Question

To the nearest tenth, find the perimeter of ∆ABC with vertices A(2,4), B(-2,1) and C(2,1).

Answer 1

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Answer 2

Answer:

The perimeter of ∆ABC is 12 units.

Step-by-step explanation:

Given : ∆ABC with vertices A(2,4), B(-2,1) and C(2,1).

To find : The perimeter of the triangle ABC?

Solution :

$A=(2,4)=(x_a,y_a)$

$B=(-2,1)=(x_b,y_b)$

$C=(2,1)=(x_c,y_c)$

Perimeter of ∆ABC: P=AB+BC+AC

$AB=\sqrt{(x_b-x_a)^2+(y_b-y_a)^2}$

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

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

$AB=\sqrt{(-4)^2+9}$

$AB=\sqrt{16+9}$

$AB=\sqrt{25}$

$AB=5$

$BC=\sqrt{(x_c-x_b)^2+(y_c-y_b)^2}$

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

$BC=\sqrt{(2+2)^2+(0)^2}$

$BC=\sqrt{(4)^2+0}$

$BC=\sqrt{16+0}$

$BC=\sqrt{16}$

$BC=4$

$AC=\sqrt{(x_c-x_a)^2+(y_c-y_a)^2}$

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

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

$AC=\sqrt{(0)^2+9}$

$AC=\sqrt{0+9}$

$AC=\sqrt{9}$

$AC=3$

Perimeter=AB+BC+AC

P=5+4+3

P=12

Therefore, The perimeter of ∆ABC is 12 units.