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
\(100\ m^{2}\)
