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Chapter 6: Problem 29

Find the area of the triangle whose sides have the given lengths. \(a=9, \quad b=12, \quad c=15\)

Short Answer

Expert verified
The area of the triangle is 54 square units.

Step by step solution

01

Verify if the Triangle is Valid

To find the area of a triangle with given side lengths, first confirm that these sides can form a triangle. Apply the triangle inequality theorem: for sides a, b, and c to form a triangle, the sum of any two sides must be greater than the third.Check:1. \(a + b > c\), i.e., \(9 + 12 > 15\). This is true since 21 > 15.2. \(a + c > b\), i.e., \(9 + 15 > 12\). This is true since 24 > 12.3. \(b + c > a\), i.e., \(12 + 15 > 9\). This is true since 27 > 9.Since all inequalities hold, the sides can form a triangle.
02

Calculate the Semi-Perimeter

The semi-perimeter \(s\) of a triangle is calculated as half the sum of its side lengths. Use the formula:\[ s = \frac{a + b + c}{2} \]Substitute the given values:\[ s = \frac{9 + 12 + 15}{2} = \frac{36}{2} = 18 \]
03

Apply Heron's Formula

Heron's formula allows us to find the area of a triangle when the lengths of all three sides are known. The formula is:\[ A = \sqrt{s(s-a)(s-b)(s-c)} \]Substitute the values for \(s\), \(a\), \(b\), and \(c\):\[ A = \sqrt{18(18-9)(18-12)(18-15)} \]\[ A = \sqrt{18 \times 9 \times 6 \times 3} \]
04

Calculate the Area

Now, compute the simplified expression within the square root:Calculate the multiplication: \(9 \times 6 = 54\) and \(54 \times 3 = 162\).Then multiply by 18: \(18 \times 162\).Instead, calculate each multiplication with proper simplification:\[ A = \sqrt{18 \times 9 \times 6 \times 3} \]\[ A = \sqrt{2916} \]Finally, calculate \(\sqrt{2916} = 54\).Thus, the area \(A\) is 54 square units.

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Key Concepts

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

Triangle Inequality Theorem
The **Triangle Inequality Theorem** is a fundamental concept in geometry that ensures three sides can indeed form a triangle. It states that for any three sides to constitute a triangle:
  • The sum of the lengths of any two sides must be greater than the length of the remaining side.
  • This must hold true for all three combinations of added side lengths.
For example, in our scenario with sides 9, 12, and 15:
  • The sum of 9 and 12 is 21, which is greater than 15.
  • The sum of 9 and 15 is 24, which is greater than 12.
  • The sum of 12 and 15 is 27, which is greater than 9.
Since all these conditions are satisfied, these sides adhere to the Triangle Inequality Theorem. Therefore, they can form a valid triangle.
Semi-Perimeter Formula
The **Semi-Perimeter Formula** is often used in triangle-related calculations, especially for finding the area using Heron's Formula. The semi-perimeter of a triangle is simply half of the triangle's perimeter, providing a key value used in further computations.To compute the semi-perimeter \(s\), use the formula:\[ s = \frac{a + b + c}{2} \]where \(a\), \(b\), and \(c\) are the side lengths of the triangle.In our case, substituting in the given values:
  • \(a = 9\), \(b = 12\), \(c = 15\)
  • Calculate the sum: \(9 + 12 + 15 = 36\)
  • Then, divide by 2: \(s = \frac{36}{2} = 18\)
Thus, the semi-perimeter \(s\) is 18, which is essential for finding the area using Heron's Formula.
Area of Triangle
To find the **Area of a Triangle** when the side lengths are known, we use **Heron's Formula**. This formula is highly effective because it calculates area using only the side lengths, avoiding the need for angles or height.Heron's Formula is expressed as:\[ A = \sqrt{s(s-a)(s-b)(s-c)} \]where \(s\) is the semi-perimeter and \(a\), \(b\), \(c\) are the sides of the triangle. Here's how to apply it:
  • Calculate \(s = 18\), as we did before.
  • Substitute: \(A = \sqrt{18(18-9)(18-12)(18-15)}\)
  • This becomes \(A = \sqrt{18 \times 9 \times 6 \times 3}\).
  • Calculate under the root: 18 times 9 is 162; 162 times 6 is 972; 972 times 3 equals 2916.
  • Finally, find \(\sqrt{2916} = 54\).
Thus, the area of the triangle is 54 square units, offering a practical example of how geometric principles synthesize into a straightforward solution.

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