Problem 41 - Algebra 2 and Trigonometry

Question

In \(\triangle A B C, A B=\sqrt{7 x+5, B C}=\sqrt{5 x+15},\) and \(A C=\sqrt{2 x}\) a. If \(A B=B C,\) find the length of each side of the triangle. b. Express the perimeter of the triangle in simplest radical form.

Step-by-Step Solution

Answer

Each side is \(AB = BC = 2\sqrt{10}\) and \(AC = \sqrt{10}\). The perimeter is \(5\sqrt{10}\).

Step 1: Equate the Given Sides

Given that \( AB = BC \), we equate the expressions for \( AB \) and \( BC \):\[\sqrt{7x+5} = \sqrt{5x+15}\]. We need to solve this equation to find the value of \(x\).

Step 2: Remove the Square Roots

Square both sides of the equation to eliminate the square roots:\[7x + 5 = 5x + 15\].

Step 3: Solve for \( x \)

Subtract \(5x\) from both sides to get:\[2x = 10\]. Then, divide both sides by 2 to find \(x\):\[x = 5\].

Step 4: Find the Length of Each Side

Substitute \(x = 5\) back into the expressions for the sides of the triangle. For \( AB \) and \( BC \):\[\sqrt{7(5)+5} = \sqrt{35+5} = \sqrt{40}\].\[\sqrt{5(5)+15} = \sqrt{25+15} = \sqrt{40}\].\[AB = BC = \sqrt{40}\]. For \( AC \), substitute as follows:\[\sqrt{2(5)} = \sqrt{10}\]. Thus, \(AB = BC = \sqrt{40}\) and \(AC = \sqrt{10}\).

Step 5: Simplify the Side Lengths

Simplify the radical expressions: \(\sqrt{40} = \sqrt{4 \cdot 10} = 2\sqrt{10}\). Thus, \(AB = BC = 2\sqrt{10}\). The length of \(AC\) remains \(\sqrt{10}\).

Step 6: Calculate the Perimeter

The perimeter \(P\) of the triangle is the sum of its sides. Compute as follows: \(P = AB + BC + AC = 2\sqrt{10} + 2\sqrt{10} + \sqrt{10} = 5\sqrt{10}\).

Key Concepts

Perimeter of a Triangle

The perimeter of a triangle is essentially the total distance around the triangle. It is calculated by adding together the lengths of its three sides. In mathematical terms, if a triangle has sides labeled as \( AB \), \( BC \), and \( AC \), then its perimeter \( P \) is the expression:\[ P = AB + BC + AC \].
For example, if you know the side lengths to be \( 2\sqrt{10} \), \( 2\sqrt{10} \), and \( \sqrt{10} \), you simply add them up. Here it would be \( 2\sqrt{10} + 2\sqrt{10} + \sqrt{10} = 5\sqrt{10} \).
This expresses how knowing the side lengths in simplest radical form can make calculations straightforward.

Solving Radical Equations

Radical equations are equations where the variable is within a radical, such as a square root. Solving them typically requires removing the radical sign by squaring both sides of the equation. Here’s how it works:

• If you have an equation like \( \sqrt{7x+5} = \sqrt{5x+15} \), the first step is to square both sides.
• This will eliminate the square roots, resulting in the equation \( 7x + 5 = 5x + 15 \).
• From here, solve for \( x \) by isolating the variable. Subtract \( 5x \) from both sides, yielding \( 2x = 10 \).
• Finally, dividing both sides by \( 2 \) gives \( x = 5 \).

Solving radical equations often involves these types of steps: squaring to remove radicals and finding the value of \( x \). This method is crucial in problems where sides of a triangle are given in radical form.

Simplifying Radical Expressions

Simplifying radical expressions is crucial for making calculations easier and more interpretable. A radical expression can often be rearranged into a simpler form.
To illustrate, consider the expression \( \sqrt{40} \). Simplification involves finding a factor of \( 40 \) that is a perfect square. This can be written as \( \sqrt{4 \cdot 10} \). Since \( 4 \) is a perfect square, it becomes \( 2 \) when simplified, giving \( 2\sqrt{10} \).
Here’s the step-by-step process:

• Identify factors of the number inside the square root that include perfect squares.
• Simplify the square root of the perfect square factor. For instance, \( \sqrt{4} = 2 \).
• Write the simplified expression as \( 2\sqrt{10} \).

By rephrasing radicals in such a manner, algebraic expressions become simpler and more manageable. This is particularly useful when calculating perimeters or when solving equations involving radicals.