, /CD/ = 9cm and ABC = 660. Calculate the area of ABCD
D 9cm C
660
A 22cm B
SPECIFIC TOPIC: AREA OF A KITE
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on kite
CONTENT: AREA OF A KITE
A kite is four- sided polygon. It looks like two triangles joined together at their bases. The diagonals bisects at right-angle B
A y b
A h1 h2 D
A y b
C
Area = Area of triangle ABC + Area of triangle BCD
= ½BC x h1 + ½ BC x h2
= ½ (y + y) x h1 + ½ (y + y) + h2
= ½ (2y)h1 + ½ (2y)h2
= yh1 + yh2
= (h1 + h2)y
Example 1; Find the area of the figure below
7cm 5cm
4.8cm 5cm 12cm
13cm
Solution
Area = (h1 + h2)y
= (4.8 + 12)5
= 16.8 x 5
= 84cm2
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
Find the area of the figure below
6cm 4cm
2.8cm 7cm 11cm
10cm
ASSIGNMENT:
Find the area of the figure below
9cm 7cm
7.9cm 8cm 14cm
16cm
SPECIFIC TOPIC: AREA OF CYCLE
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on area of a cycle
CONTENT: AREA OF A CYCLE
A cycle is also a plane shape bounded by a curved surface
Circumfrence
radius
area = ∏r2
example 1: Calculate the area of a circle radius 10.5cm
Solution
Area = ∏r2
= 22/7 x 10.5 x 10.5
= 346.5cm2
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
Calculate the area of the circles with radius
(a) 13.5cm
(b) 11.2cm
(c) 22.4cm
(d) 14.2cm
ASSIGNMENT:
Calculate the area of the circles with radius
(a) 23.1cm
(b) 22.7cm
(c) 20.5cm
(d) 12.2cm
CONTENT: AREA OF PLANE SHAPES
TEST
Find the area of the following figures
(a) ABCD is a trapezium in which AB//DC. /AB = 15cm, /BC/ = cm, /CD/ = 9cm and ABC = 260. Calculate the area of ABCD
D 9cm C
260
A 15cm B
(b) Find the area of the figure below
8cm 5cm
2.3cm 5cm 12cm
16cm
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
ASSIGNMENT:
Calculate the area of each of the following figures below
(a) 16cm 8cm (b) 1260
15cm 16cm
WEEK 5
MAIN TOPIC: Probability
SPECIFIC TOPIC: EXPERIMENTAL PROBABILITY
REFFERENCE BOOK: New General Mathematics for senior secondary school, book 2
PERFORMANCE OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on experimental probability
CONTENT: EXPERIMENTAL PROBABILITY
Example 1: A die is rolled 200 rimes. The outcomes obtained are shown in table below
number 1 2 3 4 5 6
No. Of times 25 30 45 28 40 32
Find the experimental probability of obtaining
(a) 2
(b) 5
(c) 6
Solution
(a) Probability of obtaining a 2 is
30/200 = 3/20 = 0.15
(b) Probability of obtaining a 5 is
40/200 = 1/5 = 0.2
(c) Probability of obtaining a 6 is
32/200 = 4/25 = 0.16
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems
A die is rolled 200 rimes. The outcomes obtained are shown in table below
number 1 2 3 4 5 6
No. Of times 22 34 55 18 46 22
Find the experimental probability of obtaining
(a) 2
(b) 5
(c) 6
ASSIGNMENT:
Toss a coin 200 times. Use tally to marks to record whether a head (H) or tail (T) showns up. What is the experimental probability of obtaining
(a) H
(b) T
SPECIFIC TOPIC: THEORETICAL PROBABILITY
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve some problems on theoretical probability
CONTENT: THEORETICAL PROBABILITY
Since experimental probability uses numerical records of past events to predict the future, its predictions are not absolute accurate.
However, the probability of throwing a 5 on a fair six-sided die is 1/6, since any one of the six faces is equally likely. This is an example of theoretical probability.
Example 1: Jessie throws a fair six-sided die. What is the probability that she throws
(a) a 9
(b) a 4
(c) a number greater than 2
Solution
(a) Since the faces of a six-sides die are numbers 1 to 6, it is impossible to throw a 9.
Probability = 0
(b) There is one chance in six of throwing a 4
Probability = number of required outcomes/number of possible outcomes
1/6
(c) There are four possibilities greater than 2 (3,4,5,6)
Probability = numbers of required outcomes/number of possible outcomes
4/6 = 2/3
(d) There are three possible even numbers (2,4,6)
Probability = number of required outcomes/number of possible outcomes
3/6 = ½
EVALUATION: The lesson is evaluated as the students are asked to solve the following problem.
