30. Answer: 16

Let the radii of the circles with centres $\mathrm{A}$ and $\mathrm{B}$ be $x \mathrm{~cm}$ and $y \mathrm{~cm}$ respectively, and let $\mathrm{C}$ be the point on the line segment $\mathrm{BY}$ such that $\mathrm{AC}$ is parallel to $\mathrm{XY}$. Then $\mathrm{AC}=40 \mathrm{~cm}$ and $\mathrm{BC}=(y-x) \mathrm{cm}$. Since $\triangle \mathrm{ABC}$ is a right-angled triangle, we have $(y-x)^{2}+40^{2}=41^{2}$ by Pythagorean theorem, so
$$
y-x=9 \text {. }
$$

Now consider the trapezium ABYX. We have $300=\frac{40}{2}(x+y)$, which gives
$$
x+y=15 \text {. }
$$

Solving the two simultaneous equations, we obtain $x=3$ and $y=12$. Hence
$$
\frac{b}{a}=\left(\frac{12}{3}\right)^{2}=16 .
$$
30. The figure below shows two circles with centres $\mathrm{A}$ and $\mathrm{B}$, and a line $\mathrm{L}$ which is a tangent to the circles at $X$ and $Y$. Suppose that $X Y=40 \mathrm{~cm}, A B=41 \mathrm{~cm}$ and the area of the quadrilateral $\mathrm{ABYX}$ is $300 \mathrm{~cm}^{2}$. If $a$ and $b$ denote the areas of the circles with centre $\mathrm{A}$ and centre $\mathrm{B}$ respectively, find the value of $\frac{b}{a}$.