Answer:
The length of the shadow is increasing with the rate of 1.5 feet per sec
Step-by-step explanation:
Let AB and CD represents the height of the lamppost and child respectively ( shown below )
Also, let E be a point represents the position of child.
In triangles ABE and CDE,
[tex]\angle ABE\cong \angle CDE[/tex] Â Â ( right angles )
[tex]\angle AEB\cong \angle CED[/tex] Â ( common angles )
By AA similarity postulate,
[tex]\triangle ABE\sim \triangle CDE[/tex]
∵ Corresponding sides of similar triangles are in same proportion,
[tex]\implies \frac{AB}{CD}=\frac{BE}{DE}[/tex]
We have, AB = 12 ft, CD = 4 ft, BE = BD + DE = 6 + DE,
[tex]\implies \frac{12}{4}=\frac{6+DE}{DE}[/tex]
[tex]12DE = 24 + 4DE[/tex]
[tex]8DE = 24[/tex]
[tex]DE=3[/tex]
Now, the speed of walking = 2 mph = [tex]\frac{2\times 5280}{3600}\approx 2.933\text{ ft per sec}[/tex]
Note: 1 mile = 5280 ft, 1 hour = 3600 sec
Thus, the time taken by child to reach at E Â
[tex]= \frac{\text{Walked distance}}{\text{Walking speed}}[/tex]
[tex]=\frac{6}{2.933}[/tex]
= 2.045 hours
Hence, the change rate in the length of shadow
[tex]= \frac{\text{Length of shadow}}{\text{Time taken}}[/tex]
[tex]=\frac{3}{2.045}[/tex]
= 1.5 ft per sec.
