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Chapter 5: Problem 68

A surveying team wants to calculate the length of a straight tunnel through a mountain. They form a right angle by connecting lines from each end of the proposed tunnel. One of the connecting lines is 6 miles, and the other is 8 miles. What is the length of the proposed tunnel?

Short Answer

Expert verified
The length of the proposed tunnel is 10 miles.

Step by step solution

01

Identify the Given Components

In the problem, the surveying team creates a right triangle with the proposed tunnel as the hypotenuse. The two other lines are the legs of the triangle, 6 miles and 8 miles long, respectively.
02

Apply the Pythagorean Theorem

The Pythagorean Theorem relates the sides of a right triangle with the formula \( c^2 = a^2 + b^2 \), where \( c \) is the hypotenuse and \( a \) and \( b \) are the other two sides. Here, the hypotenuse is the length of the tunnel, represented as \( c \), and the two legs are 6 and 8 miles.
03

Calculate Each Side's Square

Compute the square of the lengths of the two legs: \( 6^2 = 36 \) and \( 8^2 = 64 \).
04

Solve for the Hypotenuse

Add the squares of the two legs: \( 36 + 64 = 100 \). Then, take the square root to find the hypotenuse: \( c = \sqrt{100} = 10 \).
05

Determine the Length of the Tunnel

Based on the calculations, the length of the tunnel is 10 miles.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Right Triangle
A right triangle is a triangle that has one angle exactly equal to 90 degrees. This means two of its sides form a perfect "L" shape, with the right angle at the corner.
The two sides that form the right angle are known as the "legs" of the triangle, and the third, largest side is called the "hypotenuse." The right triangle is pivotal in geometry, often forming the basis for various calculations and applications.

In the given exercise, a right triangle is used to model the path of a tunnel through a mountain. By understanding the structure of a right triangle, students can easily apply mathematical principles鈥攍ike the Pythagorean Theorem鈥攖o solve real-world problems.
  • The legs of the right triangle in the exercise are 6 miles and 8 miles long.
  • The hypotenuse, which they need to find, represents the tunnel鈥檚 length.
Hypotenuse
The hypotenuse is the longest side of a right triangle, always opposite the right angle. It represents the direct line connecting the two points that form the right angle.
Historically, the hypotenuse has been crucial for various calculation tasks, especially in fields like architecture, engineering, and surveying.

In the context of our exercise, the hypotenuse is the length of the tunnel that the surveying team wants to construct. To find it, they will use the Pythagorean Theorem:
  • The theorem states that in a right triangle, the square of the hypotenuse (\(c\)) is equal to the sum of the squares of the other two sides (\(a\) and \(b\)).
  • Calculate using: \(c^2 = a^2 + b^2\).
  • Given: \(6^2 + 8^2 = c^2\) and therefore, \(c = 10\) miles.
This calculation tells us the shortest path through the mountain is 10 miles, demonstrating the hypotenuse鈥檚 practical utility.
Prealgebra
Prealgebra lays the foundation for more advanced mathematical learning, covering basic arithmetic concepts, simple equations, and introducing geometric concepts.
In this exercise, we're utilizing prealgebra skills to apply a fundamental formula known as the Pythagorean Theorem. Together, these concepts help simplify the process of solving problems involving right triangles.

Understanding prealgebra concepts, such as:
  • Identifying different sides of a triangle.
  • Squaring numbers (e.g., \(6^2 = 36\), \(8^2 = 64\)).
  • Adding and taking square roots to solve for unknowns heretofore.
are critical building blocks for tackling more complex mathematical challenges. Students can take these skills and apply them in various disciplines outside mathematics, like natural sciences and different aspects of engineering.

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Most popular questions from this chapter

Write each fraction as an equivalent fraction with denominator \(15 x\). $$\frac{4}{x}$$

Perform each of the following divisions. $$2 . 3 \longdiv { 0 . 1 1 5 }$$

Reduce to lowest terms. $$\frac{75}{100}$$

The problems below review some of the material on solving equations. Reviewing these problems will help you with the next section. Solve. $$6^{2}+8^{2}$$

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