Hamilton 2016 Q2

Let the radius \(DA\) be \(r\), so that \(DC\) also equals \(r\). Since arc \(AC\) is one quarter of a circle, angle \(\angle CDA\) is 90°. Hence, using Pythagoras' Theorem in the triangle \(ACD\), we obtain \(CA^2 = 2r^2\).

Now consider the segment \(S\) bounded by the arc \(AC\) and the chord \(AC\), shown shaded in the following diagram.

We shall combine this region with each of \(R\) and \(T\)—if the areas of the combined regions are equal, then the areas of \(R\) and \(T\) are equal.

The region obtained by combining \(R\) and \(S\) is one eighth of a circle with centre \(C\) and radius \(CA\). Thus its area is

The region obtained by combining \(S\)and \(T\)is one quarter of a circle with centre \(D\)and radius \(DA\). Thus its area is \(\tfrac{1}{4}\pi r^2\).

Hence the areas of the regions obtained by combining \(S\) with each of \(T\) and \(R\) are equal. Therefore the area of \(T\) is equal to the area of \(R\).