A blog about Mathematics and Mathematics Education in Singapore, as well as Mathematics Education in general. Written for students, parents, educators and other stakeholders in Singapore, and around the world.
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Friday, March 6, 2015
[Pri20150220FSA] A Fishy Shaped Area
Plan of Attack
This problem looks difficult because the shaded area does not seem to
look like any regular shape. Is it a
fish whose head is pointing in the top left direction and whose tail is in the
bottom left direction? Fortunately, this
is not a Rorschach ink-blot test.
As with all “area” problems in primary (elementary) school, we try to
break down the unfamiliar shape into regular shapes (e.g. parts of circles,
squares, triangles, rectangles). It is
basically a divide-and-conquer strategy (using heuristics H10 & H11). If we look carefully, we realise that the required
area consists of a semi-circle less a funny horn-shaped area, which I call ‘F’.
F is for funny, for want of a better description. So I am going to find the area of the
semicircle (which is half of a circle), then subtract the area of F.
We’ll worry about finding the area of F later.
semi-circle [in cm2]
= ½ ´p´ radius2
= ½ ´p´ (5) 2 = 25/2p
It is good
to leave the calculation with p until the last
OK, we are done with the first part.
[Heuristic H11] Let us us tackle
the next part, which is to find the area of F. Note that this is a 45°-45°-90° isosceles
triangle minus a 45°-degree sector (which is one-eighth of a circle, because 45°/360°
Area of F [in cm2]
= Area of triangle – area of sector
= ½ ´ 10 ´ 10 – 1/8´p´ radius2
= ½ ´ 10 ´ 10 – 1/8´p´ (10) 2
= 50 – 25/2p
Let us combine our answers. We
need to subtract 50 – 25/2p from the area of the semi-circle. If we subtracted 50 from 25/2p, we would have over-subtracted. So we need to add back 25/2p.