Question

Two radio towers have been positioned 8 miles apart at points A and B. Point D is located exactly 4 miles from point A. The third tower will be placed at point C so that all three towers are equidistant from each other.

What should be the distance from the third tower to point D? Round the answer to the nearest tenth of a mile.

Answer 1

6.9 miles

Explanation:

Since the distance between points A and B is 8 miles, and point C is equidistant from points A and B, triangle ABC is an equilateral triangle with side lengths of 8 miles.

As point D is 4 miles from point A, point D is the midpoint of AB.

Therefore, the distance between point C and point D is the altitude of the triangle.

The formula for the altitude of an equilateral triangle is:

`h=(√(3))/(2)s`

where s is the the side length.

Therefore, the altitude of an equilateral triangle with side length 8 miles is:

`\implies h=(√(3))/(2) \cdot 8`

`\implies h=4√(3)`

`\implies h=6.9\; \sf miles\;(nearest\;tenth)`

Therefore, the distance from the third tower C to point D is 6.9 miles to the nearest tenth of a mile.

Answer 2

7 miles

Explanation:

Let's call the distance from point C to point D as "x".

We know that the distance from A to B is 8 miles, and the distance from A to D is 4 miles. To ensure that all three towers are equidistant from each other, the distance from C to B must be equal to 8 miles.

Using the Pythagorean Theorem, we can find the distance from C to D:

(x)^2 + (4)^2 = (8)^2

Expanding and simplifying the equation:

x^2 + 16 = 64

x^2 = 48

x = sqrt(48) = 6.928

So the distance from the third tower to point D is approximately 6.9 miles (rounded to the nearest tenth of a mile).