Question

What is the exact perimeter of a parallelogram with vertices at (3,2), (4,4), and (6,1)?

Answer 1

We are given three vertices, but we do not know which two are adjacent. We could draw these points to find out. But that is not neccesary.

In a parallelogram we can draw four types of lines. Two types are sides and two types are diagonals. Diagonals are lngre than sides. This helps us to solve this problems. If we find distance between all points we will take only two smaller numbers because they represent sides.

Distance between two points is given by:
$d= \sqrt{ ( x_{2}- x_{1} )^{2} + (y_{2}- y_{1})^{2} }$

For points (3,2) and (4,4) distance is:
$d= \sqrt{ ( 4- 3 )^{2} + (4-2)^{2} } = \sqrt{1+4} = \sqrt{5}$
For points (3,2) and (6,1) distance is:
$d= \sqrt{ ( 6- 3 )^{2} + (1-2)^{2} } = \sqrt{9+1} = \sqrt{10}$
For points (4,4) and (6,1) distance is:
$d= \sqrt{ ( 6-4 )^{2} + (1-4)^{2} } = \sqrt{4+9} = \sqrt{13}$

Last distance is greatest and it represents a diagonal. Other two distances represent sides.
Perimeter of a parallelogram is given by:
$P=2*(a+b)=2*( \sqrt{5} + \sqrt{10} )=10.8$