A semicircle of radius $r$ is drawn with centre $V$ and diameter $U W$. The line $U W$ is then extended to the point $X$, such that the $U W$ and $W X$ are of equal length. An arc of the circle with centre $X$ and radius $4 r$ is then drawn so that the line $X Y$ is a tangent to the semicircle at $Z$, as shown. What, in terms of $r$, is the area of triangle $Y V W$ ?
A $\frac{4 r^{2}}{9}$
B $\frac{2 r^{2}}{3}$
C $r^{2}$
D $\frac{4 r^{2}}{3}$
E $2 r^{2}$
Solution: B
Since $U W=W X$, we have $W X=U W=2 r$. As $V$ is the centre of the circle with $U W$ as diameter, $V W=V Z=r$.
Hence $V X=V W+W X=3 r$. The triangles $Y V X$ and $V V W$ have the same height and $V W=\frac{1}{3} V X$. Therefore $\operatorname{area}(\triangle Y V W)=\frac{1}{3} \operatorname{area}(\triangle Y V X)$. As $V Z$ is a radius of the semicircle, and $X Y$ is a tangent to the semicircle at $Z$, the lines $V Z$ and $X Y$ are perpendicular. Therefore $\operatorname{area}(\triangle Y V X)=\frac{1}{2} V Z . X Y=\frac{1}{2}(r .4 r)=2 r^{2}$. Hence, area $(\triangle Y V W)=\frac{1}{3}$ area $(\triangle Y V X)=\frac{1}{3}\left(2 r^{2}\right)=\frac{2}{3} r^{2}$.