# The perimeter of a triangle with vertices $(0,4),(0,0)$ and $(3,0)$ is(A) 5(B) 12(C) 11(D) $7+\sqrt{5}$

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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 $12$.

Updated on 10-Oct-2022 13:28:26