Question

Two vertical posts stand side by side. One post is 8 feet tall and the other is 17 feet tall. If a 24 foot wire is stretched between the tops of the posts, how far apart are the posts?

Answer 1

Answer:

The distance between the posts is $3\sqrt{55}$ feet.

Step-by-step explanation:

In the attached figure, let DB and AC be the posts and let AD be the wire attached to the top of the posts.

BC is the distance between the posts and we need to find BC.

Note that BC = DE

Also, EC = DB = 8m and AE = AC - EC = 17 - 8 = 9 m.

Now, in the right triangle ADE,

$AD^{2} =AE^{2} +DE^{2}$

$24^{2} =9^{2} +DE^{2}$

$DE^{2} =24^{2} -9^{2}$

= (24+9) (24-9)

= 33(15)

= (3 × 11) × (3 × 5)

$DE=3\sqrt{55}$

Since DE = BC,

$BC=3\sqrt{55}$

Hence, the distance between the posts is $3\sqrt{55}$ feet.