Question

For the final exam in a scuba diving certification course, Karl navigates from one point in a lake to another. Karl begins the test \(x\) meters directly beneath the boat and swims straight down toward the bottom of the lake for 8 meters. He then turns to his right and swims in a straight line parallel to the surface of the lake and swims 24 meters, at which point he swims directly from his location, in a straight line, back to the boat. If the distance that Karl swims back to the boat is 26 meters, what is the value of \(x ?\)

Short answer

The depth from where Karl initially started from is 10 meters.

Step-by-step solution

Identify Given Values

We know that Karl swims straight down for 8 meters, then parallel to the surface for 24 meters, and finally back to the boat which forms the hypotenuse of 26 meters. We represent the initial depth where Karl started from with the variable \(x\). Hence we have: \[x= ? (unknown)\] \[a = 8 (Height of the triangle)\] \[b = 24 (Base of the triangle)\] \[c = 26 (Hypotenuse of the triangle)\]

Applying the Pythagorean theorem

We apply the Pythagorean theorem to solve for the value of \(x\), where \(x\) is represented as length \(a\), but 'downwards'. So, we will use the values of \(b = 24\) and \(c = 26\) where the equation becomes: \[x^2 = c^2 - b^2\] and substituting the given values, we get: \[x^2 = 26^2 - 24^2\]

Calculating for \(x\)

We do the subtractions and solve for \(x\). \[x^2 = 676 - 576\] \[x^2 = 100\] Taking the square root on both sides because \(x\) is the square root of 100, we find: \[x = 10\]

Key concepts

Pythagorean Theorem

The Pythagorean theorem is a cornerstone concept in geometry, especially when dealing with right triangles. A right triangle is one where one of its angles is 90 degrees. In such triangles, the side opposite the right angle is known as the hypotenuse, and the other two sides are the legs of the triangle. The Pythagorean theorem states that the square of the hypotenuse's length is equal to the sum of the squares of the lengths of the other two sides.

Expressed mathematically, if we denote the legs as \( a \) and \( b \), and the hypotenuse as \( c \), the theorem gives us:

• \( c^2 = a^2 + b^2 \)
• It's a crucial tool for calculating distances and verifying right triangles.

In our exercise, Karl's swimming forms a right triangle. He swims downward, turns, swims parallel to the lake surface, and finally swims diagonally back to the boat. By applying the Pythagorean theorem, we determine unknown distances in this right triangle using the known distances Karl swims.

Right Triangles

Right triangles have unique properties that allow us to use special mathematical rules, such as the Pythagorean theorem. In a right triangle, one angle is 90 degrees, which splits the triangle into two distinct parts, simplifying the process of calculating unknown distances and angles.

Some key features include:

• One angle is always 90 degrees.
• The hypotenuse is the longest side, across from the right angle.
• Calculations such as finding lengths are simplified with specialized formulas like the Pythagorean theorem.

In our situation, the problem involving Karl shows a right triangle formation when he navigates back to the boat. Knowing two sides (8 meters and 24 meters), we only need the hypotenuse, which is 26 meters, to solve using the theorem. This setup is classic for employing right triangle concepts to find unknowns.

Distance in Geometry

Understanding distance in geometry involves recognizing various dimensions and formations in shapes and spaces. The distance we calculate might represent length, width, depth, or a diagonal line. In many geometry problems, such as the one involving Karl, we look at these distances on different planes.

When it comes to right triangles and using the Pythagorean theorem:

• Distances on the plane of the triangle can be explored using the lengths of the sides.
• The hypotenuse represents the longest distance in a right triangle and is crucial for our calculations.

In our problem, Karl's swimming creates a right triangle where the distance to be solved includes his depth beneath the water, the parallel swim across the lake’s surface, and the diagonal back to the boat. Solving for \( x \) in this context helps students grasp how the geometry of the problem dictates the calculation of distances in space.