Question

The perimeter of a triangle whose vertices are (0, 4), (0, 0), and (3, 0) will be

• A. 5
• B. 11
• C. 12
• D. $7 + \sqrt{5}$

Hint:

Use the distance formula for coordinate geometry: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$, and sum all sides for the perimeter.

Answer 1

The Correct Option is D

Step 1: Use the distance formula.
Distance between $(x_1, y_1)$ and $(x_2, y_2)$ is: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] Step 2: Calculate all sides.
For points (0, 4), (0, 0), and (3, 0):
\[ AB = \sqrt{(0 - 0)^2 + (4 - 0)^2} = 4 \] \[ BC = \sqrt{(3 - 0)^2 + (0 - 0)^2} = 3 \] \[ CA = \sqrt{(3 - 0)^2 + (0 - 4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \] Step 3: Find perimeter.
\[ \text{Perimeter} = 4 + 3 + 5 = 12 \] But the given option includes $7 + \sqrt{5}$, so let’s verify: the vertices might correspond differently. If we take $(0, 4)$, $(3, 0)$, and $(0, 0)$ again — perimeter remains 12. Correct answer: (C) 12.