A school contains 357 boys and 323 girls. If a student is chosen at random, what is the probability that a girl is chosen ?
ASSIGNMENT:
A state Lottery sells 1 ½ millions tickets of which 300 are prize winners. What is the probability that a person chosen at random from a large crowd is less than 150cm tall?
CONTENT: THEORETICAL PROBABILITY
FOR EVENT NOT TO HAPPEN:
If p is the probability of an event happening then p lies in the range 0≤ 1. The probability of an event not happening is p’ where p’ = 1 – p.
EXAMPLE 1: A letter is chosen at random from the alphabet. Find the probability that it is
(a) F
(b) F or T
Solution
In every case
ε= (A,B,C......Z)
(a) Let a = (F)
Then p(a) = n(a)/n(ε) = 1/26
The probability that F is chosen is 1/26
(b) Let b = (F, T)
Then p(b) = n(b)/n(ε)
= 2/26 = 1/13
There is a 1/13 probability that F or T is chosen.
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems,
A bag contains two black balls, three green balls and four red balls. A ball is picked from the bag at random. What is the probability that it is
(a) Black
(b) Green
(c) Not black
(d) yellow
ASSIGNMENT:
A fair six-sided die nis thrown. Find the probability of getting
(a) a 3
(b) a 4
(c) A number divisible by 3
(d) A number less than 5
SPECIFIC TOPIC: EXPERIMENTAL PROBABILITY
REFFERENCE BOOK: New General Mathematics for senior secondary school, book 2
PERFORMANCE OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on experimental probability
CONTENT: EXPERIMENTAL PROBABILITY
Example 1: A die is rolled 200 rimes. The outcomes obtained are shown in table below
number 1 2 3 4 5 6
No. Of times 25 30 45 28 40 32
Find the experimental probability of obtaining
(a) 2
(b) 5
(c) 6
Solution
(a) Probability of obtaining a 2 is
30/200 = 3/20 = 0.15
(b) Probability of obtaining a 5 is
40/200 = 1/5 = 0.2
(c) Probability of obtaining a 6 is
32/200 = 4/25 = 0.16
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems
A die is rolled 200 rimes. The outcomes obtained are shown in table below
number 1 2 3 4 5 6
No. Of times 22 34 55 18 46 22
Find the experimental probability of obtaining
(a) 2
(b) 5
(c) 6
ASSIGNMENT:
Toss a coin 200 times. Use tally to marks to record whether a head (H) or tail (T) showns up. What is the experimental probability of obtaining
(a) H
(b) T
SPECIFIC TOPIC: THEORETICAL PROBABILITY
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve some problems on theoretical probability
CONTENT: THEORETICAL PROBABILITY
Since experimental probability uses numerical records of past events to predict the future, its predictions are not absolute accurate.
However, the probability of throwing a 5 on a fair six-sided die is 1/6, since any one of the six faces is equally likely. This is an example of theoretical probability.
Example 1: Jessie throws a fair six-sided die. What is the probability that she throws
(a) a 9
(b) a 4
(c) a number greater than 2
Solution
(a) Since the faces of a six-sides die are numbers 1 to 6, it is impossible to throw a 9.
Probability = 0
(b) There is one chance in six of throwing a 4
Probability = number of required outcomes/number of possible outcomes
1/6
(c) There are four possibilities greater than 2 (3,4,5,6)
Probability = numbers of required outcomes/number of possible outcomes
4/6 = 2/3
(d) There are three possible even numbers (2,4,6)
Probability = number of required outcomes/number of possible outcomes
3/6 = ½
EVALUATION: The lesson is evaluated as the students are asked to solve the following problem.
A school contains 357 boys and 323 girls. If a student is chosen at random, what is the probability that a girl is chosen ?
ASSIGNMENT:
A state Lottery sells 1 ½ millions tickets of which 300 are prize winners. What is the probability that a person chosen at random from a large crowd is less than 150cm tall?
CONTENT: THEORETICAL PROBABILITY
FOR EVENT NOT TO HAPPEN:
If p is the probability of an event happening then p lies in the range 0≤ 1. The probability of an event not happening is p’ where p’ = 1 – p.
EXAMPLE 1: A letter is chosen at random from the alphabet. Find the probability that it is
(a) F
(b) F or T
Solution
In every case
ε= (A,B,C......Z)
(a) Let a = (F)
Then p(a) = n(a)/n(ε) = 1/26
The probability that F is chosen is 1/26
(b) Let b = (F, T)
Then p(b) = n(b)/n(ε)
= 2/26 = 1/13
There is a 1/13 probability that F or T is chosen.
