Problem 33
Question
Find the length of the shorter leg of a right triangle if the longer leg is 12 feet more than the shorter leg and the hypotenuse is 12 feet less than twice the shorter leg.
Step-by-Step Solution
Verified Answer
The length of the shorter leg is approximately 7.4 feet.
1Step 1: Define Variables
Let the length of the shorter leg be denoted by \( x \). The problem states that the longer leg is 12 feet more than the shorter leg, so its length can be represented as \( x + 12 \). The hypotenuse is 12 feet less than twice the shorter leg, so the hypotenuse can be represented as \( 2x - 12 \).
2Step 2: Apply the Pythagorean Theorem
According to the Pythagorean Theorem, we can relate the sides of a right triangle as follows:\[x^2 + (x + 12)^2 = (2x - 12)^2\]This equation states that the sum of the squares of the legs equals the square of the hypotenuse.
3Step 3: Expand and Simplify the Equation
Expand both squared terms in the equation:\[x^2 + (x + 12)^2 = x^2 + 24x + 144\]\[(2x - 12)^2 = 4x^2 - 48x + 144\]Substitute these into the Pythagorean theorem:\[x^2 + x^2 + 24x + 144 = 4x^2 - 48x + 144\]
4Step 4: Rearrange Terms
Bring all terms to one side of the equation to set it to zero:\[0 = 4x^2 - x^2 - x^2 - 24x - 144 + 48x - 144\]Simplify:\[2x^2 + 24x - 288 = 0\]
5Step 5: Simplify Further
Simplify the equation:\[x^2 + 12x - 144 = 0\]
6Step 6: Solve the Quadratic Equation
Use the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 12 \), and \( c = -144 \):\[x = \frac{-12 \pm \sqrt{12^2 - 4 \cdot 1 \cdot (-144)}}{2 \cdot 1}\]\[x = \frac{-12 \pm \sqrt{144 + 576}}{2}\]\[x = \frac{-12 \pm \sqrt{720}}{2}\]\[x = \frac{-12 \pm 26.832}{2}\]
7Step 7: Calculate and Determine Valid Solution
Calculate the two possible solutions:\[x = \frac{-12 + 26.832}{2} \approx 7.416\]\[x = \frac{-12 - 26.832}{2} \approx -19.416\]Since the length cannot be negative, the length of the shorter leg is approximately 7.416 feet.
Key Concepts
Right TriangleQuadratic FormulaSolving Quadratic Equations
Right Triangle
A right triangle is a type of triangle that has one angle measuring 90 degrees. It's one of the most common shapes studied in geometry and has special properties related to its sides and angles. In a right triangle:
- The side opposite the right angle is called the hypotenuse. It is the longest side.
- The other two sides are known as the legs. In our problem, these are referred to as the shorter and longer legs.
Quadratic Formula
The quadratic formula is a crucial tool for solving quadratic equations, which are equations of the form \( ax^2 + bx + c = 0 \). A quadratic equation can often appear in problems involving squares and products, as in our exercise. The quadratic formula is expressed as:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here's what each part means:
- \( a \), \( b \), and \( c \) are coefficients from the quadratic equation.
- \( b^2 - 4ac \) is called the discriminant, which determines the nature of the solutions. A positive discriminant indicates two real solutions, zero implies one real solution, and if negative, the solutions are complex.
Solving Quadratic Equations
Solving quadratic equations is a key skill in algebra, useful in many mathematical problems. To solve a quadratic equation like \( x^2 + 12x - 144 = 0 \), you can:
This method shows the direct path from establishing relationships via the Pythagorean Theorem to formulating and solving a quadratic equation for precise results.
- Factor the equation, if possible: Look for two numbers that multiply to \(-144\) and add to \(12\). If this is difficult, alternative methods such as completing the square or using the quadratic formula are needed.
- Use the quadratic formula: As outlined above, it provides a straightforward path to finding the roots of the equation.
This method shows the direct path from establishing relationships via the Pythagorean Theorem to formulating and solving a quadratic equation for precise results.
Other exercises in this chapter
Problem 32
Solve.$$ (x+3)(x+8)=x $$
View solution Problem 33
Factor each completely. $$ 16 r^{2}+1 $$
View solution Problem 33
Factor each trinomial by grouping. Exercises 9 through 12 are broken into parts to help you get started. $$ 54 a^{2}-9 a-30 $$
View solution Problem 33
Factor out the GCF from each polynomial. $$ 32 x y-18 x^{2} $$
View solution