(a) In the diagram, O is the centre of the circle radius r cm and < XOY = 90°.If the area of the shaded part is 504\(cm^{2}\), calculate the value of r. [Ta...

(a) Shaded segment of the circle.

From the diagram, \(O\) is the centre, \(OX=OY=r\), \(\angle XOY=90^\circ\), and the shaded part is the segment lying between chord \(XY\) and the arc \(XKY\).

\[\text{Shaded area}=\text{area of sector } XOY-\text{area of }\triangle XOY.\]

Sector \(XOY\) subtends \(90^\circ\), so it is a quarter of the circle:

\[\text{sector}=\frac{90}{360}\pi r^{2}=\frac{1}{4}\pi r^{2}.\]

Triangle \(XOY\) is right-angled at \(O\) with legs \(r\) and \(r\):

\[\triangle XOY=\frac{1}{2}\,r\times r=\frac{1}{2}r^{2}.\]

Hence

\[504=\frac{1}{4}\pi r^{2}-\frac{1}{2}r^{2}=r^{2}\left(\frac{1}{4}\times\frac{22}{7}-\frac{1}{2}\right)=r^{2}\left(\frac{22}{28}-\frac{14}{28}\right)=r^{2}\left(\frac{8}{28}\right)=\frac{2}{7}r^{2}.\]

\[r^{2}=504\times\frac{7}{2}=1764\quad\Rightarrow\quad r=\sqrt{1764}=\boxed{42\text{ cm}}.\]

(b) Two isosceles triangles on a common base PQ.

Triangles \(PQR\) and \(PQS\) stand on opposite sides of the common base \(PQ\).

In \(\triangle PQR\) (isosceles with \(RP=RQ\)), the base angles are equal, and \(\angle PQR=66^\circ\) is a base angle, so

\[\angle RPQ=\angle RQP=66^\circ.\]

In \(\triangle PQS\) (isosceles with \(SP=SQ\)), \(\angle PSQ=109^\circ\) is the apex angle, so the base angles are

\[\angle SPQ=\angle SQP=\frac{180^\circ-109^\circ}{2}=\frac{71^\circ}{2}=35.5^\circ.\]

Since \(R\) and \(S\) are on opposite sides of \(PQ\), the angle \(RQS\) is the sum of the two base angles at \(Q\):

\[\angle RQS=\angle RQP+\angle PQS=66^\circ+35.5^\circ=\boxed{101.5^\circ}.\]