Question

The coordinate plane shows the floor plan for a swimming pool. What is the area of the pool’s border?

Answer 1

we know that

The area of a rectangle is equal to

$A=L*W$

where

L is the length side of the rectangle

W is the width side of the rectangle

see the attached figure with letters to better understand the problem  

The area of the pool's border is equal to the area of rectangle ABCD minus the area of rectangle EFGH

The formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

Step 1

Find the distance AB

Let

$A(0,12)\ B(8,18)$

substitute the values in the formula

$d=\sqrt{(18-12)^{2}+(8-0)^{2}}$

$d=\sqrt{(6)^{2}+(8)^{2}}$

$d=\sqrt{100}$

$dAB=10\ m$

Step 2

Find the distance BC

Let

$B(8,18)\ C(17,6)$

substitute the values in the formula

$d=\sqrt{(6-18)^{2}+(17-8)^{2}}$

$d=\sqrt{(-12)^{2}+(9)^{2}}$

$d=\sqrt{225}$

$dBC=15\ m$

Step 3

Find the distance EF

Let

$E(3,12)\ F(7,15)$

substitute the values in the formula

$d=\sqrt{(15-12)^{2}+(7-3)^{2}}$

$d=\sqrt{(3)^{2}+(4)^{2}}$

$d=\sqrt{25}$

$d.EF=5\ m$

Step 4

Find the distance FG

Let

$F(7,15)\ G(13,7)$

substitute the values in the formula

$d=\sqrt{(7-15)^{2}+(13-7)^{2}}$

$d=\sqrt{(-8)^{2}+(6)^{2}}$

$d=\sqrt{100}$

$dFG=10\ m$

Step 5

Find the area of the pool's border

The area of the pool's border is equal to the area of rectangle ABCD minus the area of rectangle EFGH

Area of the rectangle ABCD

$A1=AB*BC=10\ m*15\ m=150\ m^{2}$

Area of the rectangle EFGH

$A2=EF*FG=5\ m*10\ m=50\ m^{2}$

Area of the pool's border

$A=A1-A2=150\ m^{2}-50\ m^{2}=100\ m^{2}$

therefore

the answer is

$100\ m^{2}$

Answer 2

For the answer to the question above asking what is the area of the pool’s border?
A = Area 
L= Length = 15 
W + widith = 10 
A = 150 
The pool area is L = 10 
w= 5 
A= 50 
So the area of the perimeter = 100 square meters
I hope this helps