MATHEMATICS
EXTRA-CURRICULUM ACTIVITY FOR STUDENTS
BILI LEKE’S FORMULAE
At = ¼ √[(2ab + L)(2ab – L)]
L = a^2 + b^2 - c^2
At = a^2 tanα tanθ = a^2 sinαsinθ
2(tanθ + tanα) 2sin(α+θ)
a is adjacent to both θ and α
BLF GRAPH
DEVELOPMENT OF CONCEPTUAL AND MANIPULATION SKILLS AND ITS APPLICATIONS
I have discovered two formulae in mathematics: BILI LEKE’S FORMULAE and the resulting graphs for mathematical applications, artistic design and visual delight. BILI LEKE’S FORMULAE; a development of conceptual and manipulative skills and its applications.
The book I wrote on my formulae is intended to serve as introduction of Bili Leke’s formulae for area of a triangle.
The central themes of the book involve
1. Knowing Bili Leke’s formulae,
2. Learning to prove Bili Leke’s formulae, and
3. Learning to find the area of a triangle by using Bili Leke’s formulae and solve related problems.
Biliaminu Adeleke Adediran-Adegoke, discovered and presented two formulae for area of a triangle in January 1994. These formulae were not in Mathematics Encyclopaedia by world authors, and Engineering Mathematics Encyclopaedia by Head, Godman, S. V. Brain, Rutherford Cruz, in any form, in 1994. The formulae were first presented in a workshop, in the workshop on the teaching of difficult topics in Further Mathematics for Lagos state secondary schools teachers organized by Lagos state ministry of education, on April 12th 1994, by the author of these editions.
These formulae, the proofs, solutions to the questions in illustrations and worked examples and most of the questions are provided by the author. Usage and proof of these formulae would go far in helping students to develop conceptual and manipulative skills and its application.
First formula
If all sides of a triangle are known, the area of the triangle is given by:
At = ¼√(2ab + L)(2ab – L)
Where L = a^2 + b^2 – c^2
At = ¼√(2ac + L)(2ac – L)
Where L = a^2 + c^2 – b^2
At = ¼√(2bc + L)(2bc – L)
Where L = b^2 + c^2 – a^2
a, b, c are the three sides of the triangle
Second formula
If two angles and a side of a triangle are known, the area of the triangle is given by:
At = (a^2 tanαtanθ)/[2(tanθ + tanα)
Where a is the side adjacent to both angles α and θ
It can be shown that;
At = (a^2 tanαtanθ)/[2(tanθ + tanα)
= (a^2 sinαsinθ)/[2sin(θ + α)]
Question 1
Bili Leke’s formula;
At = ¼√[(2ab + L)(2ab – L)]
Where L = a^2 + b^2 – c^2
Plot the graph of logAt against log[(2ab + L)(2ab + L)] for
a + b > c and a = b, taking values of a (in cm) = 3, 4, 5, 6, 7
c = 5 cm. Find;
(i) intercept on logAt,
(ii) intercept on log[(2ab + L)(2ab – L)] axis,
(iii) slope of the graph.
Solution
(i) -0.60,
(ii) 1.20,
(iii) 0.50
GRAPH 1
Question 2
Bili Leke’s formula;
At = (a^2tanαtanθ)/[2(tanθ + tanα)]
Where a is the side adjacent to both θ and α
Given that α = 45 and θ = 45,
(a) Copy and complete the following table for
1 cm ≤ a ≤ 10 cm
a(cm) 1 2 3 4 5 6 7 8 9 10
a2(cm2)
At(cm2)
(b) Plot the graph of At against a and find the slope at
a = 5 cm.
Solution
(a) 1 cm ≤ a ≤ 10 cm, At = ¼ a2
a(cm) 1 2 3 4 5 6 7 8 9 10
a2(cm2) 1 4 9 16 25 36 49 64 81 100
At(cm2) 0.25 1 2.25 4 6.25 9 12.25 16 20.25 25
(b) 2.5 cm
GRAPH 2
Question 3
Bili Leke’s formula;
At = (a^2tanαtanθ)/[2(tanθ + tanα)]
Where a is the side adjacent to both θ and α.
