Math, asked by aishanajeeb33, 11 months ago

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

Answered by Arcel
35

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.

Answered by nitinop12
0

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

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