= Solution
The <Pythagorean theorem> gives $c^2=a^2+b^2$. Taking a <total differential> gives $2c\,dc=2a\,da+2b\,db$, so
$$
\boxed{\frac{dc}{c}=\frac{a\,da+b\,db}{a^2+b^2}=\frac{a^2}{a^2+b^2}\frac{da}{a}+\frac{b^2}{a^2+b^2}\frac{db}{b}.}
$$
The <area> of the <right triangle> is $A=ab/2$, so the <product rule> gives $dA=(b\,da+a\,db)/2$ and
$$
\boxed{\frac{dA}{A}=\frac{da}{a}+\frac{db}{b}.}
$$
The prescribed small changes therefore give $dA/A\approx0.01-0.02=-0.01$: \b[the area decreases by approximately $1\%$]. For comparison, retaining the product of the two finite percentage changes gives $(1.01)(0.98)-1=-0.0102$, an exact decrease of $1.02\%$; the <total differential> retains only the first-order terms.
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