Question

two poles of heights 6 metre and 11 metre stand vertically upright on a plane ground if the distance between is there feet is 12 metre then find the the distance between their tops​

Answer 1

GIVEN :-

• height of first pole (ab) = 6m

• height of second pole (cd) = 11m

• distance b/w feet of poles (ac) = 12m

TO FIND :-

• distance between the top of poles (bd)

SOLUTION :-

let we draw a line BE perpendicular through DC

since AC is also perpendicular to DC as pole vertical to ground

so , BE = AC = 12m

similarly , AB = EC = 6m

now ,

DE = DC - EC

DE = 11 - 6

DE = 5m

$\rm{since \: \angle \: BED = 90 \degree \:as \: BE \perp DC \: }$

$\rm{ \triangle \:BED \: is \: a \: right \: angle \: triangle}$

now we know according to Pythagoras theorem

$\implies \boxed{ \rm{ \bf{ {h}^{2} = {p}^{2} + {b}^{2} }}}$

$\implies \rm{ {(BD)}^{2} = {(DE)}^{2} + {(BE)}^{2} }$

$\implies \rm{ {(BD)}^{2} = {(5)}^{2} + {(12)}^{2} }$

$\implies \rm{ {(BD)}^{2} = 25 + 144 }$

$\implies \rm{ {(BD)}^{2} = 169 }$

$\implies \rm{ \sqrt{ {BD}^{2} } = \sqrt{169 } }$

$\implies \boxed{ \boxed{ \rm{ {BD} = 13 }}}$

hence distance between tops of pole is 13m