Answer:
The area of the smallest section is 
The area of the largest section is 
The area of the remaining section is 
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:
- Area of the second section:
(Eq.1)
- Area of the third section:
(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:
Replacing values:
Solving for x:
3. Replacing the value of x in Eq.1 and Eq.2:
- From Eq.1:
- From Eq.2:
Answer:
42 ft²
Step-by-step explanation:
Rectangle Area: L * W
Let's try plugging in some negative numbers. Let's do x=-1. 5+-1=4. So we know that if we put in a negative number for x, then n will be positive. But what if we do a number greater than -5, because 5+-5=0. So let's try x=-6. So 5+(-6)=-1. Hmm. So here it is. We know that any number under -5 will be positive and any number above -5 will be negative.
Answer:
The answer is option A) <em>r</em><em> </em>and <em>s</em>
Length: 2w + 59
width: w
diagonal: (2w + 59) + 2 = 2w + 61
Length² + width² = diagonal²
(2w + 59)² + (w)² = (2w + 61)²
(4w² + 118w + 3481) + w² = 4w² + 122w + 3721
5w² + 118w + 3481 = 4w² + 122w + 3721
w² + 118w + 3481 = 122w + 3721
w² - 4w + 3481 = 3721
w² - 4w - 240 = 0
a = 1, b = -4, c = -240
w = ![[-(b) +/- \sqrt{(b)^{2} - 4(a)(c) }]/2(a)](https://tex.z-dn.net/?f=%5B-%28b%29%20%2B%2F-%20%5Csqrt%7B%28b%29%5E%7B2%7D%20%20-%204%28a%29%28c%29%20%7D%5D%2F2%28a%29)
= ![[-(-4) +/- \sqrt{(-4)^{2} - 4(1)(-240) }]/2(1)](https://tex.z-dn.net/?f=%5B-%28-4%29%20%2B%2F-%20%5Csqrt%7B%28-4%29%5E%7B2%7D%20%20-%204%281%29%28-240%29%20%7D%5D%2F2%281%29)
= ![[4 +/- \sqrt{(16 + 960) }]/2](https://tex.z-dn.net/?f=%5B4%20%2B%2F-%20%5Csqrt%7B%2816%20%20%2B%20960%29%20%7D%5D%2F2)
= ![[4 +/- \sqrt{(976) }]/2](https://tex.z-dn.net/?f=%5B4%20%2B%2F-%20%5Csqrt%7B%28976%29%20%7D%5D%2F2)
= ![[2 +/- 4\sqrt{(61) }]/2](https://tex.z-dn.net/?f=%5B2%20%2B%2F-%204%5Csqrt%7B%2861%29%20%7D%5D%2F2)
= 
since width cannot be negative, disregard 1 - 2√61
w = 1 + 2√61 ≈ 16.62
Length: 2w + 59 = 2(1 + 2√61) + 59 = 2 + 4√61 + 59 = 61 + 4√61 ≈ 92.24
Answer: width = 16.62 in, length = 92.24 in
