= Solution
Let the semicircle have radius $R$, and write $A=|\mathbf a|$, $B=|\mathbf b|$. The <right triangle> has hypotenuse $2R$, so the <Pythagorean theorem> gives $A^2+B^2=4R^2$. Its <area> is $AB/2$, whereas half the <area> of the semicircle is $\pi R^2/4$. Equating these and eliminating $R$ gives
$$
\boxed{AB=\frac\pi8(A^2+B^2).}
$$
Equivalently, the relation between the two <norms> can be written
$$
\boxed{\frac AB=\frac{4\pm\sqrt{16-\pi^2}}{\pi}.}
$$
The two possible ratios are reciprocals, corresponding to interchanging the two sides.
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