A vertical pole of length 6cm castsa shadow 4m long on the ground

1.	A vertical pole of length 6cm castsa shadow 4m long on the ground
Answer» Let AB denoted the vertical pole of length 6m.BC is the shadow of the pole on the ground BC = 4m.Let DE denote the tower.EF is shadow of the tower on the ground.EF = 28 m.Let the height of the tower be h m.In {tex}\\triangle {/tex}ABC and {tex}\\triangle {/tex}DEF,{tex}\\angle{/tex}B = {tex}\\angle{/tex}E ......[Each equal to 90o because pole and tower are standing vertical to the ground]{tex}\\angle{/tex} C = {tex}\\angle{/tex} F .....[Same elevation]{tex}\\angle{/tex}A = {tex}\\angle{/tex}D {tex}\\because {/tex} shadows\xa0are cast at the same time{tex}\\therefore {/tex}{tex}\\triangle {/tex}ABC and {tex}\\triangle {/tex}DEF,{tex}\\angle{/tex}B= {tex}\\angle{/tex} E ............[Each equal to 90o because pole and tower are standing vertical to the ground.]{tex}\\angle{/tex}A={tex}\\angle{/tex}D ({tex}\\because {/tex}shadows\xa0are cast at the same time){tex}\\therefore {/tex}{tex}\\vartriangle {/tex}ABC ~{tex}\\vartriangle {/tex}DEF ......(AA similarity criterion){tex}\\therefore {/tex}{tex}\\frac{{AB}}{{DE}} = \\frac{{BC}}{{EF}}{/tex} ...........[{tex}\\because {/tex} corresponding sides of two similar triangles are proportional]{tex}\\Rightarrow {/tex}{tex}\\frac{6}{h} = \\frac{4}{{28}}{/tex}{tex}\\Rightarrow {/tex}{tex}h = \\frac{{6 \\times 28}}{4} \\Rightarrow h = 42{/tex}Hence, the height of the tower is 42 m

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