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems,
A bag contains two black balls, three green balls and four red balls. A ball is picked from the bag at random. What is the probability that it is
(a) Black
(b) Green
(c) Not black
(d) yellow
ASSIGNMENT:
A fair six-sided die nis thrown. Find the probability of getting
(a) a 3
(b) a 4
(c) A number divisible by 3
(d) A number less than 5
WEEK 6
CONTENT: REVISION
AREA OF PARALLELOGRAM
The area of parallelogram can be gotten in two ways, as a result of the dimensions given
a h
θ
b
fig: 1
area = b x h i.e base x perpendicular height
or
area = a x b x hsinθ i.e product of two sides and the sine of the included angle.
Example 1: Calculate the area of each of the following figures below
(a) 11cm 7cm (b) 1060
21cm 18cm
Solution
(a) Base = 21cm
Height = 7cm
Area = base x height
Area = 21 x 7
Area = 147cm2
(b) Base = 18cm
Other side 7cm
Included angle = 1060
Area = ABsinθ
Area = 18 x 7 x sin 1060
= 18 x 7 x 0.9613
= 121.1cm2
EXPERIMENTAL PROBABILITY
Example 1: A die is rolled 200 rimes. The outcomes obtained are shown in table below
number 1 2 3 4 5 6
No. Of times 25 30 45 28 40 32
Find the experimental probability of obtaining
(a) 2
(b) 5
(c) 6
Solution
(a) Probability of obtaining a 2 is
30/200 = 3/20 = 0.15
(b) Probability of obtaining a 5 is
40/200 = 1/5 = 0.2
(c) Probability of obtaining a 6 is
32/200 = 4/25 = 0.16
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
A die is rolled 200 rimes. The outcomes obtained are shown in table below
number 1 2 3 4 5 6
No. Of times 22 34 55 18 46 22
Find the experimental probability of obtaining
(a) 2
(b) 5
(c) 6
ASSIGNMENT:
Calculate the area of each of the following figures below
(a) 22cm 5.5cm (b) 1220
23cm 25cm
AREA OF PARALLELOGRAM
The area of parallelogram can be gotten in two ways, as a result of the dimensions given
a h
θ
b
fig: 1
area = b x h i.e base x perpendicular height
or
area = a x b x hsinθ i.e product of two sides and the sine of the included angle.
Example 1: Calculate the area of each of the following figures below
(a) 11cm 7cm (b) 1060
21cm 18cm
Solution
(a) Base = 21cm
Height = 7cm
Area = base x height
Area = 21 x 7
Area = 147cm2
(b) Base = 18cm
Other side 7cm
Included angle = 1060
Area = ABsinθ
Area = 18 x 7 x sin 1060
= 18 x 7 x 0.9613
= 121.1cm2
EXPERIMENTAL PROBABILITY
Example 1: A die is rolled 200 rimes. The outcomes obtained are shown in table below
number 1 2 3 4 5 6
No. Of times 25 30 45 28 40 32
Find the experimental probability of obtaining
(a) 2
(b) 5
(c) 6
Solution
(a) Probability of obtaining a 2 is
30/200 = 3/20 = 0.15
(b) Probability of obtaining a 5 is
40/200 = 1/5 = 0.2
(c) Probability of obtaining a 6 is
32/200 = 4/25 = 0.16
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
A die is rolled 200 rimes. The outcomes obtained are shown in table below
number 1 2 3 4 5 6
No. Of times 22 34 55 18 46 22
Find the experimental probability of obtaining
(a) 2
(b) 5
(c) 6
ASSIGNMENT:
Calculate the area of each of the following figures below
(a) 22cm 5.5cm (b) 1220
23cm 25cm
WEEK 7
MAIN TOPIC: MATRICES
SPECIFIC TOPIC: 2 x 2 MATRICES
REFERENCE BOOK : New General Mathematics For S.S.2
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on 2 x 2 matrices
CONTENT: 2 x 2 MATRICS
ADDITION OF 2x2 MATRICES
Example 1: Add the following matrices
(a) 8 2 + 6 6
7 9 0 7
(b) 65 42 + 15 21
111 154 19 28
Solution
To add matrices together, they must be of the same order. Corresponding elements are then added together
(a) 8 2 + 6 6 = 14 8
7 9 0 7 7 16
(b) 65 42 + 15 21 = 80 63
111 154 19 28 130 182
SUBTRACTION OF MATRICES
Example 1: Subtract this matrices
65 42 - 15 21 = 50 21
111 154 19 28 92 126
