A street light bulb is fixed on a pole 6m above the level of the street. If a woman of height 1.5m casts a shadow of 3m, find how far she is away from the base of the pole.
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In $\triangle\text{LPS}$ and $\triangle\text{NWS},$Bulb L is fixed at a height of 6m above the road SP.
Woman and pole are vertical.
$\therefore\angle1=\angle2=90^\circ$
$\angle\text{S}=\angle\text{S}$ [Common]
$\therefore\triangle\text{LPS}\sim\triangle\text{NWS}$ [By AA similarity criterion]
$\Rightarrow\frac{\text{LP}}{\text{NW}}=\frac{\text{LS}}{\text{NS}}=\frac{\text{PS}}{\text{WS}}$
$\Rightarrow\frac{6\text{m}}{15\text{m}}=\frac{\text{LS}}{\text{NS}}=\frac{3+\text{x}}{3}$
$\Rightarrow\frac{6}{15}=\frac{3+\text{x}}{3}$
$\Rightarrow6\times3=1.5(3+\text{ x})$
$\Rightarrow18=4.5+1.5\text{x}$
$\Rightarrow1.5\text{x}=4.5-18$
$\Rightarrow1.5\text{x}=13.5$
$\Rightarrow\text{x}=\frac{13.5}{1.5}$
$\Rightarrow9\text{m}$
Hence, the woman is 9m away from the pole.
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