Solving Right Triangles with the Hypotenuse and a Leg Length Difference

This article covers the process of finding the lengths of the legs of a right triangle when only the hypotenuse and the difference in leg lengths are known. We will explore how to apply the Pythagorean theorem to solve this problem and provide step-by-step instructions along with numerical examples for clarity. Additionally, I will discuss the importance of quadratic equations and how to interpret their solutions in the context of geometric lengths.

Introduction to Right Triangles

A right triangle is a triangle that contains one right angle of 90 degrees. The side opposite the right angle is called the hypotenuse, which is the longest side of the triangle. The other two sides are referred to as the legs. The Pythagorean theorem states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the legs. Mathematically, this is represented as:

a^2 b^2 c^2

Solving the Given Triangle Problem

Consider a right triangle where the hypotenuse is 8 cm long, and the longer leg is 2 cm longer than the shorter leg. Let's denote the length of the shorter leg by a. Then, the length of the longer leg can be expressed as a 2. Using the Pythagorean theorem, we have:

a^2 (a 2)^2 8^2

Expanding and simplifying this equation:

a^2 a^2 4a 4 64

2a^2 4a 4 64

2a^2 4a - 60 0

Dividing the entire equation by 2:

a^2 2a - 30 0

We can solve this quadratic equation by factoring it:

(a 6)(a - 5) 0

Setting each factor to zero gives us:

a 6 0 Rightarrow a -6 Not valid since lengths cannot be negative

a - 5 0 Rightarrow a 5

Thus, the shorter leg is a 5 cm, and the longer leg is:

a 2 5 2 7 cm

Alternative Solution Method

Let the length of the longer leg be x. Using the Pythagorean theorem:

8^2 x^2 (x - 2)^2

64 x^2 x^2 - 4x 4

2x^2 - 4x 4 64

2x^2 - 4x - 60 0

Dividing by 2:

x^2 - 2x - 30 0

Solving using the quadratic formula:

x frac{-(-2) pm sqrt{(-2)^2 - 4(1)(-30)}}{2(1)} frac{2 pm sqrt{4 120}}{2} frac{2 pm sqrt{124}}{2} frac{2 pm 2sqrt{31}}{2}

Thus:

x 1 pm sqrt{31}

Since lengths must be positive, we choose:

x 1 sqrt{31} approx 6.567764363

Summary

The lengths of the legs of the triangle are:

These solutions demonstrate the application of the Pythagorean theorem and quadratic equations to solve geometric problems.