Topics: Starter | Area | Geometry | Mensuration | Shape | Simultaneous Equations

• Transum,
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• The first three problems are in increasing order of difficulty. The first can probably be guessed, the second needs more of a strategy as the numbers are larger and the third involves a fraction so probably requires the application of a more generalised method. Refreshing the page generates three different rectangles to further consolidate any techniques learned in the first round. The fourth problem is only for Nobel prize candidates!
• Simon Fletcher, Bolton UK
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• 52 cm rectangle perimeter 105 area. Is this correct? I've checked and can't find a solution.
• Transum,
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• Hello Simon, so you are looking for two numbers (the height and the width) that add up to half the perimeter, 26cm, and multiply together to give an area of 105cm².
  We can write two simultaneous equations showing this information:
  h + w = 26
  hw = 105
  We can combine the information in these two equations and eliminate one of the unknowns:
  From the first equation we see that h = 26 – w
  Substituting this into the second equation gives (26 – w)w = 105
  Expanding the brackets gives 26w - w²=105
  Rearranging gives w² - 26w + 105 = 0
  This is a quadratic equation which can be solved by factorising
  (w – 5)(w – 21) = 0
  So w could be 5cm in which case h would be 21cm or, w could be 21cm in which case h would be 5cm.
• Audrey Macmillan, @audproctor77
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• Clarifying area/perimeter confusion using a Transum Starter. Great #mathschat on squares and rows.

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