Question

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

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

Answer 1

First find length of each sides of ∆
Let A( 4 , 0) B(0, 0) and C (0 , 3) be the vertices of the triangle.

Using distance formula we have
AB = $\sqrt{4^2+0}$ = 4

BC = $\sqrt{{0+3}^2}$ = 3

CA = $\sqrt{4^2+3^2}$ = 5

Clearly adding AB + BC + CA = perimeter of ∆
$\Rightarrow$ Perimeter of ∆ ABC = 3 + 4 + 5= 12 units