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.
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∴ 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 cm
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
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