Solving triangles!? URGENT!!!!?

Question

A light, on top of a post, is 14 feet above the ground. At night a 5 foot woman is standing 18 feet from the post. The woman's shadow is 10 feet long. Her friend is standing 16 feet away from the post.

Starting 18 feet from the post, the woman walks directly away from the post for 4 seconds at a rate of 3 feet per second. Find the average speed(average speed = total distance/total time) on the tip of her shadow during that 4 second interval

http://i37.tinypic.com/2us8y7s.jpg

How do i solve this...

Answer 1

Let:

h be the height of the woman,

x be her distance from the lamp,

H be the height of the lamp,

s be the distance from the lamp of the tip of her shadow.

From similar triangles:

(s - x) / h = s / H ...(1)

Initially:

s = 28 ft, x = 18 ft, h = 5 ft.

After walking 4 sec. at 3 ft/sec:

x = 18 + 4 * 3 = 30 ft.

Using the new value of x in (1):

(s - 30) / 5 = s / 14

14(s - 30) = 5s

14s - 420 = 5s

9s = 420

s = 420 / 9 ft.

The tip of her shadow has moved

420 / 9 - 28

= (420 - 252) / 9

= 168 / 9 ft in 4 sec.

Its speed is:

168 / (9 * 4)

14 / 3 ft/sec.

There are discrepancies with some of the measurements in your diagram. I have taken 18 ft.to be the initial distance from the lamp. However, the measurements of 16 ft., 4.74 ft. and 2 ft. in the diagram cannot all be accurate.

Answer 2

let shadow's tip total distance be d.

x / 5 = (18 + 3*4 + x) / 14

x / 5 = (x + 30) / 14

14x = 5x + 150

9x = 150

x = 150 / 9 = 50 / 3

d = (x + 30) - (10 + 18)

= 50/3 + 2

= 56/3

average speed

= d / (4 seconds)

= 56 / 3*4

= 14 / 3

= 4.667 feet per second