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Exploring Algebra with Geometry

Join Sue Scott in an exploration of algebra and geometry. Learn how to divide a square, solve equations, and discover geometric proofs. Engage in hands-on activities and deepen your understanding of these mathematical concepts.

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Exploring Algebra with Geometry

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  1. An exploration of Algebra with Geometry Sue Scott

  2. Welcome! Please take 2 small pieces of paper and 6 larger pieces. You’ll also need a ruler, pen and scissors, but not for your first task.

  3. Your first task is to divide a square of paper into 5 rectangles, each with the same dimensions. No tools other than your hands and folding skills!

  4. This is the geometry. • Make these folds, in order: • BD, • F • D A then EF, then AE. Now fold through the intersection of BD and AE. Next step is to fold AG. • Now fold through the intersection of BD and AG. • B • C E • I • G Next fold AH, and then through the intersection. • H

  5. This is the geometry. • J • F A • D • If you now fold over rectangle ABIJ, • four times, • you will have 5 rectangles, • all the same size. • C • B E • G • I • H

  6. So that was geometry. • Let’s look at the algebra. The length of each side is 1 unit. A • D • y = -2x + 1 y = x • x = -2x + 1 • x + 2x = 1 • C • 3x = 1 • B x =

  7. So that was geometry. • Let’s look at the algebra. The length of each side is 1 unit. • y = -3x + 1 y = x • x = -3x + 1 • x + 3x = 1 • 4x = 1 x =

  8. So that was geometry. • Let’s look at the algebra. The length of each side is 1 unit. • y = -4x + 1 y = x • x = -4x + 1 • x + 4x = 1 • 5x = 1 x =

  9. We can use the geometry or the algebra to find all unit fractions, and then all rational numbers. • y = -nx + k • y = -nx + 1 y = x • y = x • Solving simultaneously gives: • x = -nx + 1 • x + nx = 1 • x(1+n) = 1 • x = x =

  10. Here is some more geometry with a square. Fold along the diagonal PQ. • a • b • P Then fold corner Q onto line PQ, making a small triangle. • a • a Fold it out again. Fold horizontally and vertically through the vertices of the triangle. • b • b • Q • a • b

  11. Here is some more geometry with a square. • a • b This square has sides of length a + b. • a2 • ab • a • a • + b2 a2 • (a + b)2 = • + 2ab ab b2 Or was it algebra? • b • b • a • b

  12. Here is yet more geometry with a square. • a • a - b • b This square has sides of length a. • ab • (a - b)2 • a - b • (a - b)2 - 2ab = a2 + b2 It’s algebra again! ab • b2 b2 • b

  13. We’ll use blocks this time for our geometry. Make a 2x2x2 cube. Make a 3x3x3 cube. • Make 3 3x3x2 cuboids. Make 3 2x2x3 cuboids. Now put them together to make a cube.

  14. We’ll use blocks this time for our geometry. 2x2x2 = 8 3x3x3 = 27 • 3 x 3x3x2 = 3 x 18 = 54 3 x 2x2x3 = 3 x 12 = 36 27 + 36 + 8 + 54 = 125 = 53

  15. We have a 5 x 5 x 5 cube. • All sides are 3 + 2 long. It’s volume is (3 + 2)3 Which we have seen is the same as 33 + 3(2 x 2 x 3) + 3(3 x 3 x 2) + 23 • + • + 3 x + 3 x

  16. Let’s look at a cube of sides a + b. • Our large cube is made up of 8 parts: a b • a2b • ab2 b a a • (a + b)3 = a3 • a2b b3 3ab2 + b • a3 + • 3a2b + a • b3 b Algebra!! • ab2 b • a2b a a b

  17. And now for some more squares. Take one large and one small square, placed together. Rule two lines as shown, having lengths x the same. Label the 5 pieces. • B • D • x A • E • C • x • A • C • Cut long the lines, then rearrange the 5 pieces to make a large square. • B • E • D

  18. You will see that you can arrange your 3 squares as shown. Which will now be familiar to many of you as a geometric demonstration of Pythagoras’ Theorem. • Another demonstration involves leaving the small square intact. • A • C • B • E • D

  19. Place the 2 squares together so that straight lines AB and CD are formed. Find the centre of the large square. • G C • E • Draw line EF through the centre, parallel to AD. • F • B • A Draw line GH perpendicular to EF, through the centre. • H • D Cut along the blue lines, then put the 5 pieces together to make a larger square.

  20. This gives another demonstration of Pythagoras’ Theorem. • G C In a right angled triangle the sum of the squares on the two shorter sides is equal to the square on the hypotenuse. • E • F • B • A • J • D • AD2 = AJ2 + JD2 Henry Perigal found this proof about 1830.

  21. This exploration uses 4 identical right angled triangles. We can see that the area of the large square, side length a + b, is equal to the area of the four triangles and the area of the square with side c. • c • b • a c2 So (a + b)2 = • 4 x ½ (ab) + a2 + b2 + 2ab • = 2ab + c2 a2 + b2 = c2

  22. This exploration uses the same 4 identical right angled triangles. Make a mark the same distance from each corner of your square. Join the marks with lines to create the four triangles from your square. • c • b Label the sides of each triangle, a, b, c and then cut them out. • a • b Rearrange the triangles to make a square with side length c. • a • c

  23. This exploration uses the same 4 identical right angled triangles. We can see that the area of the large square, with sides c, is equal to the area of the 4 triangles and the area of the small inner square. (a - b)2 • 4 x ½ (ab) + • c2 = • b c2 = 2ab + a2 + b2 – 2ab • a • c c2 = a2 + b2

  24. Your next mission, should you choose to accept it, is to make the largest possible equilateral triangle out of your square. No tools other than your hands and folding skills!

  25. This creates the largest equilateral triangle from the square. • D • G Fold the square in half, at CD. • F • Fold side AB so that B lies on CD, giving points E and F. • E Fold EFG. While B is on F, fold AB back on AE, giving line AH. • H Now fold triangle AGH, the largest equilateral triangle possible in the square. • B C A

  26. And your final challenge is to make the largest possible regular octagon from your square. After using your hands and folding skills you will need scissors to cut your octagon out.

  27. This creates a regular octagon from the square. • G A • B Fold the square in half, at EF and also at GH. • Crease square EGFH. • F • E Fold BE on EG and GB on GE. Fold or cut along the lines to create your regular octagon. • D C • H

  28. Pascal’s triangle. While we call it Pascal’s triangle, after Blaise Pascal, in fact he only investigated it, 1654, but by no means discovered it. Chu Chi-Chieh c1300 tabulated the binomial coefficients in triangular form, referring to it as “the ancient method”.

  29. In maths, as in all things, there are often many ways to solve a problem. The secret to success and, more importantly, happiness, is knowing which solution is going to be most concise and also which is most enjoyable, and then choosing wisely. Sue Scott 2019

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