Answer: Option D
Let initial base = b cm and initial height = h cm
Then,
Initial area :
$$ = \left( {\frac{{1bh}}{2}} \right)c{m^2}$$
New base :
$$\eqalign{
& = \left( {140\% {\text{ of }}b} \right)cm \cr
& = \left( {\frac{{140b}}{{100}}} \right)cm \cr
& = \left( {\frac{{7b}}{5}} \right)cm \cr} $$
New height :
$$\eqalign{
& = \left( {60\% {\text{ of }}h} \right)cm \cr
& = \left( {\frac{{60h}}{{100}}} \right)cm \cr
& = \left( {\frac{{3h}}{5}} \right)cm \cr} $$
New area :
$$\eqalign{
& = \left( {\frac{1}{2} \times \frac{{7b}}{5} \times \frac{{3h}}{5}} \right)c{m^2} \cr
& = \left( {\frac{{21bh}}{{50}}} \right)c{m^2} \cr} $$
Area decreased :
$$\eqalign{
& = \left( {\frac{{1bh}}{2} - \frac{{21bh}}{{50}}} \right)c{m^2} \cr
& = \left( {\frac{{4bh}}{{50}}} \right)c{m^2} \cr} $$
Percentage decrease :
$$\eqalign{
& = \left( {\frac{{4bh}}{{50}} \times \frac{2}{{bh}} \times 100} \right)\% \cr
& = 16\% \cr} $$