If the height of a triangle is decreased by 40% and its base is increased by 40%, what will be the effect on its area ?

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 :
$$\eqalign{ & = \left( {\frac{{4bh}}{{50}} \times \frac{2}{{bh}} \times 100} \right)\% \cr & = 16\% \cr} $$