The perimeter of the triangle with vertices (0, 4), (0, 0) and (3, 0) is

Let A(0, 4), B(0, 0) and C(3, 0) be the given vertices. So

$A B=\sqrt{(0-0)^{2}+(4-0)^{2}}=\sqrt{16}=4$

$B C=\sqrt{(0-3)^{2}+(0-0)^{2}}=\sqrt{9}=3$

$A C=\sqrt{(0-3)^{2}+(4-0)^{2}}=\sqrt{9+16}=5$

Therefore
AB + BC + AC = 4 + 3 + 5 = 12
Hence, the correct answer is option (d).