To find the perimeter and area of triangle ABC, we need to follow a few mathematical steps.
First, let's find the distances between each pair of vertices to determine the lengths of the sides of the triangle.
1. **Calculate \( AB \):**
Using the distance formula \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \):
\[
AB = \sqrt{(16 - 2)^2 + (2 - 8)^2} = \sqrt{(14)^2 + (-6)^2} = \sqrt{196 + 36} = \sqrt{232} \approx 15.23
\]
2. **Calculate \( AC \):**
\[
AC = \sqrt{(6 - 2)^2 + (2 - 8)^2} = \sqrt{(4)^2 + (-6)^2} = \sqrt{16 + 36} = \sqrt{52} \approx 7.21
\]
3. **Calculate \( BC \):**
\[
BC = \sqrt{(6 - 16)^2 + (2 - 2)^2} = \sqrt{(-10)^2 + (0)^2} = \sqrt{100} = 10
\]
Now, add these distances to find the perimeter:
\[
Perimeter = AB + AC + BC = 15.23 + 7.21 + 10 = 32.44 \text{ units}
\]
Next, calculate the area of the triangle using the coordinate points and the shoelace formula:
\[
\text{Area} = \frac{1}{2} \left| x_1y_2 + x_2y_3 + x_3y_1 - (y_1x_2 + y_2x_3 + y_3x_1) \right|
\]
Substitute the coordinates \( A(2,8), B(16,2), C(6,2) \):
\[
\text{Area} = \frac{1}{2} \left| (2 \cdot 2 + 16 \cdot 2 + 6 \cdot 8) - (8 \cdot 16 + 2 \cdot 6 + 2 \cdot 2) \right|
\]
\[
= \frac{1}{2} \left| (4 + 32 + 48) - (128 + 12 + 4) \right|
\]
\[
= \frac{1}{2} \left| 84 - 144 \right|
\]
\[
= \frac{1}{2} \left| -60 \right|
\]
\[
= \frac{1}{2} \times 60 = 30 \text{ square units}
\]
Therefore, the perimeter of triangle ABC is \( 32.44 \) units, and its area is \( 30 \) square units.
Type the correct answer in each box. If necessary, round your answers to the nearest hundredth.
The vertices of triangle ABC are A(2,8), B(16,2), and C(6,2). The perimeter of triangle ABC is _____ units, and its area is ______ square units.
1 answer
1 answer