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Magazine Chapter 8: Problem 65
Find the area of triangle \(A B C\). \(a=12 \mathrm{m}, b=16 \mathrm{m}, c=25 \mathrm{m}\)
Short Answer
Expert verified
The area of triangle ABC is approximately 77.8 square meters.
Step by step solution
01
Verify if Triangle is Right-Angled
Check if the triangle is right-angled using the Pythagorean theorem. Verify if \(c^2 = a^2 + b^2\). Here, \(c = 25 \text{m}\), \(a = 12 \text{m}\), and \(b = 16 \text{m}\). Calculate \(a^2 + b^2\) and \(c^2\).
02
Calculate \(a^2\) and \(b^2\)
First compute \(a^2\) and \(b^2\). \(a^2 = 12^2 = 144\), \(b^2 = 16^2 = 256\).
03
Calculate \(c^2\)
Now compute \(c^2 = 25^2 = 625\).
04
Compare \(a^2 + b^2\) with \(c^2\)
Add \(a^2\) and \(b^2\): \(144 + 256 = 400\). Since \(400 eq 625\), the triangle is not right-angled.
05
Use Heron's Formula
Since the triangle is not right-angled, use Heron's formula to find the area. First, calculate the semi-perimeter \(s\) using \[s = \frac{a+b+c}{2}\text{.}\]
06
Calculate Semi-Perimeter
Compute \(s\) with given sides: \[s = \frac{12 + 16 + 25}{2} = \frac{53}{2} = 26.5 \text{m}\]
07
Compute Area
Use the semi-perimeter \(s\) to calculate the area with Heron's formula: \[ \text{Area} = \text{sqrt}\big(s(s-a)(s-b)(s-c)\big)\] Substitute values: \[ \text{Area} = \text{sqrt}\big(26.5(26.5-12)(26.5-16)(26.5-25)\big) \] \[= \text{sqrt}\big(26.5 \times 14.5 \times 10.5 \times 1.5\big)\]
08
Solve Area Calculation
Perform the multiplication inside the square root: \[ \text{Area} = \text{sqrt}(26.5 \times 14.5 \times 10.5 \times 1.5) \] \[= \text{sqrt}(6056.6875) \] \[ \text{Area} \thickapprox 77.8 \text{m}^2 \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Pythagorean theorem
The Pythagorean theorem helps determine if a triangle is right-angled. It states that for a triangle with sides a, b, and c (with c being the longest side or hypotenuse), the triangle is right-angled if and only if:
\[ c^2 = a^2 + b^2 \]
In our example, the sides of the triangle were given as a = 12 m, b = 16 m, and c = 25 m. To check if our triangle is right-angled, we first calculate:
\[ a^2 = 12^2 = 144 \]
Then:
\[ b^2 = 16^2 = 256 \]
Adding these up gives:
\[ a^2 + b^2 = 144 + 256 = 400 \]
For the triangle to be right-angled, this sum should equal \[ c^2 \]. Calculating \[ c^2 = 25^2 \] gives:
\[ c^2 = 625 \]
Because 400 does not equal 625, our triangle is not right-angled.
\[ c^2 = a^2 + b^2 \]
In our example, the sides of the triangle were given as a = 12 m, b = 16 m, and c = 25 m. To check if our triangle is right-angled, we first calculate:
\[ a^2 = 12^2 = 144 \]
Then:
\[ b^2 = 16^2 = 256 \]
Adding these up gives:
\[ a^2 + b^2 = 144 + 256 = 400 \]
For the triangle to be right-angled, this sum should equal \[ c^2 \]. Calculating \[ c^2 = 25^2 \] gives:
\[ c^2 = 625 \]
Because 400 does not equal 625, our triangle is not right-angled.
Heron's formula
When a triangle is not right-angled, we can use Heron's formula to find the area. Heron's formula states that the area of a triangle can be found using its side lengths and the semi-perimeter. The formula is:
\[ \text{Area} = \text{sqrt}(s(s-a)(s-b)(s-c)) \]
Where 鈥榮鈥� is the semi-perimeter of the triangle. In our case:
\[ a = 12 \text{m}, b = 16 \text{m}, c = 25 \text{m} \]
We need to calculate the semi-perimeter first (explained in our next section). Once we have 's', we plug in the values:
\[ \text{Area} = \text{sqrt}(26.5(26.5 - 12)(26.5 - 16)(26.5 - 25)) \]
\[ = \text{sqrt}(26.5 \times 14.5 \times 10.5 \times 1.5) \]
After performing the multiplication inside the square root:
\[ \text{Area} = \text{sqrt}(6056.6875) \]
\[ \text{Area} \thickapprox 77.8 \text{m}^2 \]
Thus, the area of the triangle is approximately 77.8 square meters.
\[ \text{Area} = \text{sqrt}(s(s-a)(s-b)(s-c)) \]
Where 鈥榮鈥� is the semi-perimeter of the triangle. In our case:
\[ a = 12 \text{m}, b = 16 \text{m}, c = 25 \text{m} \]
We need to calculate the semi-perimeter first (explained in our next section). Once we have 's', we plug in the values:
\[ \text{Area} = \text{sqrt}(26.5(26.5 - 12)(26.5 - 16)(26.5 - 25)) \]
\[ = \text{sqrt}(26.5 \times 14.5 \times 10.5 \times 1.5) \]
After performing the multiplication inside the square root:
\[ \text{Area} = \text{sqrt}(6056.6875) \]
\[ \text{Area} \thickapprox 77.8 \text{m}^2 \]
Thus, the area of the triangle is approximately 77.8 square meters.
semi-perimeter calculation
The semi-perimeter is a crucial step in using Heron's formula to find the area of a triangle. The semi-perimeter 's' is calculated using the formula:
\[ s = \frac{a + b + c}{2} \]
In our example, the side lengths are 12 m, 16 m, and 25 m. So, the semi-perimeter is:
\[ s = \frac{12 + 16 + 25}{2} \]
Adding these values together gives:
\[ 12 + 16 + 25 = 53 \]
Dividing by 2, we get:
\[ s = \frac{53}{2} = 26.5 \text{m} \]
This semi-perimeter value 's' is then used in the Heron's formula to find the triangle's area, as demonstrated in the previous section.
\[ s = \frac{a + b + c}{2} \]
In our example, the side lengths are 12 m, 16 m, and 25 m. So, the semi-perimeter is:
\[ s = \frac{12 + 16 + 25}{2} \]
Adding these values together gives:
\[ 12 + 16 + 25 = 53 \]
Dividing by 2, we get:
\[ s = \frac{53}{2} = 26.5 \text{m} \]
This semi-perimeter value 's' is then used in the Heron's formula to find the triangle's area, as demonstrated in the previous section.
