Problem 25
Question
Use Heron's formula to find the area of each triangle. Round to the nearest square unit. \(a=4\) feet \(, b=4\) feet \(, c=2\) feet
Step-by-Step Solution
Verified Answer
The area of the triangle is approximately \(4\) square feet.
1Step 1: Calculate the semi-perimeter (s)
First, find the semi-perimeter of the triangle. The semi-perimeter is calculated using the formula \(s = (a + b + c)/2\). In this case, \(a = 4\) feet, \(b = 4\) feet and \(c = 2\) feet. Thus \[s = (4 + 4 + 2)/2 = 5\] feet.
2Step 2: Substitute the values into Heron's Formula
Next, substitute the values of \(a\), \(b\), \(c\) and \(s\) obtained in the previous step into Heron's formula. \[A = \sqrt{5(5 - 4)(5 - 4)(5 - 2)} = \sqrt{5*1*1*3} = \sqrt{15}\] square feet.
3Step 3: Round the answer
Finally, round the answer to the nearest whole number to get the area of the triangle. Therefore, the area of this triangle is approximately \(\sqrt{15} = 3.87\) square feet, and when rounded to the nearest square unit, \(4\) square feet.
Key Concepts
Triangle Area CalculationSemi-Perimeter of a TriangleSquare Roots in Geometry
Triangle Area Calculation
Calculating the area of a triangle is fundamental in geometry. The most common formula is \frac{1}{2} \times base \times height for right-angled triangles. However, when you don't have the height or are dealing with a scalene triangle, where all sides can have different lengths, Heron's formula comes in handy. It is expressed as:
\[ A = \sqrt{s(s - a)(s - b)(s - c)} \]
where A is the area, and s is the semi-perimeter of the triangle. The semi-perimeter is half the sum of all sides, and a, b, and c are the lengths of the sides. In our exercise, the sides measure 4 feet, 4 feet, and 2 feet, leading to the use of Heron's formula after determining the semi-perimeter. The advantages of Heron's formula are that it only requires the lengths of the sides and no additional height measurements, which makes it particularly useful for complex problems.
\[ A = \sqrt{s(s - a)(s - b)(s - c)} \]
where A is the area, and s is the semi-perimeter of the triangle. The semi-perimeter is half the sum of all sides, and a, b, and c are the lengths of the sides. In our exercise, the sides measure 4 feet, 4 feet, and 2 feet, leading to the use of Heron's formula after determining the semi-perimeter. The advantages of Heron's formula are that it only requires the lengths of the sides and no additional height measurements, which makes it particularly useful for complex problems.
Semi-Perimeter of a Triangle
The semi-perimeter of a triangle is half the sum of the lengths of its sides. It’s a crucial intermediary step in various geometrical calculations, including Heron's formula for determining a triangle's area. The formula for semi-perimeter (s) is:
\[ s = \frac{a + b + c}{2} \]
In our example, with sides measuring 4, 4, and 2 feet, the semi-perimeter is calculated as 5 feet. Understanding the semi-perimeter concept is essential as it simplifies the process of using Heron's formula and leads to fewer mistakes in complex calculations. It acts as a composite measure of the triangle's size and is foundational for applying Heron's formula effectively.
\[ s = \frac{a + b + c}{2} \]
In our example, with sides measuring 4, 4, and 2 feet, the semi-perimeter is calculated as 5 feet. Understanding the semi-perimeter concept is essential as it simplifies the process of using Heron's formula and leads to fewer mistakes in complex calculations. It acts as a composite measure of the triangle's size and is foundational for applying Heron's formula effectively.
Square Roots in Geometry
Square roots play a pivotal role in geometry, especially when using Heron’s formula to calculate the area of a triangle. The square root function reverses the process of squaring a number. In Heron's formula, the area calculation involves taking the square root of the product of the semi-perimeter and its extensions.
The calculation in our example yielded \[ A = \sqrt{5(5 - 4)(5 - 4)(5 - 2)} = \sqrt{15} \]
Here, \(\sqrt{15}\) simplifies to approximately 3.87. It's critical to grasp this concept, as the accuracy of the area relies on the proper extraction of square roots. Square roots are real numbers for positive radicands, and when dealing with areas, negative results are not feasible. Therefore, familiarity with square roots is necessary for solving a wide range of geometrical problems.
The calculation in our example yielded \[ A = \sqrt{5(5 - 4)(5 - 4)(5 - 2)} = \sqrt{15} \]
Here, \(\sqrt{15}\) simplifies to approximately 3.87. It's critical to grasp this concept, as the accuracy of the area relies on the proper extraction of square roots. Square roots are real numbers for positive radicands, and when dealing with areas, negative results are not feasible. Therefore, familiarity with square roots is necessary for solving a wide range of geometrical problems.
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