Answer 1

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