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Sia ? 6 years, 6 months ago
Given: height of first pole CD = 6 m
Height of second pole AB = 11 m
Distance between feet of poles BD = 12 m
To find: distance between the tops of the pole
i.e, AC

Draw {tex}\mathrm { CE } \perp \mathrm { AB }{/tex} and join AC.
{tex}\therefore{/tex} CE = DB = 12 m
AE = AB – BE = AB – CD = (11 – 6) m = 5 m
In right angled triangle ACE, by Pythagoras theorem, we have
{tex}A C ^ { 2 } = C E ^ { 2 } + A E ^ { 2 } = 12 ^ { 2 } + 5 ^ { 2 }{/tex}
= 144 + 25 = 169
{tex}\Rightarrow{/tex} AC = 13 m
Hence, the distance between the tops of the two poles is 13 m.
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