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Question
A girl of height 90 cm is walking away from the base of a lamp-post at a speed of 1.2m/sec. If the lamp is 3.6 m above the ground, find the length of her shadow after 4 seconds.
Solution
We have,
Height of girl = 90 cm = 0.9 m
Height of lamp-post = 3.6 m
Speed of girl = 1.2 m/sec
∴ Distance moved by girl (CQ) = Speed × Time
= 1.2 × 4 = 4.8m
Let length of shadow (AC) = x cm
In ΔABC and ΔAPQ
∠ACB = ∠AQP [Each 90°]
∠BAC = ∠PAQ [Common]
Then, ΔAB ~ ΔAPQ [By AA similarity]
`therefore"AC"/"AQ"="BC"/"PQ"` [Corresponding parts of similar Δ are proportional]
`rArrx/(x+4.8)=0.9/3.6`
`rArrx/(x+4.8)=1/4`
⇒ 4x = x + 4.8
⇒ 4x – x = 4.8
⇒ 3x = 4.8
`rArrx=4.8/3=1.6` m
∴ Length of shadow = 1.6m
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