Question

Find the perimeter of a triangle whose vertices have the (3,10) (5,2) (4,12)
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Answer 1

Answer:

the answer to your question is

Step-by-step explanation:

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Answer 2

$\sf \fcolorbox{red}{pink}{ \huge{Solution :)}}$

Let , the vertices of triangle be A(3,10) , B(5,2) and C(4,12)

The distance between point A(3,10) and B (5,2) is

$\sf \mapsto AB = \sqrt{ {(5- 3)}^{2} + {(2 - 10)}^{2} } \\ \\\sf \mapsto AB = \sqrt{ {(2)}^{2} + {( - 8)}^{2} } \\ \\ \sf \mapsto AB = \sqrt{4 + 64} \\ \\\sf \mapsto AB = \sqrt{68} \: \: units$

The distance between the point B(5,2) and C(4,12) is

$\sf \mapsto BC = \sqrt{ {(4 - 5)}^{2} + {(12 - 2)}^{2} } \\ \\ \sf \mapsto BC = \sqrt{ {(1)}^{2} + {(10)}^{2} } \\ \\ \sf \mapsto BC = \sqrt{1 + 100} \\ \\ \sf \mapsto BC = \sqrt{101} \: \: units$

The distance between A(3,10) and C(4,12) is

$\sf \mapsto AC = \sqrt{ {(4 - 3)}^{2} + {(12 - 10)}^{2} } \\ \\\sf \mapsto AC = \sqrt{ {(1)}^{2} + {(2)}^{2} } \\ \\ \sf \mapsto AC = \sqrt{1 + 4} \\ \\ \sf \mapsto AC = \sqrt{5} \: \: units$

Therefore , the perimeter of triangle whose vertices are A(3,10) , B(5,2) and C(4,12) is

$\sf \hookrightarrow Perimeter = \sqrt{68} + \sqrt{101} + \sqrt{5} \\ \\ \sf \hookrightarrow Perimeter = 20.53 \: \: units$

Hence , the perimeter of triangle is 20.53 units

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