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

Question

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

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    0
    2021-09-19T04:56:44+00:00

    ANSWER

    The perimeter is 11.4 units

    EXPLANATION

    Perimeter is the distance around the figure.

    We use the distance formula,

    d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

    to determine the length of all the sides and add them.

    |AB|=\sqrt{(-2--1)^2+(1-4)^2}

    |AB|=\sqrt{(-1)^2+(-3)^2}

    |AB|=\sqrt{1+9}

    |AB|=\sqrt{10} \approx 3.162

    |AC|=\sqrt{(2--1)^2+(1-4)^2}

    |AC|=\sqrt{(2+1)^2+(1-4)^2}

    |AC|=\sqrt{(3)^2+(-3)^2}

    |AC|=\sqrt{9+9}

    |AC|=\sqrt{18} \approx 4.24

    |BC|=\sqrt{(2--2)^2+(1-1)^2}

    |BC|=\sqrt{(2+2)^2+(1-1)^2}

    |BC|=\sqrt{(4)^2+(0)^2}

    |BC|=\sqrt{14}=4

    Therefore perimeter=|AB|+|BC|+|AC|

    =4.00+3.16+4.24

    =11.40 units

    0
    2021-09-19T04:57:07+00:00

    AB= \sqrt{(-2-(-1))^2+(1-4)^2}= \sqrt{1+9}= \sqrt{10} \approx    3.2\\\\ BC= \sqrt{(2-(-2))^2+(1-1)^2}= \sqrt{16}=4\\\\ AC= \sqrt{(2-(-1))^2+(1-4)^2}= \sqrt{9+9}= \sqrt{18} \approx    4.2 \\\\ P_{ABC}=AB+BC+AC=3.2+4+4.2=11.4

    The perimeter of ∆ABC = 11.4 units

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