The perimeter of a triangle with vertices $ (0,4),(0,0) $ and $ (3,0) $ is
(A) 5
(B) 12
(C) 11
(D) $ 7+\sqrt{5} $
Given:
The vertices of a triangle are \( (0,4),(0,0) \) and \( (3,0) \).
To do:
We have to find the perimeter of the triangle,
Solution:
Let the vertices of the triangle be $A(0,4),B(0,0)$ and $C(3,0)$.
We know that,
Perimeter of a triangle $=$ Sum of the lengths of the sides of the triangle
This implies,
Perimeter $=AB+BC+CA$
Using the distance formula,
$d=\sqrt{( x_2-x_1)^2+( y_2-y_1)^2}$
Perimeter $=\sqrt{(0-0)^{2}+(0-4)^{2}}+\sqrt{(3-0)^{2}+(0-0)^{2}}+\sqrt{(3-0)^{2}+(0-4)^{2}}$
$=\sqrt{0+16}+\sqrt{9+0}+\sqrt{9=16}$
$=4+3+\sqrt{25}$
$=7+5$
$=12$
The perimeter of the triangle is $12$.
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