Question

Question #e2f2f

Answer 1

25 ft wire is attached to the ground at `color(blue)(8.75 ft)` from the base of the pole.

Explanation:

I’m using a figure, available on the net, that is somewhat relevant to this sum. All the measurements labeled in the figure may be ignored.

Given
`BC = 30 ft, BD = 25 ft, DC = 10 ft`

Let `AD = x, AB = y`

To find ‘x’

`AB^2 = BD^2 + AD^2`

`y^2 = 25^2 - x^2`

`y^2 + x^2 = 625`. Eqn (1)

`AB^2 = BC^2 + AC^2 = BC^2 + (AB + DC)^2`

`y^2 = 30^2 - (x + 10)^2`

`y^2 + x^2 + 20x = 30^2 - 10^2 = 800` Eqn (2)

Substituting Eqn (1) from Eqn(2),

`cancel(y^2 + x^2) + 20x - cancel(y^2 + x^2) = 800 - 625`

`20x = 175` or `x = 175/20 = 8.75 ft`

25 ft wire is attached to he ground at 8.75 ft from the base of the pole.