Given that α = 45,
(a) Copy and complete the following table for
α + θ < 180 and √1 cm ≤ a ≤ √5 cm
θ 300 450 600 900 1200
tanθ
At1
(a=√1 cm)
At2
(a=√2 cm)
At3
(a=√3 cm)
At4
(a=√4 cm
At5
(a=√5 cm
(b) Plot the graph of At against θ for At1, At2, At3, At4, At5
on the same axes and the graph of At = 0.00986a2θ
for √1 cm ≤ a ≤ √5 cm and 0 ≤ θ ≤ 120 on the same axes
Solution
(a) α = 45
α + θ < 180 and √1 cm ≤ a ≤ √5 cm
θ 30 45 60 90 120
tanθ 0.5774 1.0000 1.7321 ∞ -1.7321
At1
(a=√1 cm) 0.18 0.25 0.32 0.50 1.18
At2
(a=√2 cm) 0.37 0.50 0.63 1.00 2.37
At3
(a=√3 cm) 0.55 0.75 0.95 1.5 3.56
At4
(a=√4 cm 0.74 1.00 1.26 2.00 4.74
At5
(a=√5 cm 0.93 1.25 1.58 2.50 5.93
(b) At = 0.00986a^2 θ, √1 cm ≤ a ≤ √5 cm and 0 ≤ θ ≤ 120
θ 0 30 45 60 90 120
At1
(a=√1 cm) 0 0.30 0.44 0.59 0.89 1.18
At2
(a=√2 cm) 0 0.59 0.89 1.18 1.77 2.37
At3
(a=√3 cm) 0 0.89 1.33 1.77 2.66 3.55
At4
(a=√4 cm) 0 1.18 1.77 2.37 3.55 4.73
At5
(a=√5 cm) 0 1.48 2.22 2.96 4.44 5.92
GRAPH 3
EXTRA-CURRICULUM ACTIVITY FOR STUDENTS
BILI LEKE’S FORMULAE
At = ¼ √[(2ab + L)(2ab – L)]
L = a^2 + b^2 - c^2
At = a^2 tanα tanθ = a^2 sinαsinθ
2(tanθ + tanα) 2sin(α+θ)
a is adjacent to both θ and α
BLF GRAPH
DEVELOPMENT OF CONCEPTUAL AND MANIPULATION SKILLS AND ITS APPLICATIONS
I have discovered two formulae in mathematics: BILI LEKE’S FORMULAE and the resulting graphs for mathematical applications, artistic design and visual delight. BILI LEKE’S FORMULAE; a development of conceptual and manipulative skills and its applications.
The book I wrote on my formulae is intended to serve as introduction of Bili Leke’s formulae for area of a triangle.
The central themes of the book involve
1. Knowing Bili Leke’s formulae,
2. Learning to prove Bili Leke’s formulae, and
3. Learning to find the area of a triangle by using Bili Leke’s formulae and solve related problems.
Biliaminu Adeleke Adediran-Adegoke, discovered and presented two formulae for area of a triangle in January 1994. These formulae were not in Mathematics Encyclopaedia by world authors, and Engineering Mathematics Encyclopaedia by Head, Godman, S. V. Brain, Rutherford Cruz, in any form, in 1994. The formulae were first presented in a workshop, in the workshop on the teaching of difficult topics in Further Mathematics for Lagos state secondary schools teachers organized by Lagos state ministry of education, on April 12th 1994, by the author of these editions.
These formulae, the proofs, solutions to the questions in illustrations and worked examples and most of the questions are provided by the author. Usage and proof of these formulae would go far in helping students to develop conceptual and manipulative skills and its application.
