A simple rectangle... *(Re-posted with changed options)*

The increased base is $b+b\times \cfrac { 10 }{ 100 } =\cfrac { 11 }{ 10 } b$ . Similarly the increased altitude must be of the form $\left( xh \right)$ , where $h$ is the initial values of the altitude.

According to the problem we have $\left( \cfrac { 11 }{ 10 } b \right) \left( xh \right) =bh$ . Solving it for $x$ , we get $x=\cfrac { 10 }{ 11 }$ . Therefore the final altitude after the transformations is $\cfrac { 10 }{ 11 } h$ . Increase in altitude is $\cfrac { \left( final\quad altitude \right) -\left( intitial\quad altitude \right) }{ \left( intitial\quad altitude \right) } \times 100$ or $\cfrac { \left( \cfrac { 10 }{ 11 } h \right) -h }{ h } \times 100=\boxed { -9\cfrac { 1 }{ 11 } }$ %. Hence the answer.