A plane makes interceptsOA, OB and OC whose measurements are a, b and c on the OX, OY and OZ axes. The area of triangle ABC is

Options

A. `1/2 (ab+ bc+ca)`
B. `1/2 (a ^(2) b^(2) +b ^(2) c ^(2) + c ^(2) a ^(2)) ^(1//2)`
C. `1/2 abc (a+b+c)`
D. `1/2 (a+b+c)^(2)`

Answer

To find the area of triangle ABC formed by the intercepts OA, OB, and OC on the OX, OY, and OZ axes with lengths a, b, and c respectively, we can follow these steps:

### Step 1: Understand the Geometry
The points A, B, and C can be represented in three-dimensional space as follows:
- A = (a, 0, 0)
- B = (0, b, 0)
- C = (0, 0, c)

### Step 2: Calculate the Lengths of the Sides of Triangle ABC
We need to find the lengths of the sides of triangle ABC:
- Length AB = distance between points A and B
- Length BC = distance between points B and C
- Length CA = distance between points C and A

Using the distance formula:
- \( AB = \sqrt{(a - 0)^2 + (0 - b)^2 + (0 - 0)^2} = \sqrt{a^2 + b^2} \)
- \( BC = \sqrt{(0 - 0)^2 + (b - 0)^2 + (0 - c)^2} = \sqrt{b^2 + c^2} \)
- \( CA = \sqrt{(0 - a)^2 + (0 - 0)^2 + (c - 0)^2} = \sqrt{c^2 + a^2} \)

### Step 3: Use Heron's Formula to Calculate the Area
To find the area of triangle ABC, we can use Heron's formula:
1. Calculate the semi-perimeter \( s \):
\[ s = \frac{AB + BC + CA}{2} = \frac{\sqrt{a^2 + b^2} + \sqrt{b^2 + c^2} + \sqrt{c^2 + a^2}}{2} \]

2. Apply Heron's formula:
\[ \text{Area} = \sqrt{s(s - AB)(s - BC)(s - CA)} \]

### Step 4: Substitute the Values
Substituting the lengths of the sides into Heron's formula gives:
\[ \text{Area} = \sqrt{s \left(s - \sqrt{a^2 + b^2}\right) \left(s - \sqrt{b^2 + c^2}\right) \left(s - \sqrt{c^2 + a^2}\right)} \]

### Step 5: Simplify the Expression
After simplification, the area of triangle ABC can be expressed as:
\[ \text{Area} = \frac{1}{2} \sqrt{a^2b^2 + b^2c^2 + c^2a^2} \]

### Final Answer
Thus, the area of triangle ABC is:
\[ \text{Area} = \frac{1}{2} \sqrt{a^2b^2 + b^2c^2 + c^2a^2} \]