two poles of height a metre and b metre are P metres apart . prove that the height of the point of intersection of lines joining top of each poleto the foot of opposite pole is given by
metres .
Answers
Answer:So, from the diagram, is similar to ,
Now, in , ,
So, from the diagram, is similar to ,
Now, on adding equation (1) and equation (2),
is the distance between two poles.
Step-by-step explanation:
Answer:
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Step-by-step explanation:
Let ABAB and CDCD be two poles of heights aa metres and bb metres respectively such that the poles are pp metres apart i.e. AC=pAC=p metres. Suppose the lines ADAD and BCBC meet at OO such that OL=hOL=h metres.
Let CL=x and LA=y.Then,x+y=p.Let CL=x and LA=y. Then, x+y=p.
In triangle ABC△ABC and triangle LOC△LOC, we have
angle CAB=angle CLO∠CAB=∠CLO [Each equal to 90^circ90
∘
]
angle C=angle C∠C=∠C [Common]
\therefore ∴ triangle CAB similar to triangle CLO
△CAB∼△CLO [By AA-criterion of similarity]
⇒ \dfrac { CA }{ CL } =\dfrac { AB }{ LO }
CL
CA
=
LO
AB
⇒ \dfrac { p }{ x } =\dfrac { a }{ h }
x
p
=
h
a
⇒ x=\dfrac { ph }{ a } x=
a
ph
...........(i)
In \triangle ALO△ALO and triangle ACD△ACD, we have
angle ALO=angle ACD∠ALO=∠ACD [Each equal to 90^ circ90
∘
]
angle A=angle A∠A=∠A [Common]
therefore ∴ triangle ALO sim triangle ACD△ALO∼△ACD [By AA-criterion of similarity]
⇒ dfrac { AL }{ AC } =dfrac { OL }{ DC }
AC
AL
=
DC
OL
⇒ \dfrac { y }{ p } =\dfrac { h }{ b }
p
y
=
b
h
⇒ y=\dfrac { ph }{ b } y=
b
ph
[\because ∵ AC=x+y=pAC=x+y=p]........(ii)
From (i) and (ii), we have
x+y=\dfrac { ph }{ a } +\dfrac { ph }{ b } x+y=
a
ph
+
b
ph
⇒ p=ph\left( \dfrac { 1 }{ a } +\dfrac { 1 }{ b } \right) p=ph(
a
1
+
b
1
) [\because ∵ x+y=px+y=p]
⇒ 1=h\left( \dfrac { a+b }{ ab } \right) 1=h(
ab
a+b
)
⇒ h=\dfrac { ab }{ a+b } \ metres h=
a+b
ab
metres
Hence, the height of the intersection of the lines joining the top of each pole to the foot of the opposite pole is \dfrac { ab }{ a+b }
a+b
ab
metres.
