Question

A girl of height 90 cm is walking away from the base of a lamp-post at a speed of 1.2 metre per second. if the lamp is 3.6 m above the ground. find the length of eyeshadow after 4 seconds.

Answer 1

Answer:

Step-by-step explanation:

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

∴ BD = 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.

∴ ED = x cm

In ∆EDC and ∆EBA

∠EDC = ∠EBA  (90°)

∠DEC = ∠BEA (Common)

∴ ∆EDC ∼ ∆EBA  (AA similarity)

EC/EA = DC/BA = ED/BE

EC/EA = ED/BE

90cm/360cm = x cm/(480+x) cm

1/4 = x/x+480

4x = x + 480

3x = 480

x = 480/3

x = 160

(corresponding sides or proportional)

∴ Length of her shadow after 4 second is 160

Thanks

Answer 2

Answer:

The length of eyeshadow after 4 seconds is 160 m.

Step-by-step explanation:

Let the girl of height be at point D on the ground from the lamp post after 4 seconds. Therefore,

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. Therefore,

BD=x cm

In ∆BDE and ∆BAC,

(i) ∠BDE=∠BAC  (90°)

(ii)∠DBE=∠ABC   (Common)

Thus, ∆BDE∼∆BAC   (AA similarity)

  $\frac{BE}{BC} = \frac{DE}{AC} = \frac{BD}{AB}$

( Corresponding sides are proportional )

$\frac{90}{360} = \frac{x}{480 + x} \\ \frac{1}{4} = \frac{x}{480 + x} \\ 4x = 480 + x \\ 4x - x = 480 \\ 3x = 480 \\ x = \frac{480}{3} \\ x = 160$

Hence length of her shadow after 4 seconds is 160 m.