Problem 31 - Algebra 2 and Trigonometry

Question

The area of a right triangle is 6\(\sqrt{2}\) square centimeters and the length of one leg is \(\sqrt{12}\) cenimeters. What is the length of the other leg? What is the length of the hypotenuse?

Step-by-Step Solution

Answer

The other leg is \( 2\sqrt{6} \) cm, and the hypotenuse is 6 cm.

Step 1: Area Formula for Right Triangle

The area of a right triangle is calculated using the formula \( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \). In this problem, the area is given as \( 6\sqrt{2} \) and one leg of the triangle (i.e., either base or height) is \( \sqrt{12} \). We need to first find the length of the other leg (base or height).

Step 2: Solve for the Other Leg

Using the area formula we have \( \frac{1}{2} \times \sqrt{12} \times x = 6\sqrt{2} \). Solving for \( x \), we multiply both sides by 2 to get \( \sqrt{12} \times x = 12\sqrt{2} \), then divide by \( \sqrt{12} \) to find \( x = \frac{12\sqrt{2}}{\sqrt{12}} \). Simplifying this gives \( x = 2\sqrt{6} \).

Step 3: Use Pythagorean Theorem to Find Hypotenuse

For a right triangle with legs \( a \) and \( b \), and hypotenuse \( c \), the Pythagorean Theorem is \( a^2 + b^2 = c^2 \). Here, \( a = \sqrt{12} \), \( b = 2\sqrt{6} \). Substituting these values, we get \( (\sqrt{12})^2 + (2\sqrt{6})^2 = c^2 \). Calculate: \( 12 + 24 = c^2 \), so \( c^2 = 36 \).

Step 4: Calculate the Hypotenuse

Taking the square root of both sides of the equation \( c^2 = 36 \), we find \( c = \sqrt{36} = 6 \). Thus, the hypotenuse is 6 centimeters.

Key Concepts

Area of a Triangle

The area of a triangle is an important aspect to understand, especially when dealing with right triangles. For right triangles, the area can be easily calculated using the formula:

• \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]

This formula applies because a right triangle can essentially be thought of as half of a rectangle. The "base" and "height" refer to the two sides that form the right angle. In the given exercise, the area is known as \( 6\sqrt{2} \text{ cm}^2 \), and one of the legs of the triangle is given as either the base or the height, \( \sqrt{12} \text{ cm} \). By substituting these values into the formula and solving for the unknown leg, or side, you can find the length of the other leg. This approach simplifies the process and helps determine the dimensions of any two sides of a right triangle when only one is given.

Pythagorean Theorem

The Pythagorean Theorem is a fundamental principle in mathematics, especially useful in solving problems related to right triangles. It is expressed with the formula:

• \[ a^2 + b^2 = c^2 \]

In this formula, \(a\) and \(b\) represent the lengths of the two legs of a right triangle, and \(c\) is the length of the hypotenuse, which is the side opposite the right angle.

The theorem states that the square of the hypotenuse is equal to the sum of the squares of the other two sides. In the given exercise, once the lengths of both legs are known (\(\sqrt{12}\) and \(2\sqrt{6}\)), you substitute them into the formula to solve for the hypotenuse. This theorem is a powerful tool in geometry, helping to identify or verify the relationship between the sides of a right triangle.

Square Root Simplification

Simplifying square roots is an essential skill in algebra and geometry, allowing for easier calculation and manipulation of equations. When you encounter an expression under a square root, you can sometimes break it down into simpler terms. The exercise showcases a need to simplify expressions like \(\sqrt{12}\) and \(\frac{12\sqrt{2}}{\sqrt{12}}\). Specifically:

• \(\sqrt{12} = \sqrt{4 \times 3} = \sqrt{4} \times \sqrt{3} = 2\sqrt{3}\)
• For the expression \(\frac{12\sqrt{2}}{\sqrt{12}}\), simplify by recognizing that dividing by a square root can be countered by multiplying top and bottom by the conjugate or a complementary square root.

This simplification helps in clearing fractions and making calculations more manageable. As shown, simplifying square roots involves breaking down numbers into their prime factors or recognizing perfect squares within them, thereby reducing complex expressions into more familiar forms allowing easier progress through calculations.