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
(1) -6 14 + -3 -4
9 -12 10 0
(2) 5 7 - 3 -2
8 5 6 6
ASSIGNMENT:
Simplify the following matrices
(i) 62 32 + 21 20
33 44 30 32
(ii) 26 22 - 12 10
32 4 26 6
SPECIFIC TOPIC: 3 x 3 MATRICES
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on 3 x 3 matrices
CONTENT: ADDITION AND SUBTRACTION OF 3 x 3 MATRICES
Example 1: A music store sells music in three formats: records, cassettes and CDs. It also sells the following types of music: classical, rock/pop and jazz. It records the number of each type and format sold on two consecutive Saturdays. The results of this are presented in the matrices below:
28 14 46 + 24 10 51
56 91 15 35 82 24
17 5 7 15 8 6
Solution
28 14 46 + 24 10 51 52 24 97
56 91 15 35 82 24 = 91 173 39
17 5 7 15 8 6 32 13 13
EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
Simplify the following matrices
82 41 64 + 64 15 56
16 19 51 53 28 42
17 25 17 51 18 16
ASSIGNMENT:
Simplify the following matrices
95 14 64 - 64 11 56
55 44 51 63 22 42
19 53 17 31 17 16
SPECIFIC TOPIC: MULTIPLICATION OF MATRICES
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on multiplication of 2 x 2 matrices
PREVIOUS KNOWLEDGE: The Students have been taught addition and subtraction of 2 x 2
CONTENT: MULTIPLICATION OF MATRICES
Multiplying matrices together involves multiplying the elements in the rows of the first matrix by the elements in the columns of the second matrix
Example 1:
Simplify 6 3 2 4
2 4 x 1 3
0 1
Solution
(6 x 2) + (3 x 1) or (6 x 4) + ( 3 x 3) 15 33
(2 x 2) + (4 x 1) (2 x 4) + (4 x 3) = 8 20
(0 x 2) + (1 x 1) (0 x 4) + (1 x 3) 1 3
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
Simplify
4 3 2 4 2 8
1 2 7 6 1 7
7 2 0
ASSIGNMENT:
Simplify
2 -2 2 3 5 3
3 2 3 6 3 5
5 1 0
SPECIFIC TOPIC: MULTIPLYING MATRICES BY A SCALAR QUANTITY
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on scalar quantity matrices
CONTENT: MULTIPLYING MATRICES BY A SCALAR QUANTITY
Simplify the following matrices with the given scalar
2 8 6
4 2
Solution
2 8 6 = 16 12
4 2 8 4
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
Simplify the following
(i) 4 5 7
6 12 08 2
8 4 6
(ii) ½ 42 16 12
12 22 24
ASSIGNMENT:
Simplify the following matrices
(a) 3.5 6 12
4 2
(b) 3/7 28 7
7 14
SPECIFIC TOPIC: ADDITION SUBTRACTION AND MULTIPLICATION OF MATRICES
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on matrices
CONTENT: TEST
Simplify the following matrices
(i) 62 32 + 21 20
33 44 30 32
(ii) 26 22 - 12 10
32 4 26 6
(iii) 26 22 - 12 10
32 4 26 6
2. Simplify (i)
4 3 2 4 2 8
1 2 7 6 1 7
2 3 4 7 2 0
(i) ½ 42 16 12
12 22 24
ASSIGNMENT:
Simplify the following matrices
(iii) 12 32 + 12 10
43 44 20 62
(iv) 26 22 - 11 10
22 4 16 11
(iv) 16 23 - 11 20
21 14 26 1 6
2. Simplify (i)
2 2 2 4 5 8
4 2 5 4 2 7
3 3 0 7 2 2
(ii) ½ 22 9 44
4 10 14
SPECIFIC TOPIC: 2 x 2 MATRICES
REFERENCE BOOK : New General Mathematics For S.S.2
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on 2 x 2 matrices
CONTENT: 2 x 2 MATRICS
ADDITION OF 2x2 MATRICES
Example 1: Add the following matrices
(a) 8 2 + 6 6
7 9 0 7
(b) 65 42 + 15 21
111 154 19 28
Solution
To add matrices together, they must be of the same order. Corresponding elements are then added together
(a) 8 2 + 6 6 = 14 8
7 9 0 7 7 16
(b) 65 42 + 15 21 = 80 63
111 154 19 28 130 182
SUBTRACTION OF MATRICES
Example 1: Subtract this matrices
65 42 - 15 21 = 50 21
111 154 19 28 92 126
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
(1) -6 14 + -3 -4
9 -12 10 0
(2) 5 7 - 3 -2
8 5 6 6
ASSIGNMENT:
Simplify the following matrices
(i) 62 32 + 21 20
33 44 30 32
(ii) 26 22 - 12 10
32 4 26 6
SPECIFIC TOPIC: 3 x 3 MATRICES
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on 3 x 3 matrices
