Problem 102 - Algebra and Trigonometry

Question

Based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. Find the area of a triangle with sides \(6,11,\) and \(13 .\)

Step-by-Step Solution

Answer

The area of the triangle is approximately 32.86 square units.

Step 1 - Recall Heron's Formula

Heron's Formula allows us to find the area of a triangle when we know the lengths of all three sides. The formula is: \[ \text{Area} = \sqrt{s(s-a)(s-b)(s-c)} \] where \( a, b, c \) are the lengths of the sides of the triangle and \( s \) is the semi-perimeter of the triangle given by \( s = \frac{a+b+c}{2} \).

Step 2 - Calculate the Semi-perimeter

To use Heron's Formula, first calculate the semi-perimeter of the triangle: \[ s = \frac{6+11+13}{2} = 15 \]

Step 3 - Substitute Values into Heron's Formula

Now, substitute the side lengths and the semi-perimeter into Heron's Formula: \[ \text{Area} = \sqrt{15(15-6)(15-11)(15-13)} \] Simplify the expression inside the square root: \[ \text{Area} = \sqrt{15 \times 9 \times 4 \times 2} \]

Step 4 - Calculate the Area

Proceed to calculate the value inside the square root: \[ 15 \times 9 \times 4 \times 2 = 1080 \] Therefore, the area of the triangle is: \[ \text{Area} = \sqrt{1080} \approx 32.86 \]

Key Concepts

area of a triangle

Understanding how to find the area of a triangle is essential in geometry. One powerful method uses Heron's Formula, especially helpful when we know all three side lengths but not the height.

This formula works for any triangle—whether it is scalene, isosceles, or equilateral—unlike other area formulas which may require additional information. The general method to find the area via Heron's Formula involves calculating the semi-perimeter and then using a specific square root expression.

Curious how this formula works? Let's break it down: knowing the lengths of sides provides a precise way to figure out a triangle's area without direct height measurements. It's incredibly useful for solving many problems quickly and accurately.

Remember, mastering this makes solving more complex geometrical questions a piece of cake!

semi-perimeter

The semi-perimeter is a key part of Heron's Formula. It's half the perimeter of a triangle. You might wonder why we use it. This step simplifies plugging values into the formula.

To find it, simply add up all the side lengths and divide by two. Here’s an example: for a triangle with sides 6, 11, and 13, calculate the semi-perimeter as follows:

\[\begin{equation} s = \frac{6+11+13}{2} = 15 \end{equation}\]
Now, you've got 15 as the semi-perimeter.

This single value plays a crucial role in deriving the area from Heron's Formula. Once you have the semi-perimeter, you can use it within Heron's Formula by subtracting each side length, simplifying the calculation of the area.

Nailing this part sets you up for success in applying the full formula effortlessly.

algebraic calculations

Algebraic calculations form the bedrock of using Heron's Formula effectively. Once you have the semi-perimeter, you need to perform some multiplications and subtractions before calculating the square root.

First, for each side length of the triangle (a, b, and c), subtract these from the semi-perimeter (s). In our example with sides 6, 11, and 13, and semi-perimeter 15:

\[\begin{equation} s - a = 15 - 6 = 9 \end{equation}\] \[\begin{equation} s - b = 15 - 11 = 4 \end{equation}\] \[\begin{equation} s - c = 15 - 13 = 2 \end{equation}\]
Now, substitute all these values back into Heron's Formula:
\[\begin{equation} \text{Area} = \sqrt{15(9)(4)(2)} \end{equation}\]

Multiplying these inside the square root:


\[\begin{equation} 15 \times 9 = 135\end{equation}\] \[\begin{equation} 135 \times 4 = 540 \end{equation}\] \[\begin{equation} 540 \times 2 = 1080 \end{equation}\]

You get 1080. Finally, take the square root to find the area:
\[\begin{equation} \text{Area} = \sqrt{1080} \approx 32.86 \end{equation}\]

Each algebraic step moves you closer to the solution. Perfecting these calculations ensures you can confidently apply Heron's Formula.