Find the perimeter of a triangle with vertices A(3, 1), B(2, -1), and C(-3, 2). Leave

Mathematics

1 answer:

natulia [17]3 years ago

7 0

The perimeter of triangle is: 14.15 units

Step-by-step explanation:

First of all we have to find the lengths of sides of triangles

Given

A(3, 1), B(2, -1), and C(-3, 2)

The distance formula will be used to find the lengths of sides

$d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\$

So,

$AB = \sqrt{(2-3)^2+(-1-1)^2}\\=\sqrt{(-1)^2+(2)^2}\\=\sqrt{1+4}\\=\sqrt{5}\\=2.24\ unitsBC = \sqrt{(-3-2)^2+(2+1)^2}\\=\sqrt{(-5)^2+(3)^2}\\=\sqrt{25+9}\\=\sqrt{34}\\= 5.83\ units\\AC = \sqrt{(-3-3)^2+(2-1)^2}\\=\sqrt{(-6)^2+(1)^2}\\=\sqrt{36+1}\\=\sqrt{37}\\=6.08\ units$

The perimeter of triangle will be:

$P = AB+BC+AC\\= 2.24+5.83+6.08\\=14.15\ units$

Hence,

The perimeter of triangle is: 14.15 units

Keywords: Triangle Perimeter

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