CONTENT: ADDITION AND SUBTRACTION OF 3 x 3 MATRICES
Example 1: A music store sells music in three formats: records, cassettes and CDs. It also sells the following types of music: classical, rock/pop and jazz. It records the number of each type and format sold on two consecutive Saturdays. The results of this are presented in the matrices below:
28 14 46 + 24 10 51
56 91 15 35 82 24
17 5 7 15 8 6
Solution
28 14 46 + 24 10 51 52 24 97
56 91 15 35 82 24 = 91 173 39
17 5 7 15 8 6 32 13 13
EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
Simplify the following matrices
82 41 64 + 64 15 56
16 19 51 53 28 42
17 25 17 51 18 16
ASSIGNMENT:
Simplify the following matrices
95 14 64 - 64 11 56
55 44 51 63 22 42
19 53 17 31 17 16
SPECIFIC TOPIC: MULTIPLICATION OF MATRICES
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on multiplication of 2 x 2 matrices
PREVIOUS KNOWLEDGE: The Students have been taught addition and subtraction of 2 x 2
CONTENT: MULTIPLICATION OF MATRICES
Multiplying matrices together involves multiplying the elements in the rows of the first matrix by the elements in the columns of the second matrix
Example 1:
Simplify 6 3 2 4
2 4 x 1 3
0 1
Solution
(6 x 2) + (3 x 1) or (6 x 4) + ( 3 x 3) 15 33
(2 x 2) + (4 x 1) (2 x 4) + (4 x 3) = 8 20
(0 x 2) + (1 x 1) (0 x 4) + (1 x 3) 1 3
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
Simplify
4 3 2 4 2 8
1 2 7 6 1 7
7 2 0
ASSIGNMENT:
Simplify
2 -2 2 3 5 3
3 2 3 6 3 5
5 1 0
SPECIFIC TOPIC: MULTIPLYING MATRICES BY A SCALAR QUANTITY
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on scalar quantity matrices
CONTENT: MULTIPLYING MATRICES BY A SCALAR QUANTITY
Simplify the following matrices with the given scalar
2 8 6
4 2
Solution
2 8 6 = 16 12
4 2 8 4
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
Simplify the following
(i) 4 5 7
6 12 08 2
8 4 6
(ii) ½ 42 16 12
12 22 24
ASSIGNMENT:
Simplify the following matrices
(a) 3.5 6 12
4 2
(b) 3/7 28 7
7 14
SPECIFIC TOPIC: ADDITION SUBTRACTION AND MULTIPLICATION OF MATRICES
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on matrices
CONTENT: TEST
Simplify the following matrices
(i) 62 32 + 21 20
33 44 30 32
(ii) 26 22 - 12 10
32 4 26 6
(iii) 26 22 - 12 10
32 4 26 6
2. Simplify (i)
4 3 2 4 2 8
1 2 7 6 1 7
2 3 4 7 2 0
(i) ½ 42 16 12
12 22 24
ASSIGNMENT:
Simplify the following matrices
(iii) 12 32 + 12 10
43 44 20 62
(iv) 26 22 - 11 10
22 4 16 11
(iv) 16 23 - 11 20
21 14 26 1 6
2. Simplify (i)
2 2 2 4 5 8
4 2 5 4 2 7
3 3 0 7 2 2
(ii) ½ 22 9 44
4 10 14
WEEK 8
SPECIFIC TOPIC: INVERSE MATRICES
REFERENCE BOOK : New General Mathematics For S.S.2
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve inverse of matrices
CONTENT:
Inverse of a Matrix
Matrix Inverse
For a square matrix A, the inverse is written A-1. When A is multiplied by A-1 the result is the identity matrix I. Non-square matrices do not have inverses.
Note: Not all square matrices have inverses. A square matrix which has an inverse is called invertible or nonsingular, and a square matrix without an inverse is called noninvertible or singular.
AA-1 = A-1A = I
Example: For matrix , its inverse is since
AA-1 =
and A-1A = .
Here are three ways to find the inverse of a matrix:
1. Shortcut for 2x2 matrices
For , the inverse can be found using this formula:
Example:
Use Gauss-Jordan elimination to transform [ A | I ] into [ I | A-1 ].
Example: The following steps result in .
so we see that .
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
2 1 0
A = 1 -3 4 Find A-1
1 0 3
CONTENT:
Inverse of a Matrix
Matrix Inverse
2. Augmented matrix method
Use Gauss-Jordan elimination to transform [ A | I ] into [ I | A-1 ].
