A vertical pole of length 15 m standing on the point A. ABC are points on horizontal ground forming a triangle with `angleA = 90^(@)`.
If AB = 6m` and AC = 8m and a point 'D' (on horizontal plane) is equidistant from A, B, C. Find distance of D from top of the pole

Options

A. `5sqrt(10)m`
B. `10sqrt(5)m`
C. `6sqrt(10)m`
D. None of these

Explanation

To solve the problem, we need to find the distance from point D on the horizontal plane to the top of the pole (point F) which is at point A. Here are the steps to find the solution:

### Step 1: Identify the triangle and its dimensions
We have a right triangle ABC with:
- AB = 6 m
- AC = 8 m
- Angle A = 90°

### Step 2: Calculate the length of side BC
Using the Pythagorean theorem:
\[ BC = \sqrt{AB^2 + AC^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \text{ m} \]

### Step 3: Find the midpoint D of segment BC
Since point D is equidistant from points A, B, and C, it is located at the midpoint of segment BC. The length of BD and CD is half of BC:
\[ BD = CD = \frac{BC}{2} = \frac{10}{2} = 5 \text{ m} \]

### Step 4: Form a right triangle AFD
Now, we can form a right triangle AFD where:
- AD = 5 m (the distance from A to D)
- AF = 15 m (the height of the pole)

### Step 5: Calculate the distance FD using the Pythagorean theorem
In triangle AFD, we need to find the hypotenuse FD:
\[ FD = \sqrt{AF^2 + AD^2} = \sqrt{15^2 + 5^2} = \sqrt{225 + 25} = \sqrt{250} = 5\sqrt{10} \text{ m} \]

### Final Answer
The distance from point D to the top of the pole (point F) is:
\[ FD = 5\sqrt{10} \text{ m} \]

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