Question

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

(a) 7 + $\sqrt{5}$ (b) 5 (c) 10 (d) 12 [CBSE 2014]

Answer 1

Let A(0, 4), O(0, 0) and B(3, 0) be the vertices of ∆AOB.

Using distance formula, we get

OA = $\sqrt{{\left(0-0\right)}^2+{\left(4-0\right)}^2}=\sqrt{16}=4$ units

OB = $\sqrt{{\left(3-0\right)}^2+{\left(0-0\right)}^2}=\sqrt{9}=3$ units

AB = $\sqrt{{\left(3-0\right)}^2+{\left(0-4\right)}^2}=\sqrt{9+16}=\sqrt{25}=5$ units

∴ Perimeter of ∆AOB = OA + OB + AB = 4 + 3 + 5 = 12 units

Thus, the required perimeter of the triangle is 12 units.

Hence, the correct answer is option D.