Find the perimeter P of ▱JKLM with vertices J(−3,−2), K(−5,−5), L(1,−5), and M(3,−2). Round your answer to the nearest tenth, if necessary.

Given:

Vertices of JKLM are J(−3,−2), K(−5,−5), L(1,−5), and M(3,−2).

To find:

The perimeter P of a parallelogram JKLM.

Solution:

Distance formula:

[tex]D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]

Using distance formula, we get

[tex]JK=\sqrt{\left(-5-\left(-3\right)\right)^2+\left(-5-\left(-2\right)\right)^2}[/tex]

[tex]JK=\sqrt{\left(-5+3\right)^2+\left(-5+2\right)^2}[/tex]

[tex]JK=\sqrt{\left(-2\right)^2+\left(-3\right)^2}[/tex]

[tex]JK=\sqrt{4+9}[/tex]

[tex]JK=\sqrt{13}[/tex]

Similarly,

[tex]KL=\sqrt{\left(1-\left(-5\right)\right)^2+\left(-5-\left(-5\right)\right)^2}=6[/tex]

[tex]LM=\sqrt{\left(3-1\right)^2+\left(-2-\left(-5\right)\right)^2}=\sqrt{13}[/tex]

[tex]JM=\sqrt{\left(3-\left(-3\right)\right)^2+\left(-2-\left(-2\right)\right)^2}=6[/tex]

Now, perimeter P of ▱JKLM is

[tex]P=JK+KL+LM+JM[/tex]

[tex]P=\sqrt{13}+6+\sqrt{13}+6[/tex]

[tex]P=2\sqrt{13}+12[/tex]

[tex]P=2(3.61)+12[/tex]

[tex]P=7.22+12[/tex]

[tex]P=19.22[/tex]

[tex]P\approx 19.2[/tex]

Therefore, the perimeter P of ▱JKLM is 19.2 units.

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