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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.
First, note that being diagonals of the rectangle BOEC, BC
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.
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
Ans: The perimeter of the shaded
region is 42.4 cm.
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.