Question

Find the perimeter and the area of the polygon with the given vertices. A (12,2), B (12,13), C (15,13), D (15,2)

Answer 1

Alright, lets get started.

Please have a look at the diagram I have attached.

There are 4 vertices means A (12,2) , B (12,13), C (15,13) & D (15,2)

So, the side AB =
`13-2 =11`

side BC =
`15-12=3`

side CD =
`13-2= 11`

side DA =
`15-12=3`

So the perimeter =
`2 (a+b)`

Perimeter =
`2 (11+3)`

Perimeter =
`2 * 14 = 28`

Area=
`a*b`

Area =
`11*3 = 33`

So the perimeter is 28 and area is 33. : Answer

Hope it will help :)

Answer 2

Perimeter: 28, Area: 33

Explanation:

The perimeter of the rectangle is given by the sum of the length of each side:

`AB=√((x_A-x_B)^2+(y_A-y_B)^2)=√((12-12)^2+(2-13)^2)=√(11^2)=11`

`BC=√((x_B-x_C)^2+(y_B-y_C)^2)=√((12-15)^2+(13-13)^2)=√(3^2)=3`

`CD=√((x_C-x_D)^2+(y_C-y_D)^2)=√((15-15)^2+(13-2)^2)=√(11^2)=11`

`DA=√((x_D-x_A)^2+(y_D-y_A)^2)=√((15-12)^2+(2-2)^2)=√(3^2)=3`

So, the perimeter is

`p=11+3+11+3=28`

Concerning the area, for a rectangle the area is equal to the product of length and width:

`A=Lw`

In this case, the length is L=11 and the width is w=3, so the area is

`A=Lw=(11)(3)=33`