Find the perimeter of △ABC with vertices A(−5, −5), B(3, −5), and C(−5, 1).

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asked Oct 2, 2018 189k views

1 Answer

Psurikov · Answer 1 · 2018-10-06T11:25:28+0000

check the picture below.

so notice, the sides AB and AC you can pretty much count them off the grid.

now, to get the CB side.

$\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) C&({{ -5}}\quad ,&{{ 1}})\quad % (c,d) B&({{ 3}}\quad ,&{{ -5}}) \end{array}\qquad % distance value d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2} \\\\\\ CB=√([3-(-5)]^2+[-5-1]^2)\implies CB=√((3+5)^2+(-5-1)^2) \\\\\\ CB=√(8^2+(-6)^2)\implies CB=√(100)\implies CB=10$

sum all three sides up, and that's the perimeter of the triangle.

Find the perimeter of △ABC with vertices A(−5, −5), B(3, −5), and C(−5, 1).-example-1

Find the perimeter of △ABC with vertices A(−5, −5), B(3, −5), and C(−5, 1).

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