Question

Find the perimeter p of ▱jklm with vertices j(2,2), k(5,3), l(5,−3), and m(2,−4). Round your answer to the nearest tenth, if necessary.

Answer 1

Consider the vertices of parallelogram JKLM with vertices J(2,2) , K(5,3) , L(5,-3) and M(2,-4).

Perimeter JKLM = Length JK + Length KL + Length LM + Length JM

Length JK = (2,2) (5,3)

The length(or distance) between two points say $(x_{1},y_{1})$ and $(x_{2},y_{2})$ is given by the distance formula:

$\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

Now, length JK = $\sqrt{(5-2)^{2}+(3-2)^{2}}$

= $\sqrt(10)$ units

Since, JKLM is a parallelogram. In parallelogram opposite sides are equal in length.

Therefore, LM =  $\sqrt(10)$ units

Now, length KL = $\sqrt{(5-5)^{2}+(-3-3)^{2}}$

= 6 units

Since, JKLM is a parallelogram. In parallelogram opposite sides are equal in length.

Therefore, JM =  6 units

Perimeter of JKLM =  $\sqrt(10)$ +  $\sqrt(10)$ + 6 + 6

= 2 $\sqrt(10)$ + 12

= 18.324

Rounding to the nearest tenth, we get

= 18.3 units.

Therefore, the perimeter of JKLM is 18.3 units.