Question

Rectangle ABCD is graphed in the coordinate plane.The following are the vertices of the rectangle:A(-2,1),B(3,2),C(3,4),and D(-2,4), what is the perimeter of rectangle ABCD

Answer 1

It wouldn’t be a rectangle as the Y coordinates for A and B would be non-leveled while the rest of the points were however you probably meant A(-2,1),B(3,1) which would have an answer of 16 if you used the points C and D to get the perimeter.

Answer 2

Therefore,

`\textrm{Perimeter of rectangle}=14\ cm`

Explanation:

Given:

Rectangle ABCD

vertices of the rectangle:

A(-2,2),

B(3,2),C(3,4),and D(-2,4),

Let,

Length of Rectangle be AB and Width be BC

To Find:

Perimeter of rectangle ABCD = ?

Solution:

We Know ,

`\textrm{Perimeter of rectangle}=2(Length + Width)`

Lets find length and width by Distance Formula,

`l(AB) = \sqrt{((x_(2)-x_(1))^(2)+(y_(2)-y_(1))^(2) )}`

On substituting the values we get

`l(AB) = \sqrt{((3+2)^(2)+(2-2)^(2) )}`

`l(AB) = \sqrt{((5)^(2)}=5\ units`

Similarly,

`l(BC) = \sqrt{((3-3)^(2)+(4-2)^(2) )}`

`l(BC) = √(4)=2\ units`

Now substituting AB and BC in Perimeter Formula we get

`\textrm{Perimeter of rectangle}=2(AB + BC)=2(5+2)=2* 7=14\ cm`

Therefore,

`\textrm{Perimeter of rectangle}=14\ cm`

Note Here Vetex A is wrong the correct on is A(-2 ,2)