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

Answer:

Perimeter: 28, Area: 33

Step-by-step explanation:

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

$AB=\sqrt{(x_A-x_B)^2+(y_A-y_B)^2}=\sqrt{(12-12)^2+(2-13)^2}=\sqrt{11^2}=11$

$BC=\sqrt{(x_B-x_C)^2+(y_B-y_C)^2}=\sqrt{(12-15)^2+(13-13)^2}=\sqrt{3^2}=3$

$CD=\sqrt{(x_C-x_D)^2+(y_C-y_D)^2}=\sqrt{(15-15)^2+(13-2)^2}=\sqrt{11^2}=11$

$DA=\sqrt{(x_D-x_A)^2+(y_D-y_A)^2}=\sqrt{(15-12)^2+(2-2)^2}=\sqrt{3^2}=3$

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

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

Answer 2

Answer:

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 :)