Answer :
Answer:
Length = 50 units
width = 35 units
Step-by-step explanation:
Let A, B, C and D be the corner of the pools.
Given:
The points of the corners are.
[tex]A(x_{1}, y_{1}})=(-20, 25)[/tex]
[tex]B(x_{2}, y_{2}})=(30, 25)[/tex]
[tex]C(x_{3}, y_{3}})=(30, -10)[/tex]
[tex]D(x_{4}, y_{4}})=(-20, -10)[/tex]
We need to find the dimension of the pools.
Solution:
Using distance formula of the two points.
[tex]d(A,B)=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}[/tex]----------(1)
For point AB
Substitute points A(30, 25) and B(30, 25) in above equation.
[tex]AB=\sqrt{(30-(-20))^{2}+(25-25)^{2}}[/tex]
[tex]AB=\sqrt{(30+20)^{2}}[/tex]
[tex]AB=\sqrt{(50)^{2}[/tex]
AB = 50 units
Similarly for point BC
Substitute points B(-20, 25) and C(30, -10) in equation 1.
[tex]d(B,C)=\sqrt{(x_{3}-x_{2})^{2}+(y_{3}-y_{2})^{2}}[/tex]
[tex]BC=\sqrt{(30-30)^{2}+((-10)-25)^{2}}[/tex]
[tex]BC=\sqrt{(-35)^{2}}[/tex]
BC = 35 units
Similarly for point DC
Substitute points D(-20, -10) and C(30, -10) in equation 1.
[tex]d(D,C)=\sqrt{(x_{3}-x_{4})^{2}+(y_{3}-y_{4})^{2}}[/tex]
[tex]DC=\sqrt{(30-(-20))^{2}+(-10-(-10))^{2}}[/tex]
[tex]DC=\sqrt{(30+20)^{2}}[/tex]
[tex]DC=\sqrt{(50)^{2}}[/tex]
DC = 50 units
Similarly for segment AD
Substitute points A(-20, 25) and D(-20, -10) in equation 1.
[tex]d(A,D)=\sqrt{(x_{4}-x_{1})^{2}+(y_{4}-y_{1})^{2}}[/tex]
[tex]AD=\sqrt{(-20-(-20))^{2}+(-10-25)^{2}}[/tex]
[tex]AD=\sqrt{(-20+20)^{2}+(-35)^{2}}[/tex]
[tex]AD=\sqrt{(-35)^{2}}[/tex]
AD = 35 units
Therefore, the dimension of the rectangular swimming pool are.
Length = 50 units
width = 35 units