1. In each of Figure A, Figure B, and Figure $\mathrm{D}$, the circle has centre $O$ and radius 2 .
(a) Determine the area of the shaded region in Figure A.
(b) Determine the area of the shaded region in Figure B.
(c) In Figure C, $\triangle P Q R$ is equilateral and has side length of 2 . Determine the area of $\triangle P Q R$.
(d) Determine the area of the shaded region in Figure D.
1. (a) Since the circle has radius 2 , its area is $\pi 2^{2}=4 \pi$.
The central angle for the shaded sector is $90^{\circ}$, which is one-quarter of the full circle $\left(360^{\circ}\right)$.
Therefore, the shaded area is one-quarter of the area of the full circle, or $\frac{1}{4} \times 4 \pi=\pi$.
(b) We label the diagram as shown.
The area of the shaded region equals the area of the sector $O S V$ minus the area of $\triangle O S V$.
From (a), the area of the sector is $\pi$.
Since the triangle is right-angled and has base and height both equal to 2 , its area is $\frac{1}{2}(2)(2)=2$.
Therefore, the area of the shaded region is $\pi-2$.
(c) Since $\triangle P Q R$ is equilateral, then altitude $P T$ cuts base $Q R$ into two equal pieces.
Since $Q R=2$, then $Q T=T R=1$.
By the Pythagorean Theorem in $\triangle P T Q$, we have
$$
P T=\sqrt{P Q^{2}-Q T^{2}}=\sqrt{2^{2}-1^{2}}=\sqrt{3}
$$
since $P T>0$.
Therefore, $\triangle P Q R$ has base $Q R=2$ and height $P T=\sqrt{3}$.
Thus, the area of $\triangle P Q R$ is $\frac{1}{2}(2)(\sqrt{3})=\sqrt{3}$.
(d) We label the diagram as shown.
The area of the shaded region equals the area of the sector $W O X$ minus the area of $\triangle O W X$.
From (c), the area of $\triangle O W X$ is $\sqrt{3}$, since $\triangle O W X$ is congruent to $\triangle P Q R$.
Since $\triangle O W X$ is equilateral, then $\angle W O X=60^{\circ}$; since $\frac{60^{\circ}}{360^{\circ}}=\frac{1}{6}$, then this is one-sixth of the full circle.
Therefore, the area of sector $W O X$ is one-sixth of the area of the entire circle, or $\frac{1}{6}(4 \pi)=\frac{2}{3} \pi$.
Thus, the area of the shaded region is $\frac{2}{3} \pi-\sqrt{3}$.