6+ 5+ 2- 6+ 3- 1- 3+ 2- 7+ 2- 2÷ 4 3 2 1 4 1 2 3 2 3 1 4 1 3 4 2 4÷ 36x 8X 8+ 16x 24x 1 5+ 2 1 4 3 2 4 3 1 1 4 2 3 4 1 2 3 1 1- 2- 3÷ 1 3- 1 4 3 2 1 3 4 2 3 1 4 2 3 2 1 4 6x 1 1- 2- 2- 2 7+ 3 2 1 4 3 2 1 4 4 2 3 1 2 4 3 1 TWO SPECIAL FRAMES Q1 The area of a trapezium is the sum of the parallel edge lengths, divided by 2, multiplied by the width. Join corresponding points on each shape to get a set of trapezia. To sum their areas, you need not do a separate calculation for each, you can just add all the edge lengths, divide by 2 and multiply by the width. A B. But you know from what you are given that the edge length total is the same for and A B. And you are also given that the blue area is the same for and Therefore their widths must be the same. Q2 Equating areas: (I) Equating perimeters: (II) Using the width information: (III) = 3. s l Dividing (I) by (II) and substituting for from (III), you find This gives you the required dimensions. B’s If you do not believe the answer to Q1, you can check by substituting back in (I) to find the area, and equating to area written as the sum of its trapezia, i.e. the perimeter total AREA AND PERIMETER PROBLEM Small triangle: area perimeter . Large triangle: area perimeter . SQUARES FROM CROSSES Now you can make a rectangle by putting the three squares together. Use Pythagoras’ Theorem to work out the perimeter of each piece by concentrating on the dotted lines. QUICKIE 33 ANSWERS FROM ISSUE 64 SYMKEN – 9 Easy Quite easy Moderate Moderate 2- 1 1- 1- 2- 10+ 3- 1- 2- 5+ 8+ 4 1 2 3 1 3 4 2 4 1 2 3 3 1 2 4
6+ 4 5+ 3 2 2- 1 6+ 4 3- 1 1- 2 3 3+ 2 2- 3 1 7+ 4 2- 1 3 2÷ 4 2 2 4÷ 1 4 3 2 4 36x 3 8X 1 1 8+ 4 16x 2 3 24x 4 1 1 5+ 2 3 1 1 1- 4 3 2- 2 3÷ 1 3 4 2 3 1 4 2 3 2 1 1 3- 4 6x 3 2 1 1 4 3 1- 2 1 4 2- 4 2 2- 3 1 2 2 7+ 4 3 1 TWO SPECIAL FRAMES Q1 The area of a trapezium is the sum of the parallel edge lengths, divided by 2, multiplied by the width. Join corresponding points on each shape to get a set of trapezia. To sum their areas, you need not do a separate calculation for each, you can just add all the edge lengths, divide by 2 and multiply by the width. But you know from what you are given that the edge length total is the same for A and B. And you are also given that the blue area is the same for A and B. Therefore their widths must be the same. Q2 Equating areas: (I) Equating perimeters: (II) Using the width information: (III) Dividing (I) by (II) and substituting for s from (III), you find l = 3. This gives you the required dimensions. If you do not believe the answer to Q1, you can check by substituting back in (I) to find the area, and equating to B’s area written as the sum of its trapezia, i.e. the perimeter total AREA AND PERIMETER PROBLEM Small triangle: area perimeter . Large triangle: area perimeter . SQUARES FROM CROSSES Now you can make a rectangle by putting the three squares together. Use Pythagoras’ Theorem to work out the perimeter of each piece by concentrating on the dotted lines. QUICKIE 33 ANSWERS FROM ISSUE 64 SYMKEN – 9 Easy Quite easy Moderate Moderate 2- 4 1 1 1- 2 3 1- 1 2- 3 10+ 4 2 3- 4 1 1- 2 2- 3 5+ 3 1 8+ 2 4
SYMmetryplus-Index-Answers-Sep-17.docx