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Andrew [12]
3 years ago
13

To the nearest tenth, find the perimeter of ABC with vertices A(-1,4), B(-2,1) and C(2,1). show your work

Mathematics
1 answer:
liubo4ka [24]3 years ago
7 0

we have to firstly apply the distance formula to find the length of sides of the triangle

distance formula = \sqrt{(x_{2} -x_{1})^2 +(y_{2}- y_{1})^2}

So length AB = \sqrt{(-2-(-1))^2+(1-4)^2}= \sqrt{1 +9}=\sqrt{10}

BC= \sqrt{(2-(-2))^2+(1-1)^2}= \sqrt{16+0}=4

AC = \sqrt{(2-(-1))^2+ (1-4)^2}= \sqrt{9+9}=\sqrt{18}=3\sqrt{2}

Now perimeter = AB+BC+AC = \sqrt{10}+4+3\sqrt{2}

Plug in calculator

perimeter= 11.4 units

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