Example: The following steps result in .
so we see that .
3. Adjoint method
A-1 = (adjoint of A) or A-1 = (cofactor matrix of A)T
Example: The following steps result in A-1 for .
The cofactor matrix for A is , so the adjoint is . Since det A = 22, we get
.
EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
Find the inverse of the matrices using Gauss John method
2 3 1 -2 5 1
1. 3 1 0 2. 5 0 2
4 2 1 1 2 1
ASSIGNMENT: find the inverse of the matrix below
3 7 4
4 2 1
5 6 2
SPECIFIC TOPIC: TRANSPOSE OF MATRICES
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on transpose of matrices
CONTENT:
Transpose of a Matrix
A matrix which is formed by turning all the rows of a given matrix into columns and vice-versa. The transpose of matrix A is written AT.
Example 2 : Find the transpose of matrix B = 2 3 6
5 2 1
2 1 4
SOLUTION
BT = 2 5 2
3 2 1
6 1 4
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
Find the transpose of the following matrices
the matrices using Gauss John method
2 3 1 -2 5 1
1. 3 1 0 2. 5 0 2
4 2 1 1 2 1
SPECIFIC TOPIC: DETERMINANT OF MATRICES
OBJECTIVE: At the end of the lesson, the students should be able to:
Find the determinant of matrices
CONTENT: DETERMINANT OF MATRICES
Example 1: Find the determinant of the matrices below
a. 5 2
A = 2 2
b.
B= 6 3
4 4
Solution
a. A = 5 x 2 – 2 x 2 = 10 – 4 = 6
b. B = 6 x 4 – 3 x 4 = 24 – 12 = 12
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
Find the determinant of the following matrices
1. 2 9
1 6
2.
3 7
5 6
SPECIFIC TOPIC: INVERSE, TRANSPOSE AND DETERMINANT
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on matrices inverse, transpose and determinant
CONTENT: TEST
I. Find the inverse of the matrices using Gauss John method
2 3 1 -2 5 1
1. 3 1 0 2. 5 0 2
4 2 1 1 2 1
II. Find the determinant of the following matrices
1. 2 9
1 6
2.
3 7
5 6
III. Find the transpose of matrix
B = 2 3 6
5 2 1
2 1 4
ASSIGNMENT:
New General mathematics book 2 page 204, Exercise 4b number 1 - 8
REFERENCE BOOK : New General Mathematics For S.S.2
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve inverse of matrices
CONTENT:
Inverse of a Matrix
Matrix Inverse
For a square matrix A, the inverse is written A-1. When A is multiplied by A-1 the result is the identity matrix I. Non-square matrices do not have inverses.
Note: Not all square matrices have inverses. A square matrix which has an inverse is called invertible or nonsingular, and a square matrix without an inverse is called noninvertible or singular.
AA-1 = A-1A = I
Example: For matrix , its inverse is since
AA-1 =
and A-1A = .
Here are three ways to find the inverse of a matrix:
1. Shortcut for 2x2 matrices
For , the inverse can be found using this formula:
Example:
Use Gauss-Jordan elimination to transform [ A | I ] into [ I | A-1 ].
Example: The following steps result in .
so we see that .
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
2 1 0
A = 1 -3 4 Find A-1
1 0 3
CONTENT:
Inverse of a Matrix
Matrix Inverse
2. Augmented matrix method
Use Gauss-Jordan elimination to transform [ A | I ] into [ I | A-1 ].
Example: The following steps result in .
so we see that .
3. Adjoint method
A-1 = (adjoint of A) or A-1 = (cofactor matrix of A)T
Example: The following steps result in A-1 for .
The cofactor matrix for A is , so the adjoint is . Since det A = 22, we get
.
EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
Find the inverse of the matrices using Gauss John method
2 3 1 -2 5 1
1. 3 1 0 2. 5 0 2
4 2 1 1 2 1
ASSIGNMENT: find the inverse of the matrix below
3 7 4
4 2 1
5 6 2
SPECIFIC TOPIC: TRANSPOSE OF MATRICES
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on transpose of matrices
CONTENT:
Transpose of a Matrix
A matrix which is formed by turning all the rows of a given matrix into columns and vice-versa. The transpose of matrix A is written AT.
Example 2 : Find the transpose of matrix B = 2 3 6
5 2 1
2 1 4
SOLUTION
BT = 2 5 2
3 2 1
6 1 4
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
Find the transpose of the following matrices
the matrices using Gauss John method
2 3 1 -2 5 1
1. 3 1 0 2. 5 0 2
4 2 1 1 2 1
SPECIFIC TOPIC: DETERMINANT OF MATRICES
OBJECTIVE: At the end of the lesson, the students should be able to:
Find the determinant of matrices
CONTENT: DETERMINANT OF MATRICES
Example 1: Find the determinant of the matrices below
a. 5 2
A = 2 2
b.
