a girl of height 90cm is walking away from the base of a lamp post at a speed of 1.2m/s. if the lamp is 3.6 above the ground find the length of her shadow after 4 second
Answers
Answer:
Length of her shadow after 4 second is 160 cm.
Step-by-step explanation:
Let the girl mentioned be standing at point D on the ground from the lamp post after 4 seconds.
∴ AD = 1.2 m/sec × 4 sec = 4.8 m = 480 cm
Assume the length of the shadow of the girl be x cm when she is at the position D.
∴ CD = x cm
Consider ∆CDE and ∆CAB:
∠CDE = ∠CAB (90° right angles)
∠DCE = ∠ACB (Common Angles)
∴ ∆CDE ∼ ∆CAB (By A.A. Similarity)
EC/EA = DC/BA = ED/BE (Corresponding Parts Of Similar Triangles)
EC/EA = ED/BE
=> 90 cm/360 cm = x cm/(480+x) cm
=> 1/4 = x/x+480
=> 4x = x + 480
=> 3x = 480
=> x = 480/3
∴ x = 160 cm
∴ Length of her shadow after 4 second is 160 cm.
Answer:
Step-by-step explanation:
Length of her shadow after 4 second is 160 cm.
Step-by-step explanation:
Let the girl mentioned be standing at point D on the ground from the lamp post after 4 seconds.
∴ AD = 1.2 m/sec × 4 sec = 4.8 m = 480 cm
Assume the length of the shadow of the girl be x cm when she is at the position D.
∴ CD = x cm
Consider ∆CDE and ∆CAB:
∠CDE = ∠CAB (90° right angles)
∠DCE = ∠ACB (Common Angles)
∴ ∆CDE ∼ ∆CAB (By A.A. Similarity)
EC/EA = DC/BA = ED/BE (Corresponding Parts Of Similar Triangles)
EC/EA = ED/BE
=> 90 cm/360 cm = x cm/(480+x) cm
=> 1/4 = x/x+480
=> 4x = x + 480
=> 3x = 480
=> x = 480/3
∴ x = 160 cm
∴ Length of her shadow after 4 second is 160 cm