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 ?

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 :
$$\begin{array}{rl}& =\left(140\%\text{ of }b\right)cm\\ & =\left(\frac{140b}{100}\right)cm\\ & =\left(\frac{7b}{5}\right)cm\end{array}$$
New height :
$$\begin{array}{rl}& =\left(60\%\text{ of }h\right)cm\\ & =\left(\frac{60h}{100}\right)cm\\ & =\left(\frac{3h}{5}\right)cm\end{array}$$
New area :
$$\begin{array}{rl}& =\left(\frac{1}{2}\times \frac{7b}{5}\times \frac{3h}{5}\right)c{m}^2\\ & =\left(\frac{21bh}{50}\right)c{m}^2\end{array}$$
Area decreased :
$$\begin{array}{rl}& =\left(\frac{1bh}{2}-\frac{21bh}{50}\right)c{m}^2\\ & =\left(\frac{4bh}{50}\right)c{m}^2\end{array}$$
Percentage decrease :
$$\begin{array}{rl}& =\left(\frac{4bh}{50}\times \frac{2}{bh}\times 100\right)\%\\ & =16\%\end{array}$$