# A parallelogram is formed by vectors vectors OA=(2,3) and OB=(1,1). Find:Find the perimeter of the parallelogram.

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You need to find the length of sides of parallelogram to find the perimeter.

Notice that magnitudes of vectors provides the lengths of sides, hence you need to evaluate the magnitudes of vectors `bar OA ` and `bar OB` such that:

`bar OA = sqrt((2-0)^2 + (3-0)^2)`

`bar OA = sqrt(4+9) `

`bar OA = sqrt 13`

`bar OB = sqrt((1-0)^2 + (1-0)^2)`

`bar OB = sqrt 2`

You need to remember that facing sides of parallelogram are of equal lengths, hence evaluating the perimeter of paralleogram yields:

`P = 2(sqrt13 + sqrt2)`

**Hence, evaluating the perimeter of parallelogram yields `P = 2(sqrt13 + sqrt2).` **

Take o as the origin which is (0,0). get a graph paper and go plotting the points itz much easier to find the 4th vertex and the perimeter

Theres not much of a explanation here but just go to the below site this might come in handy to you.............

Let OACB be the parallelogram, where O is the origin, A is (2,1) , B is (1,1) and D is the fourth vertex.

Let M be the point of intersection of the diagonals.

Then M will be ( 2+1 /2 , 1+1 /2)

that is, (3/2,1).

Hence D will have co.ords ( 2*3/2 -0 , 2*1 - 0)

ie.,D (3,2)

Lengths of diagonals:

OC = sqrt.(3^2 +2 ^2) =sqrt. (9+4) =sqrt. 13 units

AB =sqrt( 2-1 ^2 + 1-1 ^2)= 1 unit

Lengths of the sides:

OA = sqrt.(2^2 + 1^2 ) =sqrt.5 units

AC = sqrt.( 3-2 ^2 + 2-1 ^2) =sqrt2 units

Hence, perimeter = 2( sqrt.5 + sqrt. 2 ) units.