Question

asked Mar 23, 2018 187k views

1 Answer

Anton Kiselev · Answer 1 · 2018-03-29T18:36:30+0000

see the attached figure to better understand the problem

we know that the distance between two points is equal to

$d=\sqrt{(y2-y1)^(2)+(x2-x1)^(2)}$

Let

$A(-3,3)\ B(3,4))\ C(3,-3)$

Step 1

Find the distance AB

$A(-3,3)\ B(3,4)$

substitute in the formula

$d=\sqrt{(4-3)^(2)+(3+3)^(2)}$

$d=\sqrt{(1)^(2)+(6)^(2)}$

$dAB=√(37)\ units$

Step 2

Find the distance BC

$B(3,4))\ C(3,-3)$

substitute in the formula

$d=\sqrt{(-3-4)^(2)+(3-3)^(2)}$

$d=\sqrt{(-7)^(2)+(0)^(2)}$

$dBC=7\ units$

Step 3

Find the distance AC

$A(-3,3)\ C(3,-3)$

substitute in the formula

$d=\sqrt{(-3-3)^(2)+(3+3)^(2)}$

$d=\sqrt{(-6)^(2)+(6)^(2)}$

$dAC=√(72)\ units$

Step 4

Find the perimeter of the triangle

we know that

the perimeter of the triangle is the sum of the length sides of the triangle

P=AB+BC+AC

substitute the values

$P=√(37)\ units+7\ units+√(72)\ units=21.6\ units$

therefore

the answer is

the perimeter of the triangle is
$21.6\ units$

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