Monday, May 4, 2015

[Pri20150504PER] Pythagoras Precludes Perimeter Problem?

Question

Introduction
     Just like this problem, this primary school problem involving lengths seems to require Pythagoras’ Theorem, which is not taught until secondary school.  But with some clever observations, we can solve it by looking at it in another way.

Observations

     First,  note that being diagonals of the rectangle BOECBC  is the same as  OE,  which is the same as  13 cm, the radius of the quadrant.  Secondly, we know that the sum of  BO  and  OC  is 17 cm,  being half of the perimeter of the rectangle BOEC.  We do not have to know the individual lengths  BO  and  OC,  and we do not need Pythagoras’ Theorem.  Since  AO  and  OD  are radii of  13 cm  each,  their total length is  26 cm.  We can just subtract  17 cm  from this to get the total of  AB  and  CD.  We do not need to know the individual lengths of  AB  and  CD.  Once we understand these points, we are ready to solve the problem.

Solution
    Perimeter / cm   = (AB + CD) + BC + arc AED
                              = [(OA + OD) – (BO + OC)] + OE + ¼ ´ 2 ´ p  ´ 13
                              = [  13  +  13   –     ½ ´ 34  ] +  13  + ½ ´ 3.14 ´ 13
                              = 42.41
Ans:  The perimeter of the shaded region is  42.4  cm.

Remarks
     Your answer can never be more precise than the precision you use for your calculation.  When we use  3  significant figures for the value of  p,  our answer is at best correct to  3  significant figures.

H02. Use a diagram / model
H04. Look for pattern(s)
H09. Restate the problem in another way
H10. Simplify the problem
H11. Solve part of the problem


Suitable Levels
Primary School Mathematics
* other syllabuses that involve perimeters and lengths




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