B= 6 3
4 4
Solution
a. A = 5 x 2 – 2 x 2 = 10 – 4 = 6
b. B = 6 x 4 – 3 x 4 = 24 – 12 = 12
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
Find the determinant of the following matrices
1. 2 9
1 6
2.
3 7
5 6
SPECIFIC TOPIC: INVERSE, TRANSPOSE AND DETERMINANT
OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on matrices inverse, transpose and determinant
CONTENT: TEST
I. Find the inverse of the matrices using Gauss John method
2 3 1 -2 5 1
1. 3 1 0 2. 5 0 2
4 2 1 1 2 1
II. Find the determinant of the following matrices
1. 2 9
1 6
2.
3 7
5 6
III. Find the transpose of matrix
B = 2 3 6
5 2 1
2 1 4
ASSIGNMENT:
New General mathematics book 2 page 204, Exercise 4b number 1 - 8
WEEK 9
SPECIFIC TOPIC: INDICIAL EQUATION
REFFERENCE BOOK: Further Mathematics for senior secondary school, book 1
PERFORMANCE OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on indicial equation
CONTENT: INDICIAL EQUATION
Example 1: Solve (1/2)X = 8
Solution
(1/2)X = 8
2-X = 23
-x = 3
X =-3
Example 2: Solve 3(3)x = 27
Solution
3(3)x = 27
3x+1 = 33
X + 1 = 3
X = 3 – 1
X = 2
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
Solve the following questions
(i) 10x = 1/10,000
(ii) 2x = 0.125
(iii) 25(5x) = 625
(iv) 3-x = 243
ASSIGNMENT: solve the following
(i) 10x = 1/0.001
(ii) 3x = 1/81
MAIN TOPIC: INDICIAL EQUATION
SPECIFIC TOPIC: EXPONENTIAL EQUATION
PERFORMANCE OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on exponential equation.
CONTENT: EXPONENTIAL EQUATION
Example 1: Solve 52x – 26(5x) + 25 = 0
Solution
52x – 26(5x) + 25 =0
Therefore, (5x)2 – 26(5x) + 25 = 0
Let p = 5x then
P2 – 26P + 25 =0
(P – 1) (P – 25) =0
P = 1 or p = 25
5x = 1 = 50
X = 0
Or 5x = 25 = 52
X = 2
Hence x =0 or 2
Example 2: solve 52x+1 – 26(5X) + 5 =0
Solution
52x +1 -26(5x) + 5 = 0
(5x)2 x 51 – 26(5x) + 5 = 0
Let P = 5x
5p2 – 26p + 5 = 0
(5p – 1)(p – 5) = 0
P= 1/5 or p = 5
5x = 5-1
X = -1
Or 5x = 51
X = 1 Hence x = -1, or 1
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
Solve the following exponential equations:
(i) (3x)2 + 2(3x) – 3 = 0
(ii) 22x + 2x+1 – 8 =0
ASSIGNMENT: Solve the following exponential equations;
(i) 22x - 6(2x) + 8 = 0
(ii) 32x – 4(3x + 1) + 27 = 0
SPECIFIC TOPIC: INDICIAL EQUATION
CONTENT: Example 1: Solve (1/2)X = 8
Solution
(1/2)X = 8
2-X = 23
-x = 3
X =-3
Example 2: Solve 3(3)x = 27
Solution
3(3)x = 27
3x+1 = 33
X + 1 = 3
X = 3 – 1
X = 2
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
(i) 25(5x) = 625
(ii) 3-x = 243
ASSIGNMENT:
(i) 22x - 6(2x) + 8 = 0
(ii) 32x – 4(3x + 1) + 27 = 0
CONTENT: Example 1: Solve 52x – 26(5x) + 25 = 0
Solution
52x – 26(5x) + 25 =0
Therefore, (5x)2 – 26(5x) + 25 = 0
Let p = 5x then
P2 – 26P + 25 =0
(P – 1) (P – 25) =0
P = 1 or p = 25
5x = 1 = 50
X = 0
Or 5x = 25 = 52
X = 2
Hence x =0 or 2
Example 2: solve 52x+1 – 26(5X) + 5 =0
Solution
52x +1 -26(5x) + 5 = 0
(5x)2 x 51 – 26(5x) + 5 = 0
Let P = 5x
5p2 – 26p + 5 = 0
(5p – 1)(p – 5) = 0