First formula
If all sides of a triangle are known, the area of the triangle is given by:
At = ¼√(2ab + L)(2ab – L)
Where L = a^2 + b^2 – c^2
At = ¼√(2ac + L)(2ac – L)
Where L = a^2 + c^2 – b^2
At = ¼√(2bc + L)(2bc – L)
Where L = b^2 + c^2 – a^2
a, b, c are the three sides of the triangle
Second formula
If two angles and a side of a triangle are known, the area of the triangle is given by:
At = (a^2 tanαtanθ)/[2(tanθ + tanα)
Where a is the side adjacent to both angles α and θ
It can be shown that;
At = (a^2 tanαtanθ)/[2(tanθ + tanα)
= (a^2 sinαsinθ)/[2sin(θ + α)]
Question 1
Bili Leke’s formula;
At = ¼√[(2ab + L)(2ab – L)]
Where L = a^2 + b^2 – c^2
Plot the graph of logAt against log[(2ab + L)(2ab + L)] for
a + b > c and a = b, taking values of a (in cm) = 3, 4, 5, 6, 7
c = 5 cm. Find;
(i) intercept on logAt,
(ii) intercept on log[(2ab + L)(2ab – L)] axis,
(iii) slope of the graph.
Solution
(i) -0.60,
(ii) 1.20,
(iii) 0.50
GRAPH 1
Question 2
Bili Leke’s formula;
At = (a^2tanαtanθ)/[2(tanθ + tanα)]
Where a is the side adjacent to both θ and α
Given that α = 45 and θ = 45,
(a) Copy and complete the following table for
1 cm ≤ a ≤ 10 cm
a(cm) 1 2 3 4 5 6 7 8 9 10
a2(cm2)
At(cm2)
(b) Plot the graph of At against a and find the slope at
a = 5 cm.
Solution
(a) 1 cm ≤ a ≤ 10 cm, At = ¼ a2
a(cm) 1 2 3 4 5 6 7 8 9 10
a2(cm2) 1 4 9 16 25 36 49 64 81 100
At(cm2) 0.25 1 2.25 4 6.25 9 12.25 16 20.25 25
(b) 2.5 cm
GRAPH 2
Question 3
Bili Leke’s formula;
At = (a^2tanαtanθ)/[2(tanθ + tanα)]
Where a is the side adjacent to both θ and α.
Given that α = 45,
(a) Copy and complete the following table for
α + θ < 180 and √1 cm ≤ a ≤ √5 cm
θ 300 450 600 900 1200
tanθ
At1
(a=√1 cm)
At2
(a=√2 cm)
At3
(a=√3 cm)
At4
(a=√4 cm
At5
(a=√5 cm
(b) Plot the graph of At against θ for At1, At2, At3, At4, At5
on the same axes and the graph of At = 0.00986a2θ
for √1 cm ≤ a ≤ √5 cm and 0 ≤ θ ≤ 120 on the same axes
Solution
(a) α = 45
α + θ < 180 and √1 cm ≤ a ≤ √5 cm
θ 30 45 60 90 120
tanθ 0.5774 1.0000 1.7321 ∞ -1.7321
At1
(a=√1 cm) 0.18 0.25 0.32 0.50 1.18
At2
(a=√2 cm) 0.37 0.50 0.63 1.00 2.37
At3
(a=√3 cm) 0.55 0.75 0.95 1.5 3.56
At4
(a=√4 cm 0.74 1.00 1.26 2.00 4.74
At5
(a=√5 cm 0.93 1.25 1.58 2.50 5.93
(b) At = 0.00986a^2 θ, √1 cm ≤ a ≤ √5 cm and 0 ≤ θ ≤ 120
θ 0 30 45 60 90 120
At1
(a=√1 cm) 0 0.30 0.44 0.59 0.89 1.18
At2
(a=√2 cm) 0 0.59 0.89 1.18 1.77 2.37
At3
(a=√3 cm) 0 0.89 1.33 1.77 2.66 3.55
At4
(a=√4 cm) 0 1.18 1.77 2.37 3.55 4.73
At5
(a=√5 cm) 0 1.48 2.22 2.96 4.44 5.92
GRAPH 3
