Question

What is the perimeter of parallelogram JKLM, whose coordinates are $J\left( -5,2 \right),K\left( -2,6 \right),L\left( 5,6 \right),M\left( 2,2 \right)$?
A. 30 units
B. 24 units
C. 28 units
D. 21 units

Answer 1

Hint: Draw a rough diagram of parallelogram JKLM. Now, perimeter is the sum of the path of any closed figure, so add the lengths of the sides of the parallelogram. Use the distance formula for calculating distance between two points as
$\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$
Where one point has $\left( {{x}_{1}},{{y}_{1}} \right)$ coordinate and second point has $\left( {{x}_{2}},{{y}_{2}} \right)$ coordinate.

Complete Step-by-Step solution:
We have coordinates of the parallelogram JKLM as
$J\left( -5,2 \right)$, $K\left( -2,6 \right)$, $L\left( 5,6 \right)$, $M\left( 2,2 \right)$.
So, we can draw diagram of the parallelogram JKLM as

As we need to determine the perimeter of the parallelogram JKLM as shown above with the help of the given coordinates of points J, K, L and M.
Now, as we know, the perimeter is the continuous line forming the boundary of a closed geometric figure. So, we can calculate the perimeter of the parallelogram by summing the sides of them.
Hence, we get
Perimeter of parallelogram JKLM $=MJ+JK+kL+ML$ ……………………….(i)
Now, we know that opposite sides of a parallelogram are equal and parallel to each other. So, sides ‘MJ’ and ‘KL’ should be equal and sides ‘MJ’ and ‘KL’ should be equal to each other.
So, we can replace KL and MJ and ML by JK in the equation (i). So, we get perimeter of parallelogram as
Perimeter of parallelogram JKLM $=MJ+JK+kL+ML$
$=2MJ+2JK$
Perimeter $=2\left( MJ+JK \right)$ ……………………………..(ii)
Now, we need to calculate the length of the sides MJ and JK from the given coordinates of the point J, K, L, M.
So, we know that distance between two points $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ can be calculated by distance formula, which is given as
Distance between points $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ $=\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$ ……………………….(iii)
So, we can calculate the distance M and JK using the distance formula expressed in equation (iii). So, we get
MJ $=\sqrt{{{\left( 2-\left( -5 \right) \right)}^{2}}+{{\left( 2-2 \right)}^{2}}}=\sqrt{{{\left( 7 \right)}^{2}}+{{\left( 0 \right)}^{2}}}$
MJ $=\sqrt{{{\left( 7 \right)}^{2}}}=7$
So, distance MJ is given as MJ $=7units$
Now, distance JK can be calculated by the same formula written in equation (iii). So, we get distance JK as
JK $=\sqrt{{{\left( -5-\left( -2 \right) \right)}^{2}}+{{\left( 2-6 \right)}^{2}}}=\sqrt{{{\left( -5+2 \right)}^{2}}+{{\left( -4 \right)}^{2}}}$
JK $=\sqrt{{{\left( -3 \right)}^{2}}+{{\left( -4 \right)}^{2}}}=\sqrt{9+16}$
JK $=\sqrt{25}=\sqrt{{{\left( 5 \right)}^{2}}}=5units$
So, we get the lengths of sides JK and MJ as 5 units and 7 units respectively. Hence, we can put values of JK and MJ to the equation (ii) to get the perimeter of the given parallelogram. So, we get
Perimeter of parallelogram $=2\left( 5+7 \right)=2\times 12$
$=24units$
Hence, option (B) is the correct answer.

Note: One may go wrong if he/she will not represent the coordinates on a diagram. As one may calculate the distance JL or MK as well. So, be careful and observe the names of sides very carefully. Else one may add distance between the coordinates which are lying on the diagonals. So, first represent them roughly on the parallelogram and then proceed further for the perimeter.
One may apply direct formula of perimeter of a rectangle which is given as
Perimeter of rectangle $=2\left( length+breadth \right)$
It is possible because opposite sides are equal in rectangle and parallelogram both.
Don’t confuse it with the distance formula. Subtract ‘x’ coordinate of one point to another and ‘y’ coordinates of one to other. As one may go wrong if he/she applies the identity as $\sqrt{{{\left( {{x}_{1}}-{{y}_{1}} \right)}^{2}}+{{\left( {{x}_{2}}-{{y}_{2}} \right)}^{2}}}$, which is wrong. So, be clear with the positions of the coordinates.