 Teacher resources and professional development across the curriculum

Teacher professional development and classroom resources across the curriculum           A B C Solutions for Session 10, Part B

See solutions for Problems: B1 | B2 | B3 | B4 | B5    Problem B1 The largest possible perimeter is 22 units; the smallest is 14 units. This is because all but one pentomino have a perimeter of 12 units. One pentomino has a perimeter of 10 units. A method for determining perimeter when combining two pentominoes is to add the perimeters of the two pentominoes and then subtract the number of sides touching when the pentominoes are combined. To get the largest possible perimeter, combine two 12-unit pentominoes so that the least number of sides touch, which is two. So 12 + 12 - 2 = 22. To get the smallest perimeter, combine one 12-unit pentomino with the 10-unit pentomino so that as many sides as possible are touching. The maximum number of sides touching is eight, making the smallest possible combined perimeter 12 + 10 - 8, or 14 units.    Problem B2 One advantage of recording on grid paper is that it allows students to trace the perimeters of the shapes they've formed. Recording on grid paper also enables students to keep track of the different shapes. Recording all of their findings also helps them notice patterns. When tracing their combined shapes, however, if students do not also include the detail of all the squares in the pentomino, they will not be able to see the number of sides that are touching to make their new combined shape. This will lessen the power of recording their findings. Some of the problem-solving strategies students used included the guess-and-check strategy, as well as analyzing shapes and drawings for patterns (e.g., all perimeters must be even, and the combined perimeter will be the sum of the perimeters of both pentominoes less the total number of sides touching).   Problem B3 To determine the perimeter without counting, the students first added the perimeters of the two pentominoes being combined for the new shape and then subtracted the number of sides touching between the two pentominoes. To help students see this analytic method, Mr. Belber asked them questions that focused on the number of sides that touched in the new shape and how that number decreased the sum of the two pentomino perimeters. Additionally, you could ask students the following questions to confirm that they understand the concept: What patterns do you notice in the perimeters of the combined figure when the two pentominoes have two sides touching? Four sides touching? etc. Does your answer depend on which two pentominoes you choose? Can you explain why this pattern makes sense, or why increasing the number of sides touching makes the perimeter smaller?   Problem B4 This lesson has a clear mathematical purpose, which is the understanding that if area remains constant, the perimeter of shapes constructed having that area can vary. Using manipulatives helps students see that area stays constant. The tiles also allow students to see and feel the number of sides touching, and how that number affects the perimeter.   Problem B5 Mr. Belber extended learning by asking students to explain why they knew they had found the smallest or largest perimeter, how they knew they had found all possible perimeters, and why they got only even perimeters. Some generalizations he might expect students to make include the fact that only even perimeters can be made and that a rule for determining perimeter is that the perimeters of two pentominoes, minus the number of sides touching in the combined new shape, equals the perimeter of the new shape. Mr. Belber's homework assignment of repeating the task with three pentominoes provides additional generalizing opportunities and could ultimately lead to a rule for determining the maximum and minimum perimeters for any number of pentominoes.     