**Question**

**Introduction**

This is
another one of those tricky primary school mathematics questions involving
areas. A perfunctory glance at the area
seems to suggest there are four pieces.
Later you might realise that you can think of it as two quarter-circles
with two little 45°-45°-90° isosceles triangles removed.

**Plan**

Our plan
will be to first find the areas of the two quarter circles and then to subtract
the areas of the isosceles triangles.
This is our usual divide-and-conquer strategy [
Heuristics H10 & H11 ].
Notice that the two quarter-circles can be rearranged [H09] into a semi-circle with radius 10 cm. Simple enough.

What about the two exised triangles? Notice that the longest sides (the sloping sides) of the triangles
(highlighted in green) are each equal to the radius 10 cm
of the quarter-circles, simply because they, by touching the arcs, are
themselves also radii of the quarter-circles.

However, the problem seems to be that we do not know
the base and the height of each triangle.

**Examiners for Singapore Primary School mathematics like to set this sort of questions involving areas of isosceles right-angled triangles, in which you are given only the length of the hypotenuse**(the longest side). How to tackle this kind of situation? By using our imagination!
Imagine that the two triangles are brought
together. This forms a larger
right-angled isosceles triangle.
However, now you realise it is half of a 10 cm by 10 cm square. You can also imagine turning the triangle
around until one of the 10 cm sides is horizontal. Treating this as the base, the height of the
triangle is 10 cm. Either way, you are
able to solve it and get the same answer.

All that is
left now is to subtract this from your area of the semi-circle found earlier.

**Solution**

Shaded Area
[in cm

^{2}]
= Area of two quarter-circles – area of two triangles

= Area of semi-circle – area of combined triangle

= ½ ´

*p*´ (10)^{ 2}– ½ ´ 10 ´ 10
= 50

*p*– 50**Ans:**Shaded area = 107.08 cm

^{2}.

H09. Restate the problem in another way

H10. Simplify the problem

H11. Solve part of the problem

**Commentary**

We solved
this problem by breaking it down into smaller problems. Since

Please refer to this similar problem.

**areas are unchanged when you shift them, or turn them, or reflect them**, we are able to arrange the two quarter-circle pieces into one semi-circle. We can also combine the two triangles into a larger triangle for which we know the base and the height. By breaking down the problem and transmuting these smaller problems into equivalent problems, our task becomes much simpler, allowing us to get the solution quickly.Please refer to this similar problem.

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