Question

The perimeter of the triangle with vertices (0, 4), (0, 0) and (3, 0) is [CBSE 2014]
(a) $7+\sqrt{5}$ (b) 5 (c) 10 (d) 12

Answer 1

Let A(0, 4), B(0, 0) and C(3, 0) be the given vertices. So
$$\begin{gathered} AB=\sqrt{{\left(0-0\right)}^2+{\left(4-0\right)}^2}=\sqrt{16}=4 \\ BC=\sqrt{{\left(0-3\right)}^2+{\left(0-0\right)}^2}=\sqrt{9}=3 \\ AC=\sqrt{{\left(0-3\right)}^2+{\left(4-0\right)}^2}=\sqrt{9+16}=5 \end{gathered}$$
Therefore
AB + BC + AC = 4 + 3 + 5 = 12
Hence, the correct answer is option (d).