**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 ,
is the same as

*BOEC*

*BC*

*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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