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Using Heron's Area Formula use Heron's Area Formula to find the area of the triangle. $$ a=\frac{3}{5}, \quad b=\frac{5}{8}, \quad c=\frac{3}{8} $$

Short Answer

Expert verified
So, the area of the triangle, given the sides, using Heron's formula is \( \frac{23}{3200} \).

Step by step solution

01

Calculate Semi-Perimeter

We start by determining the semi-perimeter of the triangle, denoted as \(s\). This is calculated as \(s = \frac{a + b + c}{2}\). For the given sides, this would be calculated as \(s = \frac{\frac{3}{5} + \frac{5}{8} + \frac{3}{8} }{2} = \frac{23}{40}\).
02

Apply Heron's Formula

Next, we substitute the semi-perimeter and the given sides into Heron's formula. So, the area (A) of the triangle will be calculated as: \( A = \sqrt{s \, (s - a) \, (s - b) \, (s - c)} = \sqrt{ \frac{23}{40} \, ( \frac{23}{40} - \frac{3}{5}) \, ( \frac{23}{40} - \frac{5}{8}) \, ( \frac{23}{40} - \frac{3}{8}) } \)
03

Simplify the Solution

Finally, compute the above expression. This would end up being: \( A = \sqrt{ \frac{23}{40} * \frac{1}{40} * \frac{1}{40} * \frac{17}{40} } = \frac{23}{3200} \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Semi-Perimeter
The semi-perimeter of a triangle is a crucial concept when using Heron's Formula to calculate the area of a triangle. It is essentially half of the triangle's perimeter. Calculating the semi-perimeter is the first step before finding the area. To find the semi-perimeter, you simply add up all three side lengths of the triangle and then divide the result by two.
In the provided example, the sides of the triangle are given as fractions:
  • \( a = \frac{3}{5} \)
  • \( b = \frac{5}{8} \)
  • \( c = \frac{3}{8} \)
The calculation is: \[ s = \frac{a + b + c}{2} = \frac{3}{5} + \frac{5}{8} + \frac{3}{8} = \frac{23}{40} \]
This step is vital since the semi-perimeter \( s \) is then used in Heron's area formula to simplify the process. If you correctly calculate the semi-perimeter, you're well-prepared for the following steps in finding the area.
Triangle Area Calculation Using Heron's Formula
Heron's formula provides a way to calculate the area of a triangle when the side lengths are known, without needing to compute any angles. It's a simple but powerful tool in geometry. Once the semi-perimeter \( s \) is determined, you can dive into the calculation of the area. Heron's formula is expressed as: \\[ A = \sqrt{s \cdot (s-a) \cdot (s-b) \cdot (s-c)} \] Using the semi-perimeter and side lengths from the example:
  • \( s = \frac{23}{40} \)
  • \( a = \frac{3}{5} \), \( s-a = \frac{23}{40} - \frac{24}{40} = \frac{1}{40} \)
  • \( b = \frac{5}{8} \), \( s-b = \frac{23}{40} - \frac{25}{40} = \frac{-1}{40} \)
  • \( c = \frac{3}{8} \), \( s-c = \frac{23}{40} - \frac{15}{40} = \frac{8}{40} \)

Heron's formula allows for the calculation:\[ A = \sqrt{ \frac{23}{40} \cdot \frac{1}{40} \cdot \frac{-1}{40} \cdot \frac{8}{40} } \] This calculates to the area of \( \frac{23}{3200} \) after simplifying.
Fractions in Geometry
Working with fractions in geometry can sometimes be tricky, especially in formulas involving multiplication and simplification. It's important to handle fractions carefully to ensure accurate results. In the context of Heron's area formula, all calculations must rigorously follow the rules of fractions involving addition, subtraction, multiplication, and division.
Here are some tips for working with fractions in geometry:
  • Always find a common denominator for addition and subtraction.
  • Break down complex fraction multiplications step by step.
  • When multiplying fractions, multiply the numerators together and the denominators together separately.
  • Reduce fractions to their simplest form whenever possible.
In the example, the care taken with fractions ensures that each step builds correctly on the previous one. This involves correctly determining the semi-perimeter \( s \), and deconstructing Heron's formula into manageable steps, which makes the seemingly complex process easier to handle.

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