Question

A triangle has vertices (1,4), (1,-2), and (-3,-2). what is the approximate perimeter of the triangle

Answer 1

Answer:

• Perimeter of Traingle is 10 + 2√5

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Step-by-step explanation:

Let,

• A = (1,4)
• B = (1,-2)
• C = (-3,-2)

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To find the perimeter of the traingele with vertices of A (1,4), B(1, -2) and C(-3,-2). We have to first find the distance between each pair of points, which will give length of the sides.

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Using distance formula,

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`\dashrightarrow \sf \: \: AB = \sqrt{( x_(2) - x_(1))^2 + (y_(2) - y_(1))^2} \\ \\`

`\dashrightarrow \sf \: \: AB = \sqrt{{ {(1 - 1)}^(2) } + {( - 2 - 4)}^(2) } \\ \\`

`\dashrightarrow \sf \: \: AB = \sqrt{0 + {( - 6)}^(2) } \\ \\`

`\dashrightarrow \sf \: \: AB = √(36) \\ \\`

`\pink{ \dashrightarrow\sf \: \: {\underline{AB = 6 }}}\\ \\ \\`

`\dashrightarrow \sf \: \: BC = \sqrt{{ {( - 3 - 1)}^(2) } + {( - 2 + 2)}^(2) } \\ \\`

`\dashrightarrow \sf \: \:BC = \sqrt{ {( - 4)}^(2) + 0} \\ \\`

`\dashrightarrow \sf \: \:BC = √(16) \\ \\`

`\pink{ \dashrightarrow \sf \: \:{ \underline{ BC = 4 }}}\\ \\ \\`

`{\dashrightarrow { \sf \: \: {AC = \sqrt{{ {( - 3 - 1)}^(2) } + {(2 - 4)}^(2)}}} } \\ \\`

`\dashrightarrow \sf \: \:AC = \sqrt{ {( - 4)}^(2) + {( - 2)}^(2) } \\ \\`

`\dashrightarrow\sf \: \: AC = √(16 + 4) \\ \\`

`\dashrightarrow \sf \: \:AC = √(20) \\ \\`

`\pink{ \dashrightarrow \sf \: \: { \underline{AC = 2 √(5) }}}`

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Now,

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`\dashrightarrow\sf \: \: Perimeter = AB + BC + AC \\ \\`

`\dashrightarrow\sf \: \: 6 + 4 + 2√5 \\ \\`

`\dashrightarrow\sf \: \: {\pink{ \underline{ \boxed { \pmb{ \sf{ \: 10 + 2√5 \: \: }}}}}}`

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Hence,

• Perimeter of Traingle is 10 + 2√5