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The vertices of [tex]\(\Delta ABC\)[/tex] are [tex]\(A(2, 8)\)[/tex], [tex]\(B(16, 2)\)[/tex], and [tex]\(C(6, 2)\)[/tex]. The perimeter of [tex]\(\Delta ABC\)[/tex] is [tex]\(\_\_\_\_\)[/tex] units, and its area is [tex]\(\_\_\_\_\)[/tex] square units.

Sagot :

GinnyAnswer
Certainly! Let's solve the problem step-by-step. Given the vertices of triangle [tex]\( \triangle ABC \)[/tex]:
- [tex]\( A(2, 8) \)[/tex]
- [tex]\( B(16, 2) \)[/tex]
- [tex]\( C(6, 2) \)[/tex]

### 1. Calculate the lengths of the sides of the triangle

Step 1.1: Calculate the distance [tex]\( AB \)[/tex]

The distance formula between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is:

[tex]\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \][/tex]

For [tex]\( A(2, 8) \)[/tex] and [tex]\( B(16, 2) \)[/tex]:

[tex]\[ AB = \sqrt{(16 - 2)^2 + (2 - 8)^2} \][/tex]

[tex]\[ AB = \sqrt{14^2 + (-6)^2} \][/tex]

[tex]\[ AB = \sqrt{196 + 36} \][/tex]

[tex]\[ AB = \sqrt{232} \][/tex]

[tex]\[ AB \approx 15.23 \][/tex]

Step 1.2: Calculate the distance [tex]\( BC \)[/tex]

For [tex]\( B(16, 2) \)[/tex] and [tex]\( C(6, 2) \)[/tex]:

[tex]\[ BC = \sqrt{(6 - 16)^2 + (2 - 2)^2} \][/tex]

[tex]\[ BC = \sqrt{(-10)^2 + 0^2} \][/tex]

[tex]\[ BC = \sqrt{100} \][/tex]

[tex]\[ BC = 10.00 \][/tex]

Step 1.3: Calculate the distance [tex]\( AC \)[/tex]

For [tex]\( A(2, 8) \)[/tex] and [tex]\( C(6, 2) \)[/tex]:

[tex]\[ AC = \sqrt{(6 - 2)^2 + (2 - 8)^2} \][/tex]

[tex]\[ AC = \sqrt{4^2 + (-6)^2} \][/tex]

[tex]\[ AC = \sqrt{16 + 36} \][/tex]

[tex]\[ AC = \sqrt{52} \][/tex]

[tex]\[ AC \approx 7.21 \][/tex]

### 2. Calculate the perimeter of the triangle

The perimeter [tex]\( P \)[/tex] of the triangle is the sum of its side lengths:

[tex]\[ P = AB + BC + AC \][/tex]

[tex]\[ P \approx 15.23 + 10.00 + 7.21 \][/tex]

[tex]\[ P \approx 32.44 \][/tex]

### 3. Calculate the area of the triangle

To calculate the area of a triangle given its vertices, use the formula:

[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]

For [tex]\( A(2, 8) \)[/tex], [tex]\( B(16, 2) \)[/tex], and [tex]\( C(6, 2) \)[/tex]:

[tex]\[ \text{Area} = \frac{1}{2} \left| 2(2 - 2) + 16(2 - 8) + 6(8 - 2) \right| \][/tex]

[tex]\[ \text{Area} = \frac{1}{2} \left| 0 + 16(-6) + 6(6) \right| \][/tex]

[tex]\[ \text{Area} = \frac{1}{2} \left| -96 + 36 \right| \][/tex]

[tex]\[ \text{Area} = \frac{1}{2} \left| -60 \right| \][/tex]

[tex]\[ \text{Area} = \frac{1}{2} \times 60 \][/tex]

[tex]\[ \text{Area} = 30.00 \][/tex]

### Final Answer:
Hence, the perimeter of [tex]\( \triangle ABC \)[/tex] is approximately [tex]\( 32.44 \)[/tex] units, and its area is [tex]\( 30.00 \)[/tex] square units.
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