BRICK BLOCK BOO.pptx Mathematical investigation

MAthematical investigation

1. BRICK, BLOCK, BOO! A Mathematical Investigation by Gornes , Iris T. Larroza , Shera Marie S. Lorania , Jecky Hanes C.

2. INTRODUCTION Mathematical investigation refers to the sustained exploration of mathematical situation. It allows us to learn about mathematics, especially the nature of mathematical activity and thinking. In addition it allows us to seek patterns, formulate new conjectures, and establish truth by rigorous deduction from appropriate chosen axioms and definitions. One of the patterns that found is the patterns in brick walls. Bricks are rectangular blocks typically made from clay, shale, or other materials such as concrete. They are commonly used in construction for building walls, pavements, and other structures. Mostly bricks are laid in an alternate pattern, also known as stretcher bond, to have a balance of strength, stability and aesthetic appeal, making it a common choice in bricklaying for various construction projects. 2 INTRODUCTION

3. INTRODUCTION 2 INTRODUCTION

4. SITUATION An engineer is planning to create a brick wall where bricks measures 2 square units. He wants to place the bricks alternately to form a pattern with nxn dimension. Investigate… 3 SITUATION

5. EXPLORATION OF THE SITUATION 4 EXPLORATION OF THE SITUATION

6. EXPLORATION OF THE SITUATION 5 EXPLORATION OF THE SITUATION 3, 4, 5, 6, 7, . . . 3, 6, 10, 15, 21, . . .

7. EXPLORATION OF THE SITUATION 6 EXPLORATION OF THE SITUATION

8. STATEMENT OF THE PROBLEM 7 STSATEMENT OF THE PROBLEM This investigation aimed to know the number of rectangular blocks in an (n x n) dimension . Specifically, this investigation sought answers to the following questions : Given (n x n) dimension starts at 3x3, How many rectangular blocks can fit in an (n x n) dimension? How many excess rectangular blocks are in (n x n) dimension?

9. Data Gathering and Conjecture 8 DATA GATHERING AND CONJECTURE

10. 9 DATA GATHERING AND CONJECTURE The first differences are not equal, but the second differences are equal to 1. So, it is a quadratic function such that R(n)= an 2 + bn +c = 0 . an 2 +an +c = if n=3 if n=4 if n=5 an 2 + bn +c = 0 an 2 + bn +c = 0 an 2 + bn +c = 0 a(3) 2 + b(3) +c= 3 a(4) 2 + b(4) +c= 6 a(5) 2 + b(5) +c= 10 eq.1 9a + 3b + c = 3 eq.2 16a + 4b + c = 6 eq.3 25a + 5b + c = 10   eq.2 16a + 4b + c = 6 ----------------------- 16a + 4b + c = 6 eq.1 9a + 3b + c = 3 ------------------------ 9a - 3b - c = -3 7a +b = 3 eq.4 eq.3 25a + 5b + c = 10 ----------------------- 25a + 5b+c = 10 eq.2 16a + 4b + c = 6 ----------------------- -16a – 4b – c = - 6 9a +b = 4 eq.5 eq.5 9a +b = 4 ----------------------- 9a + b = 4 eq.4 7a +b = 3 ----------------------- -7a – b = - 3 2a = 1 a= ½ Substitution Substitute a= ½ to equation 4 7a +b = 3 7(½) +b = 3 b= - ½ Substitute a= ½, b= - ½ to equation 1 9a + 3b + c = 3 9 (½) +3(-½) +c = 3 c= Since a= ½, b= - ½ and c=0 then an 2 + bn +c = 0 ½ n 2 - ½n =0 =0 R(n ) =   Data Gathering and Conjecture

11. 10 DATA GATHERING AND CONJECTURE Data Gathering and Conjecture

12. 11 DATA GATHERING AND CONJECTURE Data Gathering and Conjecture

13. 12 DATA GATHERING AND CONJECTURE Data Gathering and Conjecture

14. 13 DATA GATHERING AND CONJECTURE Data Gathering and Conjecture Since the common difference occur at the first differences the equation is in the form of a linear equation an + b=0   if n=3 if n=4 an + b = 0 an + b = 0 eq.1 3a + b = 3/2 eq.2 4a + b = 2   eq.1 3a + b = 3/2 ----------------------- -3a – b= -3/2 eq.2 4a + b = 2 ----------------------- 4a + b = 2 a = 1/2 Substitution Substitute a=1/2 to equation 1 3a +b = 3/2 3(1/2) +b = 3/2 b= Since a= ½, b= 0 then an + b = 0 1/2n =0 =0 E(n) =    

15. 14 DATA GATHERING AND CONJECTURE Data Gathering and Conjecture

16. 15 DATA GATHERING AND CONJECTURE Data Gathering and Conjecture

17. 16 TESTING/ VERIFYING CONJECTURES TESTING / VERIFYING CONJECTURES

18. 17 TESTING/ VERIFYING CONJECTURES TESTING / VERIFYING CONJECTURES

19. 18 TESTING/ VERIFYING CONJECTURES TESTING / VERIFYING CONJECTURES

20. 19 TESTING/ VERIFYING CONJECTURES TESTING / VERIFYING CONJECTURES

21. 20 PROVING/ JUSTIFYING CONJECTURES PROVING/ JUSTIFYING CONJECTURES

22. 21 PROVING/ JUSTIFYING CONJECTURES PROVING/ JUSTIFYING CONJECTURES

23. 22 PROVING/ JUSTIFYING CONJECTURES PROVING/ JUSTIFYING CONJECTURES Conjecture 2 The number of excess rectangular block in (n x n) dimension is generated through the formula: E(n)= , n ≥ 3. Proof: Using the principle of Geometric: Prove that the number of excess rectangular block in (n x n) dimension is E(n )= .  

24. 23 PROVING/ JUSTIFYING CONJECTURES PROVING/ JUSTIFYING CONJECTURES When n= (3x3) dimension: When n= (4x4) dimension: R(3) = R(4) = R(3 ) = R(4) = R(4 ) = 2     The excess rectangular block can be formed by joining together two squares that does not included in the fixed rectangular blocks, as shown by the animation. The measure of rectangular block is 2 square units, so that’s why it divides by 2 . If the (n x n) is odd, there is one square that do not have a pair and it count as 1/2 block . Since the block is rectangular, then the excess is generated .  

25. 24 SUMMARY SUMMARY This mathematical investigation is about finding how many rectangular blocks can be fit in an (n x n) dimension when the blocks lay alternately and also its excess is been explored. The following conjectures were generated through the patterns, and relationships that we observed based on the data presented in the investigation : The number of rectangular blocks that can fit in (n x n) dimension is generated through the formula: R(n)= , where n≥3. The number of excess rectangular block in (n x n) dimension is generated through the formula: E(n)= , n≥ 3 . The stated conjectures were tested, verified to be true and proven by using Algebraic rule using 2 nd difference, Mathematical Induction and Geometric.  

26. 25 POSSIBLE EXTENSION POSSIBLE EXTENSION In the light of the results and findings, it is recommended that further investigation be conducted to find out : Using the same situation, find the formula for the odd and even (n x n) dimension, find the formula sum of the number of blocks in three consecutive ( n x n) dimension . f ind the total vertices that can be found in an ( nxn ) dimension.

27. Thank you for listening! BRICK, BLOCK, BOO MATHEMATICAL INVESTIGATION AND MODELING 2 GROUP 3