Question

The vertices of ∆ABC are A(2, 8), B(16, 2), and C(6, 2). what is the perimeter of ∆ABC

Answer 1

Answer:

Perimeter of ABC = 32.4 units.

Step-by-step explanation:

We have a ΔABC with vertices A(2, 8), B(16, 2) and C(6, 2), and we are asked to find it's perimeter.

$\boxed{\sf \ Perimeter \ of \ \triangle ABC = AB + BC + AC \ }$

So, we'll individually find the distances of each side using the distance formula.

$\boxed{\sf \ Distance \ formula = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \ }$

Distance of AB:

x₁ = 2

x₂ = 16

y₁ = 8

y₂ = 2

$\sf AB= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\ \\\sf AB= \sqrt{(16-2)^2+(2-8)^2}\\ \\\sf AB= \sqrt{(14)^2+(-6)^2}\\ \\\sf AB= \sqrt{196+36 \ }\\ \\\sf AB= \sqrt{232}\\ \\\sf AB= 2\sqrt{58}\\ \\ \sf AB = 15.2 \ units. \ (approx)$

Distance of BC.

x₁ = 16

x₂ = 6

y₁ = 2

y₂ = 2

$\sf BC=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\ \\\sf BC=\sqrt{(6-16)^2+(2-2)^2}\\ \\\sf BC=\sqrt{(-10)^2+(0)^2}\\ \\\sf BC=\sqrt{100}\\ \\\sf BC= 10 \ units.$

Distance of AC.

x₁ = 2

x₂ = 6

y₁ = 8

y₂ = 2

$\sf AC=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\ \\\sf AC=\sqrt{(6-2)^2+(2-8)^2}\\ \\\sf AC=\sqrt{(4)^2+(-6)^2}\\ \\\sf AC=\sqrt{16+36}\\ \\\sf AC=\sqrt{52}\\ \\\sf AC=2\sqrt{13} \ units.\\ \\\sf AC = 7.2 \ units. \ (approx)$

Perimeter of ΔABC.

$\sf P(\triangle ABC) = AB + BC + AC\\ \\\sf P(\triangle ABC) = 15.2 \ + \ 10 \ + \ 7.2\\ \\\sf P(\triangle ABC) = 25.2 \ + \ 7.2\\ \\\underline{\underline{\sf P(\triangle ABC) = 32.4 \ units.}}$

Answer 2

Given: The vertices of Δ ABC are A(2, 8), B(16, 2) and C(6, 2).

To find: The perimeter of the so formed triangle.

Answer:

(Diagram for reference attached below.)

In order to do so, we'll have to find the length/distance between the two points for each side.

Formula to do so (distance formula): For any two given points

$\tt \sqrt{\bigg(x_2\ -\ x_1\bigg)^2\ +\ \bigg(y_2\ -\ y_1\bigg)^2}$

Let's first find the length of AB.

From the points A(2, 8) and B(16, 2), we have:

$\tt x_1\ =\ 2\\\\x_2\ =\ 16\\\\y_1\ =\ 8\\\\y_2\ =\ 2$

Using them in the formula,

$\tt \ \ \ \ \ \ \sqrt{\bigg(16\ -\ 2\bigg)^2\ +\ \bigg(2\ -\ 8\bigg)^2}\\\\\\\\=\ \ \sqrt{\bigg(14\bigg)^2\ +\ \bigg(-6\bigg)^2}\\\\\\\\=\ \ \sqrt{\bigg(196\ +\ 36\bigg)}\\\\\\\\=\ \ \sqrt{\bigg(232\bigg)}\\\\\\\\=\ \ \bf 15.2\ units\ \tt [est]$

Now, let's find the length of AC.

From the points A(2, 8) and C(6, 2), we have:

$\tt x_1\ =\ 2\\\\x_2\ =\ 6\\\\y_1\ =\ 8\\\\y_2\ =\ 2$

Using them in the formula,

$\tt \ \ \ \ \ \ \sqrt{\bigg(6\ -\ 2\bigg)^2\ +\ \bigg(2\ -\ 8\bigg)^2}\\\\\\\\=\ \ \sqrt{\bigg(4\bigg)^2\ +\ \bigg(-6\bigg)^2}\\\\\\\\=\ \ \sqrt{\bigg(16\ +\ 36\bigg)}\\\\\\\\=\ \ \sqrt{\bigg(52\bigg)}\\\\\\=\ \ \bf 7.2\ units\ \tt [est]$

Now, the length of BC.

From the points B(16, 2) and C(6, 2), we have:

$\tt x_1\ =\ 16\\\\x_2\ =\ 6\\\\y_1\ =\ 2\\\\y_2\ =\ 2$

Using them in the formula,

$\tt \ \ \ \ \ \ \sqrt{\bigg(6\ -\ 16\bigg)^2\ +\ \bigg(2\ -\ 2\bigg)^2}\\\\\\\\=\ \ \sqrt{\bigg(-10\bigg)^2}\\\\\\\\=\ \ \sqrt{\bigg(100\bigg)}\\\\\\\\=\ \ \bf 10\ units\ \tt [est]$

Now, perimeter of Δ ABC = AB + BC + AC.

Adding up all the sides,

Perimeter (Δ ABC) = 15.2 + 10 + 7.2 = 32.4

Therefore, perimeter of Δ ABC = 32.4 units.