What is the perimeter of a triangle with corners at $(1, 4)$ $(1 ,4 )$ , $(6, 7)$ $(6 ,7 )$ , and $(4, 2)$ $(4 ,2 )$ ?

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Answer 1 · 2016-09-17T17:01:36

Perimeter $= \sqrt{34} + \sqrt{29} + \sqrt{13} = 3.60555$ $=sqrt(34)+sqrt(29)+sqrt(13)=3.60555$

$A (1, 4)$ $A(1,4)$ and $B (6, 7)$ $B(6,7)$ and $C (4, 2)$ $C(4,2)$ are the vertices of the triangle.

Compute for the length of the sides first.

Distance AB

$d_{A B} = \sqrt{{(x_{A} - x_{B})}_{A}^{2} + {(y_{A} - y_{B})}_{A}^{2}}$ $d_(AB)=sqrt((x_A-x_B)^2+(y_A-y_B)^2)$

$d_{A B} = \sqrt{{(1 - 6)}^{2} + {(4 - 7)}^{2}}$ $d_(AB)=sqrt((1-6)^2+(4-7)^2)$

$d_{A B} = \sqrt{{(- 5)}^{2} + {(- 3)}^{2}}$ $d_(AB)=sqrt((-5)^2+(-3)^2)$

$d_{A B} = \sqrt{25 + 9}$ $d_(AB)=sqrt(25+9)$

$d_{A B} = \sqrt{34}$ $d_(AB)=sqrt(34)$

Distance BC

$d_{B C} = \sqrt{{(x_{B} - x_{C})}_{B}^{2} + {(y_{B} - y_{C})}_{B}^{2}}$ $d_(BC)=sqrt((x_B-x_C)^2+(y_B-y_C)^2)$

$d_{B C} = \sqrt{{(6 - 4)}^{2} + {(7 - 2)}^{2}}$ $d_(BC)=sqrt((6-4)^2+(7-2)^2)$

$d_{B C} = \sqrt{{(2)}^{2} + {(5)}^{2}}$ $d_(BC)=sqrt((2)^2+(5)^2)$

$d_{B C} = \sqrt{4 + 25}$ $d_(BC)=sqrt(4+25)$

$d_{B C} = \sqrt{29}$ $d_(BC)=sqrt(29)$

Distance BC

$d_{A C} = \sqrt{{(x_{A} - x_{C})}_{A}^{2} + {(y_{A} - y_{C})}_{A}^{2}}$ $d_(AC)=sqrt((x_A-x_C)^2+(y_A-y_C)^2)$

$d_{A C} = \sqrt{{(1 - 4)}^{2} + {(4 - 2)}^{2}}$ $d_(AC)=sqrt((1-4)^2+(4-2)^2)$

$d_{A C} = \sqrt{{(- 3)}^{2} + {(2)}^{2}}$ $d_(AC)=sqrt((-3)^2+(2)^2)$

$d_{A C} = \sqrt{9 + 4}$ $d_(AC)=sqrt(9+4)$

$d_{A C} = \sqrt{13}$ $d_(AC)=sqrt(13)$

Perimeter $= \sqrt{34} + \sqrt{29} + \sqrt{13} = 3.60555$ $=sqrt(34)+sqrt(29)+sqrt(13)=3.60555$

God bless....I hope the explanation is useful.

What is the perimeter of a triangle with corners at (1 ,4 )(1,4)(1 ,4 ), (6 ,7 )(6,7)(6 ,7 ), and (4 ,2 )(4,2)(4 ,2 )?