Problem 24 - Algebra 2 and Trigonometry

Question

The altitude, \(\overline{C D}\) , to the hypotenuse, \(\overline{A B}\) , of right triangle \(A B C\) separates the hypotenuse into two segments, \(\overline{A D}\) and \(\overline{D B}\) . If \(A D=D B+4\) and \(C D=12\) centimeters, find \(D B, A D,\) and \(A B\) . Recall that the length of the altitude to the hypotenuse of a right triangle is the mean proportional between the lengths of the segments into which the hypotenuse is separated, that is, \(\frac{A D}{C D}=\frac{C D}{D B}\)

Step-by-Step Solution

Answer

DB = 10 cm, AD = 14 cm, AB = 24 cm.

Step 1: Set up the Proportional Equation

According to the given mean proportional relationship, we have the equation \( \frac{AD}{CD} = \frac{CD}{DB} \). Substitute the values: \( CD = 12 \), giving \( \frac{AD}{12} = \frac{12}{DB} \).

Step 2: Cross Multiply to Solve for AD

Cross multiply the proportion to solve for \( AD \):\[ AD \cdot DB = 12 \cdot 12 \]\[ AD \cdot DB = 144 \]

Step 3: Use the Given Relationship Between AD and DB

We know \( AD = DB + 4 \). Substitute this into the equation from Step 2:\[ (DB + 4) \cdot DB = 144 \]

Step 4: Expand and Rearrange the Equation

Expand and rearrange the equation into a quadratic form:\[ DB^2 + 4DB - 144 = 0 \]

Step 5: Solve the Quadratic Equation

Use the quadratic formula \( DB = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to solve:Here, \( a = 1 \), \( b = 4 \), and \( c = -144 \).\( DB = \frac{-4 \pm \sqrt{16 + 576}}{2} \)\( DB = \frac{-4 \pm 24}{2} \)This gives two possible solutions: \( DB = 10 \) or \( DB = -14 \).Since length cannot be negative, use \( DB = 10 \).

Step 6: Solve for AD Using DB

Substitute \( DB = 10 \) back into the relationship \( AD = DB + 4 \):\[ AD = 10 + 4 = 14 \]

Step 7: Find AB Using Segments AD and DB

The hypotenuse \( AB \) is the sum of segments \( AD \) and \( DB \):\[ AB = AD + DB = 14 + 10 = 24 \]

Key Concepts

Right Triangle

A right triangle is a type of triangle that includes one angle precisely equal to 90 degrees. This special angle is often useful because it allows us to apply various mathematical properties and theorems, such as the Pythagorean theorem.

In our exercise, triangle ABC is a right triangle, meaning that one of its angles—the one at point C—is a right angle. Because of this, the side opposite the right angle is known as the hypotenuse. The hypotenuse is the longest side in a right triangle.

• The side opposite the right angle
• Commonly involved in Pythagorean theorem calculations
• Often associated with other geometric properties, such as mean proportionals

This characteristic sets the stage for various calculations and proofs, like finding segment lengths or relating elements through ratios.

Mean Proportional

The mean proportional in a right triangle is a fascinating concept connecting geometry with algebra. It's an integral part of solving problems dealing with altitudes drawn to the hypotenuse of a right triangle.

Basically, if you drop an altitude from the right angle to the hypotenuse, you create two smaller segments on the hypotenuse. In our problem, segment CD is the altitude, and it divides hypotenuse AB into segments AD and DB. The mean proportional theorem states:

• The altitude is the mean proportional between the two segments.
• Mathematically, this is given by: \( \frac{AD}{CD} = \frac{CD}{DB} \).
• This relationship allows us to set up certain quadratic equations to find missing lengths.

Understanding the role of the mean proportional is crucial for solving these kinds of geometry problems. It helps in setting up the appropriate equations that allow you to figure out lengths that might not be otherwise evident.

Quadratic Equation

A quadratic equation is a polynomial equation of degree two. It takes the form: \( ax^2 + bx + c = 0 \). In the context of geometric problems, quadratic equations often come into play when dealing with relationships involving squares.

In this exercise, once the proportional relationship was established, solving for the lengths involved required setting up and solving a quadratic equation. By expanding and rearranging terms like \((DB + 4) \cdot DB = 144\), we derive the quadratic equation:

• \( DB^2 + 4DB - 144 = 0 \)
• Use of the quadratic formula: \( DB = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
• Resulting in the realistic length of segment DB being positive

Quadratic equations are vital for finding real, applicable solutions to geometry problems, allowing a deeper understanding of inherent relationships.

Hypotenuse

The hypotenuse is one of the most recognizable features in a right triangle. It stands as the longest side, positioned opposite the right angle.

In geometric problems like ours, the hypotenuse can be dissected, creating relationships that lead to solutions for otherwise hidden elements. Specifically, once the altitude is dropped, the hypotenuse gets split into segments AD and DB.

• The altitude and the segments maintain a unique mean proportional relationship.
• The overall length of the hypotenuse (AB) is the sum of these two segments: \( AB = AD + DB \).
• This sum helps close the loop on finding lengths once you establish their individual measures.

Understanding the properties of the hypotenuse plays a crucial role in solving right triangle problems effectively. It's from here that many fundamental insights into the triangle's structure stem.