Chapter 9: Problem 25
Find the area of each triangle. Round answers to two decimal places. $$a=4, \quad b=4, \quad c=4$$
Short Answer
Expert verified
The area of the triangle is approximately 6.93 square units.
Step by step solution
01
- Identify the formula for the area of a triangle using Heron's formula
Heron's formula is used to find the area of a triangle when the lengths of all three sides are known. The formula is: \[ A = \sqrt{s(s-a)(s-b)(s-c)} \] where \( s \) is the semi-perimeter of the triangle.
02
- Calculate the semi-perimeter
The semi-perimeter \( s \) is calculated as: \[ s = \frac{a + b + c}{2} \] Given \( a = 4 \), \( b = 4 \), \( c = 4 \), we have: \[ s = \frac{4 + 4 + 4}{2} = 6 \]
03
- Substitute values into Heron's formula
Using the values of \( s \) and the sides \( a, b, c \), we substitute into Heron's formula: \[ A = \sqrt{6(6-4)(6-4)(6-4)} = \sqrt{6 \cdot 2 \cdot 2 \cdot 2} = \sqrt{48} \]
04
- Compute the square root and round to two decimal places
The area \( A \) is calculated as: \[ A = \sqrt{48} \approx 6.93 \] Rounded to two decimal places, the area of the triangle is approximately 6.93.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Area of a Triangle
To find the area of a triangle when you know the lengths of all three sides, you can use Heron's formula. This formula is very handy and works for any triangle, whether it's right-angled, obtuse, or acute. The general form of Heron's formula is:
\( A = \sqrt{s(s-a)(s-b)(s-c)} \)
where \( A \) is the area and \( s \) is the semi-perimeter of the triangle. First, you calculate the semi-perimeter, then plug all known values into the formula to find the area. This method breaks down a complex process into easy, manageable steps, making it possible to compute the area using basic arithmetic and square root operations.
For example, if each side of the triangle is 4 units (an equilateral triangle), you would first determine the semi-perimeter and then use it to find the area as detailed in the exercise solution.
\( A = \sqrt{s(s-a)(s-b)(s-c)} \)
where \( A \) is the area and \( s \) is the semi-perimeter of the triangle. First, you calculate the semi-perimeter, then plug all known values into the formula to find the area. This method breaks down a complex process into easy, manageable steps, making it possible to compute the area using basic arithmetic and square root operations.
For example, if each side of the triangle is 4 units (an equilateral triangle), you would first determine the semi-perimeter and then use it to find the area as detailed in the exercise solution.
Semi-Perimeter Calculation
The semi-perimeter \( s \) is half of the triangle's perimeter. It's a crucial value needed in Heron's formula for calculating the area. The formula for the semi-perimeter is given by:
\[ s = \frac{a + b + c}{2} \]
Here, \( a, b, \) and \( c \) are the lengths of the sides of the triangle. Breaking down the calculation:
Using the given example:
\ a = 4, b = 4, c = 4\
\[ s = \frac{4 + 4 + 4}{2} = 6 \]
With this semi-perimeter value, you can move on to find the area of the triangle using Heron's formula. This concept simplifies the initial steps toward calculating the area, transforming a seemingly tough geometry problem into basic arithmetic.
\[ s = \frac{a + b + c}{2} \]
Here, \( a, b, \) and \( c \) are the lengths of the sides of the triangle. Breaking down the calculation:
- Add up the lengths of all three sides.
- Divide the sum by 2 to get the semi-perimeter.
Using the given example:
\ a = 4, b = 4, c = 4\
\[ s = \frac{4 + 4 + 4}{2} = 6 \]
With this semi-perimeter value, you can move on to find the area of the triangle using Heron's formula. This concept simplifies the initial steps toward calculating the area, transforming a seemingly tough geometry problem into basic arithmetic.
Square Root
The square root function is used in Heron's formula to finalize the area calculation. After finding the semi-perimeter and substituting all values into the Heron’s formula, the process involves calculating the inner product and then taking the square root of the result. Let's break it down further:
When using Heron's formula \( A = \sqrt{s(s-a)(s-b)(s-c)} \), you arrive at:
\[ A = \sqrt{6(6-4)(6-4)(6-4)} = \sqrt{6 \cdot 2 \cdot 2 \cdot 2} = \sqrt{48} \]
Finally, to find \( \sqrt{48} \), use a calculator or estimate by finding the nearest perfect squares. You obtain:
\[ \sqrt{48} \approx 6.93 \]
This result is the area of the triangle, rounded to two decimal places. Understanding how to handle the square root operation is key, as it transforms the summed products into a manageable result, giving you the precise area you need. Whether you use a calculator or do it manually, this concept ensures you can complete the problem accurately.
When using Heron's formula \( A = \sqrt{s(s-a)(s-b)(s-c)} \), you arrive at:
\[ A = \sqrt{6(6-4)(6-4)(6-4)} = \sqrt{6 \cdot 2 \cdot 2 \cdot 2} = \sqrt{48} \]
Finally, to find \( \sqrt{48} \), use a calculator or estimate by finding the nearest perfect squares. You obtain:
\[ \sqrt{48} \approx 6.93 \]
This result is the area of the triangle, rounded to two decimal places. Understanding how to handle the square root operation is key, as it transforms the summed products into a manageable result, giving you the precise area you need. Whether you use a calculator or do it manually, this concept ensures you can complete the problem accurately.