Question

A girl of heigh 90 cm is walking away from the base of a lamp-post at a speed of 1.2 m/sec. If the lamp is 3.6 m above the ground, find the length of her shadow after 4 seconds.

Answer 1

SOLUTION :  

Given : Girl’s height = 90 cm = 90/100= 0.9 m, speed = 1.2m/sec , time = 4 sec and height of lamp post = 3.6 m

Let, AB be the lamp post , CD be the girl

Let, DE is the length of her shadow = x .

Distance moved by girl (BD) = speed × time

1.2 x 4

BD = 4.8 m

In ΔABE and ΔCDE,

∠B = ∠D  (each 90°)

∠E = ∠E  (common)

ΔABE∼ΔCDE

[By A-A similarity criterion]

BE/DE = AB/CD

[Corresponding Sides of a similar triangle are proportional]

BD + DE / DE = AB/CD

(4.8+x)/ x = 3.6/0.9  

4.8+x/ x = 4

4.8 +x  = 4x

4x - x = 4.8

3x = 4.8

x = 4.8/3

x = 1.6

Hence, the length of her shadow after 4 sec. is 1.6 m.

HOPE THIS ANSWER WILL HELP YOU…

Answer 2

Hey mate hope you understand this

Then the solution is:

Let the girl be at point D on the ground from the lamppost after 4 sec.

∴ AD = 1.2 m/sec × 4 sec = 4.8 m = 480 cm

Suppose the length of the shadow of the girl be x cm when she at position D.

∴ CD = x cm

In ∆CDE and ∆CAB

∠CDE = ∠CAB  (90°)

∠DCE = ∠ACB  (Common)

∴ ∆CDE ∼ ∆CAB  (AA similarity)


∴ Length of her shadow after 4 second is 160 cm