P= 1/5 or p = 5
5x = 5-1
X = -1
Or 5x = 51
X = 1 Hence x = -1, or 1
EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
(i) 22x - 6(2x) + 8 = 0
(ii) 32x – 4(3x + 1) + 27 = 0
ASSIGNMENT:
(iii) 32x - 9(2x) + 8 = 0
(iv) 33x – 6(3x + 1) + 27 = 0
CONTENT: TEST
1. Solve 3(3)x = 27
2. 10x = 1/10,000
3. 2x = 0.125
4. 25(5x) = 625
5. 3-x = 243
REFFERENCE BOOK: Further Mathematics for senior secondary school, book 1
PERFORMANCE OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on indicial equation
CONTENT: INDICIAL EQUATION
Example 1: Solve (1/2)X = 8
Solution
(1/2)X = 8
2-X = 23
-x = 3
X =-3
Example 2: Solve 3(3)x = 27
Solution
3(3)x = 27
3x+1 = 33
X + 1 = 3
X = 3 – 1
X = 2
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
Solve the following questions
(i) 10x = 1/10,000
(ii) 2x = 0.125
(iii) 25(5x) = 625
(iv) 3-x = 243
ASSIGNMENT: solve the following
(i) 10x = 1/0.001
(ii) 3x = 1/81
MAIN TOPIC: INDICIAL EQUATION
SPECIFIC TOPIC: EXPONENTIAL EQUATION
PERFORMANCE OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on exponential equation.
CONTENT: EXPONENTIAL EQUATION
Example 1: Solve 52x – 26(5x) + 25 = 0
Solution
52x – 26(5x) + 25 =0
Therefore, (5x)2 – 26(5x) + 25 = 0
Let p = 5x then
P2 – 26P + 25 =0
(P – 1) (P – 25) =0
P = 1 or p = 25
5x = 1 = 50
X = 0
Or 5x = 25 = 52
X = 2
Hence x =0 or 2
Example 2: solve 52x+1 – 26(5X) + 5 =0
Solution
52x +1 -26(5x) + 5 = 0
(5x)2 x 51 – 26(5x) + 5 = 0
Let P = 5x
5p2 – 26p + 5 = 0
(5p – 1)(p – 5) = 0
P= 1/5 or p = 5
5x = 5-1
X = -1
Or 5x = 51
X = 1 Hence x = -1, or 1
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
Solve the following exponential equations:
(i) (3x)2 + 2(3x) – 3 = 0
(ii) 22x + 2x+1 – 8 =0
ASSIGNMENT: Solve the following exponential equations;
(i) 22x - 6(2x) + 8 = 0
(ii) 32x – 4(3x + 1) + 27 = 0
SPECIFIC TOPIC: INDICIAL EQUATION
CONTENT: Example 1: Solve (1/2)X = 8
Solution
(1/2)X = 8
2-X = 23
-x = 3
X =-3
Example 2: Solve 3(3)x = 27
Solution
3(3)x = 27
3x+1 = 33
X + 1 = 3
X = 3 – 1
X = 2
EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
(i) 25(5x) = 625
(ii) 3-x = 243
ASSIGNMENT:
(i) 22x - 6(2x) + 8 = 0
(ii) 32x – 4(3x + 1) + 27 = 0
CONTENT: Example 1: Solve 52x – 26(5x) + 25 = 0
Solution
52x – 26(5x) + 25 =0
Therefore, (5x)2 – 26(5x) + 25 = 0
Let p = 5x then
P2 – 26P + 25 =0
(P – 1) (P – 25) =0
P = 1 or p = 25
5x = 1 = 50
X = 0
Or 5x = 25 = 52
X = 2
Hence x =0 or 2
Example 2: solve 52x+1 – 26(5X) + 5 =0
Solution
52x +1 -26(5x) + 5 = 0
(5x)2 x 51 – 26(5x) + 5 = 0
Let P = 5x
5p2 – 26p + 5 = 0
(5p – 1)(p – 5) = 0
P= 1/5 or p = 5
5x = 5-1
X = -1
Or 5x = 51
X = 1 Hence x = -1, or 1
EVALUATION: The lesson is evaluated as the students are asked to solve the following questions.
(i) 22x - 6(2x) + 8 = 0
(ii) 32x – 4(3x + 1) + 27 = 0
ASSIGNMENT:
(iii) 32x - 9(2x) + 8 = 0
(iv) 33x – 6(3x + 1) + 27 = 0
CONTENT: TEST
1. Solve 3(3)x = 27
2. 10x = 1/10,000
3. 2x = 0.125
4. 25(5x) = 625
5. 3-x = 243
