Question

Find the perimeter of triangle ABC with


vertices A(1,1) B(,7,1) C(1,9)
​

Answer 1

Question

Find the perimeter of triangle ABC with

vertices A(1,1) B(,7,1) C(1,9)

Solution

Given:-

• Vertices are A(1,1) , B(7,1) and C(1,9)

Find:-

• Perimeter of triangle

Explanation

Distance between two point

☛ AB = √[(x'-x")²+(y'-y")²]

Let,

In ∆ ABC , First calculate length of all sides.

Side AB

Where,

• x' = 1, x" = 7
• y' = 1 , y" = 1

AB = √[(1-7)²+(1-1)²]

➠ AB = √[(-6)²]

➠AB = √36

➠AB = 6 unit.

Side BC

BC = √[(7-1)²+(1-9)²]

➠ BC = √[6²+(-8)²]

➠BC = √[36+64]

➠BC = √100

➠BC = 10 unit

Side CA

CA = √[(1-1)²+(9-1)²]

➠ CA = √[8²

➠CA = 8 unit

So, perimeter of ∆ABC

☛ Perimeter of ∆ABC = AB + BC + CA

➠ Perimeter of ∆ABC = 6 + 10 + 8

➠ Perimeter of ∆ABC = 24 units .

_________________

Answer 2

Given: A triangle formed by the points A(1, 1), B(7, 1) and C(1, 9).

To find: The perimeter.

Answer:

To do so, we'll have to find the length (distance between two points) of each side.

Distance formula:

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

Let's first find the length of side AB.

From the points A(1, 1) and B(7, 1), we have:

$\tt x_1\ =\ 1\\\\x_2\ =\ 7\\\\y_1\ =\ 1\\\\y_2\ =\ 1$

Using them in the formula,

$\tt AB\ =\ \sqrt{\Bigg(7\ -\ 1\Bigg)^2\ +\ \Bigg(1\ -\ 1\Bigg)^2}\\\\\\AB\ =\ \sqrt{\Bigg(6\Bigg)^2\ +\ \Bigg(0\Bigg)^2}\\\\\\\bf AB\ =\ 6\ units$

Now, let's find the distance of BC.

From the points B(7, 1) and C(1, 9), we have:

$\tt x_1\ =\ 7\\\\x_2\ =\ 1\\\\y_1\ =\ 1\\\\y_2\ =\ 9$

Using them in the formula,

$\tt BC\ =\ \sqrt{\Bigg(1\ -\ 7\Bigg)^2\ +\ \Bigg(9\ -\ 1\Bigg)^2}\\\\\\BC\ =\ \sqrt{\Bigg(-6\Bigg)^2\ +\ \Bigg(8\Bigg)^2}\\\\\\BC\ =\ \sqrt{\Bigg(100\Bigg)}\\\\\\\bf BC\ =\ 10\ units$

Now, the distance of AC.

From the points A(1, 1) and C(1, 9), we have:

$\tt x_1\ =\ 1\\\\x_2\ =\ 1\\\\y_1\ =\ 1\\\\y_2\ =\ 9$

Using them in the formula,

$\tt AC\ =\ \sqrt{\Bigg(1\ -\ 1\Bigg)^2\ +\ \Bigg(9\ -\ 1\Bigg)^2}\\\\\\AC\ =\ \sqrt{\Bigg(0\Bigg)^2\ +\ \Bigg(8\Bigg)^2}\\\\\\\bf AC\ =\ 8\ units$

Perimeter of Δ ABC = Length of AB + Length of BC + Length of AC.

Perimeter of Δ ABC = 6 units + 10 units + 8 units

Perimeter of Δ ABC = 24 units.