Answer :
Given:
area of square, A = 16 [tex]cm^{2}[/tex]
error in area, dA = 0.03 cm^{2}
Step-by-Step Explanation:
Let 'a' be the side of the square
area of square, A = [tex]a^{2}[/tex] (1)
A = 16 = [tex]a^{2}[/tex]
Therefore, a = 4 cm
for max tolerable error in length 'da', differentiate eqn (1) w.r.t 'a':
dA = 2a da
[tex]0.03 = 2\times 4\times da[/tex]
da = [tex]\frac{0.03}{8}[/tex]
da = 0.0375 cm