Problem 42 - Algebra and Trigonometry with Analytic Geometry

Question

A triangular field has sides of lengths \(a, b\), and \(c\) (in yards). Approximate the number of acres in the field \(\left(1\right.\) acre \(\left.=4840 \mathrm{yd}^{2}\right)\) $$a=320, \quad b=350, \quad c=500$$

Step-by-Step Solution

Answer

The field is approximately 10.84 acres.

Step 1: Calculate the Semi-Perimeter

The semi-perimeter \(s\) of a triangle is the sum of its sides divided by 2. We use the formula: \[ s = \frac{a + b + c}{2} \] Substituting the given side lengths: \[ s = \frac{320 + 350 + 500}{2} = 585 \text{ yards} \]

Step 2: Use Heron's Formula to Find the Area

Heron's formula gives the area \(A\) of a triangle with sides \(a, b,\) and \(c\) as follows: \[ A = \sqrt{s(s-a)(s-b)(s-c)} \] Using the semi-perimeter \(s = 585\), compute: \[ A = \sqrt{585(585 - 320)(585 - 350)(585 - 500)} \] \[ A = \sqrt{585 \times 265 \times 235 \times 85} \] After calculation, this equals approximately \(A \approx 52,497.26 \text{ square yards} \).

Step 3: Convert Area from Square Yards to Acres

Since 1 acre equals 4840 square yards, convert the area into acres using the conversion factor: \[ \text{Acres} = \frac{\text{Area in square yards}}{4840} \] \[ \text{Acres} = \frac{52,497.26}{4840} \approx 10.84 \text{ acres} \]

Key Concepts

Semi-Perimeter of a Triangle

The semi-perimeter of a triangle is a useful concept in geometry, especially when you want to determine the area of a triangle using Heron's formula. The semi-perimeter is simply half the sum of the lengths of the triangle's sides. This value is crucial in simplifying area calculations for triangles. It's represented by the symbol \( s \). If you know the side lengths \( a, b, \text{ and } c \), you can compute the semi-perimeter with this formula:

\[ s = \frac{a + b + c}{2} \]
In our exercise example, with sides measuring 320, 350, and 500 yards, the semi-perimeter will be half of the total sum of these side lengths. This calculation will give you:

\[ s = \frac{320 + 350 + 500}{2} = 585 \text{ yards} \]

Understanding the semi-perimeter is crucial because it sets the groundwork for applying Heron's formula to find the triangle's area effectively.

Area Conversion

Converting between different units of area is an important skill in many real-world applications, especially when dealing with large spaces like fields and gardens. In the context of our exercise, the area needs to be converted from square yards to acres, which is common in land measurement.

Acres are used to measure land because they provide a more manageable figure than expressing large areas in square yards. The conversion between square yards and acres involves using a conversion factor. Specifically, 1 acre is equivalent to 4840 square yards. To convert square yards into acres, use the formula:

\[ \text{Acres} = \frac{\text{Area in square yards}}{4840} \]

For example, if the calculated area of your triangular field is 52,497.26 square yards, as in our exercise:

\[ \text{Acres} = \frac{52,497.26}{4840} \approx 10.84 \text{ acres} \]

Understanding area conversion is especially beneficial for activities related to agriculture, land development, and real estate.

Triangle Area Calculation Using Heron's Formula

Calculating the area of a triangle can be challenging if you don't have the height measurement. That’s where Heron's formula becomes extremely useful. It's especially handy because it only requires the knowledge of the side lengths of the triangle. With Heron's formula, the area \( A \) is derived from the semi-perimeter \( s \) and the lengths of the sides \( a, b, \text{ and } c \). The formula is:

\[ A = \sqrt{s(s-a)(s-b)(s-c)} \]

For a triangle with sides measuring 320, 350, and 500 yards, and a semi-perimeter of 585 yards, you would substitute these values into Heron's formula:

\[ A = \sqrt{585(585 - 320)(585 - 350)(585 - 500)} \]

This computation, which involves multiplication and finding the square root, gives the area as approximately 52,497.26 square yards. Heron's formula is not only a powerful tool in theoretical mathematics but also highly practical in fields such as engineering and architecture, where precise area calculations are critical.