**Question**

**Introduction**

The diagram
looks a bit like the Yin and Yang symbol, doesn’t it? This problem can be solved easily using the
correct insights. I present two solutions:
the first one is by direct calculation (in terms of

*p*), and the second solution uses the powerful concept of ratio of similar figures.
Whichever
method is used, first we must make observations. Can you see that there are three types of
semicircles (small, medium and large)? [H02, H04]

Let

*S*= area of small semicircle,*M*= area of medium-sized semicircle, and*L*= area of large semicircle.
Note that area of

*A*= area of*C*=*L*–*M*+*S*, and area of*B*= 2´(*M*–*S*). [H10, H11]**Solution 1**(by direct calculation)

*L*= ½

*p*(3)

^{2}=

^{9}

*/*

^{p}_{2}.

*M*= ½

*p*(2)

^{2}=

^{4}

*/*

^{p}_{2}.

*S*= ½

*p*(1)

^{2}=

*/*

^{p}_{2}.

area of

*A*= area of*C*=^{9}*/*^{p}_{2}–^{4}*/*^{p}_{2}+*/*^{p}_{2}= 3*p*
area of

*B*= 2´(^{4}*/*^{p}_{2}–*/*^{p}_{2}) = 3*p*
\ area of

Many pupils feel more
comfortable using concrete approximations like
*A*: area of*B*: area of*C*= 1 : 1 : 1. (The areas are all the same)*p*»

^{22}/

_{7}or

*p*» 3.14, but this tends to obscure relationships between entities, and makes the calculations messier.

**Solution 2**(using similar shapes)

An
powerful idea is that the ratio of areas of similar shapes is the square of the
ratios of their lengths. When a figure
is enlarged by a factor of (say) 5, we get a similar figure and the area
becomes 5

area of

^{2}= 25 times as large. So*M*= (2)^{2}*S*= 4*S*because the radius of the medium-sized semicircle is twice that of the small semicircle. Likewise,*L*= (3)^{2}*S*= 9*S*.area of

*A*= area of*C*= 9*S*– 4*S*+*S*= 6*S*
area of

*B*= 2´(4*S*–*S*) = 6*S*
\ area of

*A*: area of*B*: area of*C*= 1 : 1 : 1. (The areas are all the same)**Commentary**

The second
solution is neater because we do not need to deal with fractions or with

*p*. The ratio of areas of similar shapes is the square of the ratios of their lengths. This concept is in fact required knowledge in secondary school mathematics, including GCE ‘O’ level “Elementary” Mathematics.
H02. Use a
diagram / model

H04. Look for
pattern(s)

H10. Simplify
the problem

H11. Solve part
of the problem

**Suitable Levels**

*****GCE ‘O’ Level “Elementary” Mathematics (“similar figures”)

* Primary School Mathematics (“areas”)

* other syllabuses that involve areas, ratios and or similar shapes.

## No comments:

## Post a Comment

Note: Only a member of this blog may post a comment.