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1. Ms. Walker has a rectangular pool that is 23 feet long and 20 feet wide. Find the length of the diagonal of the pool to the nearest tenth.

2. You are trying to determine the distance across a pond. You make a triangle [tex]\( ABC \)[/tex] where [tex]\( B \)[/tex] is the right angle. You measure to find that the distance [tex]\( AB \)[/tex] is 18 feet and the distance [tex]\( AC \)[/tex] is 33 feet. What is the distance [tex]\( CB \)[/tex] to the nearest foot?

Sagot :

GinnyAnswer
To solve the given problems, we'll use the Pythagorean Theorem, which is applicable in right-angled triangles. The theorem states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides.

### 1. Length of the diagonal of Ms. Walker's pool

Ms. Walker's pool is a rectangle where the length is 23 feet and the width is 20 feet. The diagonal of the rectangle forms the hypotenuse of a right-angled triangle with sides equal to the length and the width of the pool.

Let's denote:
- [tex]\(A\)[/tex] as the length of the pool: 23 feet
- [tex]\(B\)[/tex] as the width of the pool: 20 feet
- [tex]\(D\)[/tex] as the diagonal of the pool

The diagonal [tex]\(D\)[/tex] is found using the Pythagorean Theorem:

[tex]\[ D = \sqrt{A^2 + B^2} \][/tex]

Plugging in the provided values:

[tex]\[ D = \sqrt{23^2 + 20^2} \][/tex]
[tex]\[ D = \sqrt{529 + 400} \][/tex]
[tex]\[ D = \sqrt{929} \][/tex]
[tex]\[ D \approx 30.5 \][/tex]

Therefore, the length of the diagonal of the pool is approximately 30.5 feet.

### 2. Distance across the pond (distance CB)

In triangle [tex]\(ABC\)[/tex], [tex]\(B\)[/tex] is the right angle. We are given:
- [tex]\(AB = 18\)[/tex] feet (one leg of the triangle)
- [tex]\(AC = 33\)[/tex] feet (the hypotenuse of the triangle)

We need to find the distance [tex]\(CB\)[/tex], which is the other leg of the triangle.

By the Pythagorean Theorem:

[tex]\[ AC^2 = AB^2 + CB^2 \][/tex]

Solving for [tex]\(CB\)[/tex]:

[tex]\[ CB = \sqrt{AC^2 - AB^2} \][/tex]

Plugging in the given values:

[tex]\[ CB = \sqrt{33^2 - 18^2} \][/tex]
[tex]\[ CB = \sqrt{1089 - 324} \][/tex]
[tex]\[ CB = \sqrt{765} \][/tex]
[tex]\[ CB \approx 28 \][/tex]

Therefore, the distance from [tex]\(CB\)[/tex] across the pond is approximately 28 feet.
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