\[h = \ 5\ см;\ \ S_{осн} = 4\ см^{2}.\]
\[1)\ V_{1} = \frac{1}{3} \bullet 5 \bullet 4 = \frac{20}{3}\ \left( см^{3} \right).\]
\[2)\ Высота\ после\ увеличения:\]
\[h = 5 + 5 \bullet \frac{10}{100} = 5 + \frac{5}{10} =\]
\[= 5 + \frac{1}{2} = \frac{11}{2}\ (см).\]
\[3)\ Площадь\ после\ увеличения:\]
\[S_{осн} = 4 + 4 \bullet \frac{10}{100} = 4 + \frac{4}{10} =\]
\[= 4 + \frac{2}{5} = \frac{22}{5}\ (см).\]
\[4)\ Объем\ увеличенной\ \]
\[пирамиды:\]
\[V_{2} = \frac{1}{3} \bullet \frac{11}{2} \bullet \frac{22}{5} = \frac{242}{30} = \frac{121}{15}\ \left( см^{3} \right).\]
\[5)\ V_{1} + V_{1} \bullet p\% = V_{2}\]
\[\frac{20}{3} + \frac{20}{3} \bullet \frac{p}{100} = \frac{121}{15}\]
\[\frac{20}{3} + \frac{p}{15} = \frac{121}{15}\]
\[100 + p = 121\]
\[p = 21.\]
\[Ответ:\ \ на\ 21\%.\]