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Mathematics Involving Shape and Space a Algebra. The 9-Dot Problem.

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Presentation on theme: "Mathematics Involving Shape and Space a Algebra. The 9-Dot Problem."— Presentation transcript:

1 Mathematics Involving Shape and Space a Algebra

2 The 9-Dot Problem

3 Objectives Solve geometrical problems Identify the properties of 2-d shapes

4 Have a go at Triangles Draw as many different triangles by joining three of the nine dots

5 Have a go at Quadrilaterals Draw as many different quadrilaterals by joining four of the nine dots

6 Algebra Walls Constructing and solving equations using Algebra Walls

7 Algebra walls 5 = 27 n7 ? 12 ? n + 7 ? n + 19 n + 19 = 27 n = 27 - 19 n = 8

8 Algebra walls 5 = 27 n7 ? 12 ? 15 ? 27 n = 8 Back substitute to check 8

9 Algebra walls 2n = 35 n + 37 ? 2n + 7 ? n + 10 ? 3n + 17 3n + 17 = 35 3n = 35 - 17 3n = 18 n = 6

10 Algebra walls = 35 ? 19 ? 16 ? 35 n = 6 Back substitute to check 9 2nn + 37 12 12

11 Algebra walls 5 = 42 n = 27 12 n + 5n 26 162n n + 26 = 73= 32 n -16n + 14 n + 26 24 - n3n + 4 4n + 6 = 88 4 - 2n2n + 8 6n + 7 = 33 Solve for n and back substitute to check n = 5 n = 2 n = 5 n = 6 n = 2.5

12 Constructing and solving equations using the perimeter of shapes

13 Perimeter and algebra n + 2 3n 6n The perimeter of this shape is 22cm. Find the length of each side Total perimeter = ? 3n + n + 2 + 6n =10n + 2 10n + 2 = 2210n = 20n = 2 ? ? 6 12 4 Check: 6 + 4 + 12 = 22 so n = 2 is correct The length of each side is 6, 4 and 12 Substitute n = 2 into the expressions for each length

14 Perimeter and algebra 12 - n 3n - 4 4n The perimeter of this shape is 38cm. Find the length of each side Total perimeter = ? 3n - 4 + 12 - n + n + 2 + 4n =7n + 10 7n + 10 = 387n = 28n = 4 ? ? 8 16 8 Check: 8 + 8 + 6 + 16 = 38 so n = 2 is correct The length of each side is 8, 8, 6 and 16 Substitute n = 4 into the expressions for each length n + 2 ? 6

15 Perimeter and algebra 3n - 5 2n - 4 n + 6 The perimeter of each shape is inside the shape. Find the length of each side n + 2 41cm a) n +5 n + 6 n + 30 50cm b) 15 - n 2n + 8 2n + 6 n + 2 63cm c) 2n + 8 15 - n 8n - 8 n + 6 24 – 2n 115cm d) n + 8 n + 6 Solve for n and back substitute to check n = 3 n = 6 n = 4 n = 8

16 THE END!


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