Answer :
Let's find the perimeter and area of the triangle \( \triangle ABC \) with vertices \( A(2, 8) \), \( B(16, 2) \), and \( C(6, 2) \).
### Step-by-Step Solution:
1. Find the lengths of the sides:
- We use the distance formula to calculate the lengths of the sides \( AB \), \( BC \), and \( AC \).
- The distance formula between two points \((x_1, y_1)\) and \((x_2, y_2)\) is:
[tex]\[ \text{distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
- Length of \( AB \)
[tex]\[ AB = \sqrt{(16 - 2)^2 + (2 - 8)^2} = \sqrt{14^2 + (-6)^2} = \sqrt{196 + 36} = \sqrt{232} \approx 15.23 \][/tex]
- Length of \( BC \)
[tex]\[ BC = \sqrt{(16 - 6)^2 + (2 - 2)^2} = \sqrt{10^2 + 0^2} = \sqrt{100} = 10 \][/tex]
- Length of \( AC \)
[tex]\[ AC = \sqrt{(6 - 2)^2 + (2 - 8)^2} = \sqrt{4^2 + (-6)^2} = \sqrt{16 + 36} = \sqrt{52} \approx 7.21 \][/tex]
2. Calculate the perimeter:
- The perimeter \( P \) of the triangle is the sum of the lengths of its sides:
[tex]\[ P = AB + BC + AC \approx 15.23 + 10 + 7.21 \approx 32.44 \][/tex]
3. Calculate the area:
- We can use the formula for the area of a triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), \((x_3, y_3)\):
[tex]\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \][/tex]
- Plugging in the coordinates of points \( A(2, 8) \), \( B(16, 2) \), \( C(6, 2) \):
[tex]\[ \text{Area} = \frac{1}{2} \left| 2(2 - 2) + 16(2 - 8) + 6(8 - 2) \right| = \frac{1}{2} \left| 2 \cdot 0 + 16 \cdot (-6) + 6 \cdot 6 \right| \][/tex]
[tex]\[ = \frac{1}{2} \left| 0 - 96 + 36 \right| = \frac{1}{2} \left| -60 \right| = \frac{1}{2} \cdot 60 = 30 \][/tex]
Therefore, the perimeter of [tex]\( \triangle ABC \)[/tex] is approximately [tex]\( 32.44 \)[/tex] units, and its area is [tex]\( 30.0 \)[/tex] square units.
### Step-by-Step Solution:
1. Find the lengths of the sides:
- We use the distance formula to calculate the lengths of the sides \( AB \), \( BC \), and \( AC \).
- The distance formula between two points \((x_1, y_1)\) and \((x_2, y_2)\) is:
[tex]\[ \text{distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]
- Length of \( AB \)
[tex]\[ AB = \sqrt{(16 - 2)^2 + (2 - 8)^2} = \sqrt{14^2 + (-6)^2} = \sqrt{196 + 36} = \sqrt{232} \approx 15.23 \][/tex]
- Length of \( BC \)
[tex]\[ BC = \sqrt{(16 - 6)^2 + (2 - 2)^2} = \sqrt{10^2 + 0^2} = \sqrt{100} = 10 \][/tex]
- Length of \( AC \)
[tex]\[ AC = \sqrt{(6 - 2)^2 + (2 - 8)^2} = \sqrt{4^2 + (-6)^2} = \sqrt{16 + 36} = \sqrt{52} \approx 7.21 \][/tex]
2. Calculate the perimeter:
- The perimeter \( P \) of the triangle is the sum of the lengths of its sides:
[tex]\[ P = AB + BC + AC \approx 15.23 + 10 + 7.21 \approx 32.44 \][/tex]
3. Calculate the area:
- We can use the formula for the area of a triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), \((x_3, y_3)\):
[tex]\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \][/tex]
- Plugging in the coordinates of points \( A(2, 8) \), \( B(16, 2) \), \( C(6, 2) \):
[tex]\[ \text{Area} = \frac{1}{2} \left| 2(2 - 2) + 16(2 - 8) + 6(8 - 2) \right| = \frac{1}{2} \left| 2 \cdot 0 + 16 \cdot (-6) + 6 \cdot 6 \right| \][/tex]
[tex]\[ = \frac{1}{2} \left| 0 - 96 + 36 \right| = \frac{1}{2} \left| -60 \right| = \frac{1}{2} \cdot 60 = 30 \][/tex]
Therefore, the perimeter of [tex]\( \triangle ABC \)[/tex] is approximately [tex]\( 32.44 \)[/tex] units, and its area is [tex]\( 30.0 \)[/tex] square units.
