Answer:
The area of the smallest section is [tex]A_{1}=100yd^{2}[/tex]
The area of the largest section is [tex]A_{2}=625yd^{2}[/tex]
The area of the remaining section is [tex]A_{3}=250yd^{2}[/tex]
Step-by-step explanation:
Please see the picture below.
1. First we are going to name the side of the larger square as x.
As the third section shares a side with the larger square and the four sides of a square are equal, we have the following:
- Area of the first section:
[tex]A_{1}=10yd*10yd[/tex]
[tex]A_{1}=100yd^{2}[/tex]
- Area of the second section:
[tex]A_{2}=x^{2}[/tex] (Eq.1)
- Area of the third section:
[tex]A_{3}=width*length[/tex]
[tex]A_{3}=10yd*x[/tex] (Eq.2)
2. The problem says that the total area of the enclosed field is 975 square yards, and looking at the picture below, we have:
[tex]A_{1}+A_{2}+A_{3}=975yd^{2}[/tex]
Replacing values:
[tex]100+x^{2}+10x=975[/tex]
Solving for x:
[tex]x^{2}+10x-875=0[/tex]
[tex]x=\frac{-10+\sqrt{100+(4*875)}}{2}[/tex]
[tex]x=\frac{-10+\sqrt{3600}}{2}[/tex]
[tex]x=\frac{-10+60}{2}[/tex]
[tex]x=25[/tex]
3. Replacing the value of x in Eq.1 and Eq.2:
- From Eq.1:
[tex]A_{2}=25^{2}[/tex]
[tex]A_{2}=625yd^{2}[/tex]
- From Eq.2:
[tex]A_{3}=10*25[/tex]
[tex]A_{3}=250yd^{2}[/tex]
