Question

Triangle ABC has vertices at (-4,0) (-1,6) and (3,-1) What is the perimeter of triangle ABC rounded to the nearest tenth.

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Answer 1

Here, Vertices: A(-4, 0) B(-1, 6) C(3, -1)

Length of AB = √(6 - 0)² + (-1 + 4)²
= √6² + 3²
= √45

Length of BC = √(-1-6)² + (3+1)²
= √-7² + 4²
= √65

Length of AC = √(0+ 1)² + (-4 - 3)²
= √1² + 7²
= √50

Perimeter = sum of all sides:
So, P = √45 + √65 + √50 = √160 = 4√10

In short, Your Answer would be 4√10 units

Hope this helps!

Answer 2

Answer:

The perimeter of the triangle is 21.8 unit.

Step-by-step explanation:

Given : Triangle ABC has vertices at (-4,0) (-1,6) and (3,-1).

To find : What is the perimeter of triangle ABC?

Solution :

Vertices of the triangle : A(-4, 0) ,B(-1, 6), C(3, -1)

First we find the distance between the points or length of the sides of triangle.

Length of AB - A(-4, 0) ,B(-1, 6)

$AB=\sqrt{(6 - 0)^2 + (-1 + 4)^2}$

$AB=\sqrt{(6)^2 + (3)^2}$

$AB=\sqrt{36+9}$

$AB=\sqrt{45}$

Length of BC - B(-1, 6), C(3, -1)

$BC=\sqrt{(-1-6)^2 + (3+1)^2}$

$BC=\sqrt{(-7)^2 + (4)^2}$

$BC=\sqrt{49+16}$

$BC=\sqrt{65}$

Length of AC - A(-4, 0), C(3, -1)

$AC=\sqrt{(0+1)^2 + (-4-3)^2}$

$AC=\sqrt{(1)^2 + (-7)^2}$

$AC=\sqrt{1+49}$

$AC=\sqrt{50}$

Perimeter of the triangle is sum of all sides of the triangle,

P=Length of (AB+BC+AC)

$P = \sqrt{45} + \sqrt{65} + \sqrt{50}$

$P =6.708 + 8.062 +7.071$

$P =21.841$

Nearest tenth, The perimeter of the triangle is 21.8 unit.