Question
Find the Area of the triangle
K
11 in
H
5 in
P
7 in
K
11 in
H
5 in
P
7 in
Answers
Answered by
GPT-4o mini
To find the area of a triangle when the lengths of all three sides are known, we can use Heron's formula.
First, let's denote the side lengths:
- \( a = 11 \) inches
- \( b = 5 \) inches
- \( c = 7 \) inches
### Steps:
1. **Calculate the semi-perimeter (s)**:
\[
s = \frac{a + b + c}{2} = \frac{11 + 5 + 7}{2} = \frac{23}{2} = 11.5 \text{ inches}
\]
2. **Apply Heron's formula**:
The area \( A \) of the triangle can be calculated using the formula:
\[
A = \sqrt{s(s-a)(s-b)(s-c)}
\]
First, calculate \( s-a \), \( s-b \), and \( s-c \):
- \( s - a = 11.5 - 11 = 0.5 \)
- \( s - b = 11.5 - 5 = 6.5 \)
- \( s - c = 11.5 - 7 = 4.5 \)
3. **Substituting into Heron's formula**:
\[
A = \sqrt{11.5 \times 0.5 \times 6.5 \times 4.5}
\]
4. **Calculating the product**:
Now, let's compute the product inside the square root:
\[
11.5 \times 0.5 = 5.75
\]
\[
5.75 \times 6.5 = 37.375
\]
\[
37.375 \times 4.5 = 168.1875
\]
5. **Finally, calculate the square root**:
\[
A = \sqrt{168.1875} \approx 12.95 \text{ square inches}
\]
Thus, the area of the triangle is approximately **12.95 square inches**.
First, let's denote the side lengths:
- \( a = 11 \) inches
- \( b = 5 \) inches
- \( c = 7 \) inches
### Steps:
1. **Calculate the semi-perimeter (s)**:
\[
s = \frac{a + b + c}{2} = \frac{11 + 5 + 7}{2} = \frac{23}{2} = 11.5 \text{ inches}
\]
2. **Apply Heron's formula**:
The area \( A \) of the triangle can be calculated using the formula:
\[
A = \sqrt{s(s-a)(s-b)(s-c)}
\]
First, calculate \( s-a \), \( s-b \), and \( s-c \):
- \( s - a = 11.5 - 11 = 0.5 \)
- \( s - b = 11.5 - 5 = 6.5 \)
- \( s - c = 11.5 - 7 = 4.5 \)
3. **Substituting into Heron's formula**:
\[
A = \sqrt{11.5 \times 0.5 \times 6.5 \times 4.5}
\]
4. **Calculating the product**:
Now, let's compute the product inside the square root:
\[
11.5 \times 0.5 = 5.75
\]
\[
5.75 \times 6.5 = 37.375
\]
\[
37.375 \times 4.5 = 168.1875
\]
5. **Finally, calculate the square root**:
\[
A = \sqrt{168.1875} \approx 12.95 \text{ square inches}
\]
Thus, the area of the triangle is approximately **12.95 square inches**.
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