Question

) If (3,10), (5,2), (14,12) are coordinates of the vertices of a triangle, then find

the perimeter of the triangle.
​

Answer 1

Question:

If (3,10), (5,2), (14,12) are coordinates of the vertices of a triangle, then find the perimeter of the triangle.

Solution:

We've been given the coordinates of a triangle, and we have to find it's perimeter. We can do so using the Distance formula.

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

Let the points (3, 10), (5, 2) & (14, 12) be named A, B & C respectively.

We'll find the distance of AB, BC & AC, then add them up.

Finding AB.

x₁ → 3

x₂ → 5

y₁ → 10

y₂ → 2

$\sf AB = \sqrt{\big(x_2 - x_1\big)^2 + \big(y_2 - y_1\big)^2}\\ \\ \\\sf AB = \sqrt{\big(5 - 3\big)^2 + \big(2 - 10\big)^2}\\ \\ \\\sf AB = \sqrt{\big(2\big)^2 + \big(-8\big)^2}\\ \\ \\\sf AB = \sqrt{4 + 64}\\ \\ \\\sf AB = \sqrt{68} \ units$

Finding BC.

x₁ → 5

x₂ → 14

y₁ → 2

y₂ → 12

$\sf BC = \sqrt{\big(x_2 - x_1\big)^2 + \big(y_2 - y_1\big)^2}\\ \\ \\\sf BC = \sqrt{\big(14 - 5\big)^2 + \big(12 - 2\big)^2}\\ \\ \\\sf BC = \sqrt{\big(9\big)^2 + \big(10\big)^2}\\ \\ \\\sf BC = \sqrt{81 + 100}\\ \\ \\\sf BC = \sqrt{181} \ units$

Finding CA.

x₁ → 14

x₂ → 3

y₁ → 12

y₂ → 10

$\sf CA = \sqrt{\big(x_2 - x_1\big)^2 + \big(y_2 - y_1\big)^2}\\ \\ \\\sf CA = \sqrt{\big(3 - 14\big)^2 + \big(10 - 12\big)^2}\\ \\ \\\sf CA = \sqrt{\big(11\big)^2 + \big(-2\big)^2}\\ \\ \\\sf CA = \sqrt{121 + 4}\\ \\ \\\sf CA = \sqrt{125} \ units$

Perimeter of ABC:

⇒ AB + BC + AC

⇒ √68 + √181 + √125 units.

Answer: √68 + √181 + √125 units.

Answer 2

Answer:

(√68 + √181 + √125) unit

Step-by-step explanation:

Let,  A≡(3,10), B≡(5,2) and C≡(14,12),

Then, the perimeter of triangle ABC = AB + BC + CA,

By the distance formula

Hence, the perimeter of the triangle ABC = (√68 + √181 + √125) units..