Question

The perimeter of a triangle with vertices (0,6) , (0,0) , (8,0) is ______

Answer 1

Answer:

Hope it's help you bro.

Step-by-step explanation:

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Answer 2

Given :

• Vertices of triangle = (0,6), (0,0), (8,0)

To Find :

• Perimeter of triangle?

Solution :

• Let vertices of ∆ be A(0,6), B(0,0), C(8,0)

Formula used :

$\footnotesize{\underline{\boxed{\bf{Distance\:formula = \sqrt{\Big(x_{2} - x_{1}\Big)^2 + \Big(y_{2} - y_{1}\Big)^2}\:\:\:}}}}$

Firstly lets find the length of side AB of given triangle i.e, distance between vertices A(0,6) and B(0,0).

We have,

• x₁ = 0
• x₂ = 0
• y₁ = 6
• y₂ = 0

Putting all values in distance formula,

$\\ :\implies\:\sf AB = \sqrt{\Big(0 - 0\Big)^2 + \Big(0 - 6\Big)^2}$

$\\ :\implies\:\sf AB = \sqrt{\Big(0\Big)^2 + \Big(-6\Big)^2}$

$\\ :\implies\:\sf AB = \sqrt{\Big(0\:\times\:0\Big) + \Big(\cancel{-}6\:\times\:\cancel{-}6\Big)}$

$\\ :\implies\:\sf AB = \sqrt{0 + 36}$

$\\ :\implies\:\sf AB = \sqrt{36}$

$\\ :\implies\:\sf AB = \sqrt{6\:\times\:6}$

$\\ :\implies\:\large{\underline{\boxed{\bf{AB = 6}}}}$

Hence, length of side AB is 6 units.

Now, lets find the length of side BC of given triangle i.e, distance between vertices B(0,0), C(8,0).

We have,

• x₁ = 0
• x₂ = 8
• y₁ = 0
• y₂ = 0

Putting all values in distance formula,

$\\ :\implies\:\sf BC = \sqrt{\Big(8 - 0\Big)^2 + \Big(0 - 0\Big)^2}$

$\\ :\implies\:\sf BC = \sqrt{\Big(8\Big)^2 + \Big(0\Big)^2}$

$\\ :\implies\:\sf BC = \sqrt{\Big(8\:\times\:8\Big) + \Big(0\:\times\:0\Big)}$

$\\ :\implies\:\sf BC = \sqrt{64 + 0}$

$\\ :\implies\:\sf BC = \sqrt{64}$

$\\ :\implies\:\sf BC = \sqrt{8\:\times\:8}$

$\\ :\implies\:\large{\underline{\boxed{\bf{BC = 8}}}}$

Hence, length of side BC is 8 units.

Now, lets find length of last side AC of given triangle i.e, distance between vertices A(0,6) and C(8,0).

We have,

• x₁ = 0
• x₂ = 8
• y₁ = 6
• y₂ = 0

Putting all values in distance formula,

$\\ :\implies\:\sf AC = \sqrt{\Big(8 - 0\Big)^2 + \Big(0 - 6\Big)^2}$

$\\ :\implies\:\sf AC = \sqrt{\Big(8\Big)^2 + \Big(-6\Big)^2}$

$\\ :\implies\:\sf AC = \sqrt{\Big(8\:\times\:8\Big) + \Big(\cancel{-}6\:\times\:\cancel{-}6\Big)}$

$\\ :\implies\:\sf AC = \sqrt{64 + 36}$

$\\ :\implies\:\sf AC = \sqrt{100}$

$\\ :\implies\:\sf AC = \sqrt{10\:\times\:10}$

$\\ :\implies\:\large{\underline{\boxed{\bf{BC = 10}}}}$

Now, we know that perimeter of any figure is calculated by sum of its all sides. So, perimeter of triangle will be,

$\\ :\implies\:\sf Perimeter_{(triangle)} = AB + BC + AC$

$\\ :\implies\:\sf Perimeter_{(triangle)} = 6 + 8 + 10$

$\\ :\implies\:\sf Perimeter_{(triangle)} = 14 + 10$

$\\ :\implies\:\large{\underline{\boxed{\bf{Perimeter_{(triangle)} = 24}}}}$

Hence, perimeter of triangle is 24 units.