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To find the perimeter of the parallelogram, we use the formula of distance between points:

[tex] d = \sqrt{(x2-x1)^2 + (y2-y1)^2} [/tex]

We have then:

For WX:

[tex] WX = \sqrt{(2+2)^2 + (4-4)^2}

WX = \sqrt{(4)^2 + (0)^2}

WX = \sqrt{16 + 0}

WX = \sqrt{16}

WX = 4 [/tex]

For XY:

[tex] XY = \sqrt{(2-1)^2 + (4+1)^2}

XY = \sqrt{(1)^2 + (5)^2}

XY = \sqrt{1 + 25}

XY = \sqrt{26}
[/tex]

Then, since the other two sides are parallel and have the same length, the perimeter of the parallelogram is given by:

[tex] P = 2WX + 2XY
[/tex]

Substituting values we have:

[tex] P = 2(4) + 2\sqrt{26}

P = 8 + 2\sqrt{26}

[/tex]

Answer:

The perimeter of parallelogram WXYZ is 2 √ 26 + 8 units.

The perimeter of parallelogram WXYZ is 2 [tex]\sqrt{26}[/tex] + 8 units

Given that,

The perimeter of parallelogram WXYZ is 2√ 26 + ______,units.

Co-ordinate are = w(-2,4), x(2,4), z(-3,-2),x(2,4),

To find; The perimeter of the parallelogram, we use the formula of distance between points:

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

Then:

For WX The co-ordinate are w(-2,4), x(2,4),

[tex]d = \sqrt{(x_2-x_1) ^{2} + (y_2-y_1)^{2} }\\d = \sqrt{(2-(-2))^{2} + (4-4)}^{2} \\d = \sqrt{16-0} \\d = \sqrt{16} \\d = 4[/tex]

And for XY: the co-ordinate are; x(2,4),x(2,4),

[tex]d = \sqrt{(x_2-x_1) ^{2} + (y_2-y_1)^{2} }\\ d = \sqrt{(1-(2){ ^{2} + (-1-(4))^{2} } \\\\[/tex]

[tex]d = \sqrt{1^{2} + 5^{2} } \\d = \sqrt{1+25} \\d = \sqrt{26}[/tex]

Then, since the other two sides are parallel and have the same length, the perimeter of the parallelogram is given by:

Perimeter = 2Wx + 2xY

Substituting values we have:

Perimeter = 2(4) + 2[tex]\sqrt{26}[/tex]

The perimeter of parallelogram WXYZ is 2  [tex]\sqrt{26}[/tex] + 8 units.

For the more information about Parallelogram click the link given below.

https://brainly.com/question/2773823

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