A pole has to be erected at a point on the boundary of a circular of diameter 13 metres in such a way that the differences o

2 thoughts on “<br />A pole has to be erected at a point on the boundary of a circular of diameter 13 metres in such a way that the differences o”

1. [tex] {\fcolorbox{aqua}{black}{\orange{ Provided\: Question\:» }}}[/tex]

   A pole has to be erected at a point on the boundary of a circular of diameter 13 metres in such a way that the differences of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 metres. Is it possible to do so? If yes, at what distances from the two gates should the pole be erected ?

   ★ Diagram in attachment*

   [tex] {\fcolorbox{green}{black}{\blue{ Here’s\: the\: Answer\:» }}}[/tex]

   [tex] \boxed{\sf{\underline{Sol^{n}\: \leadsto}}}[/tex] Suppose ,

   C is the position of the pole at the boundary of the circular park

   Given ,

   • AB is the diameter of the park

   Where

   • AB = 13 metre

   Let ,

   • BC = x metre
   • AC = ( x + 7 ) metre

   [tex]\because[/tex] Angle at the semi-circle is Right Angle

   [tex]\therefore[/tex] [tex] \angle[/tex] ACB = 90⁰

   [tex]\therefore[/tex] By Pythagoras Theorem

   ➪ [tex]\sf{\orange{ BC² + AC² \:=\: AB²}}[/tex]

   ➪ [tex]\sf{\red{ x² + ( x + 7 )² \:=\: 13² }} [/tex]

   ➪ [tex]\sf{ x² + x² + 14x + 49 \:=\: 169 } [/tex]

   ➪ [tex]\sf{ 2x² + 14x + 49 – 169 \:=\: 0 } [/tex]

   ➪ [tex]\sf{ 2x² + 14x – 120 \:=\: 0 } [/tex]

   ➪ [tex]\sf{\green{ x² + 7x – 60 \:=\: 0 }} [/tex]

   Here ,

   • a = 1
   • b = 7
   • c = –60

   [tex] \therefore[/tex] [tex]\boxed{\sf{\purple{b² \: -4ac}}} [/tex]

   ㅤㅤ= ( 7 )² – 4 ( 1 ) ( –60 )

   ㅤㅤ= 49 + 240

   ㅤㅤ= 289 > 0

   [tex] \therefore[/tex] The given design is possible

   Again ,

   ㅤㅤ = [tex]\sf{ x² + 7x – 60 \:=\: 0 } [/tex]

   ㅤㅤ = [tex]\sf{ x² + ( 12 + 5 )x – 60 \:=\: 0 } [/tex]

   ㅤㅤ = [tex]\sf{ x² + 12x + 5x – 60 \:=\: 0 } [/tex]

   ㅤㅤ = [tex]\sf{ x ( x + 12 ) – 5 ( x + 12 ) \:=\: 0 } [/tex]

   ㅤㅤ = [tex]\sf{\blue{ ( x + 12 ) ( x – 5 ) \:=\: 0 }} [/tex]

   [tex] \therefore[/tex] Either

   ㅤㅤ( x + 12 ) = 0ㅤorㅤ( x – 5 ) = 0

   ㅤㅤㅤ➪ x = – 12ㅤorㅤㅤ➪ x = 5

   ㅤㅤㅤ( not valid )

   ㅤㅤㅤ

   • [tex]\therefore[/tex] BC = 5 metre

   • ㅤAC = ( 5 + 7 ) metre
   • ㅤ ㅤ = 12 metre

   ㅤㅤㅤ

   ㅤㅤㅤ꧁ ʙʀᴀɪɴʟʏ×ᴋɪᴋɪ ꧂

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2. [tex]\huge\fbox{\green{\underline{Question:}}}\\\\[/tex]

   A pole has to be erected at a point on the boundary of a circular park of diameter 13 meters in such a way that the difference of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 meters. Is it the possible to do so? If yes, at what distances from the two gates should the pole be erected?

   [tex]\\\\\huge\fbox{\green{\underline{Answer:}}}\\\\[/tex]

   Let P be the required location on the boundary of a circular park such that its distance from gate B is x metre that is BP x metres.

   Then, AP = x + 7

   In the right triangle ABP we have by using Pythagoras theorem

   AP² + BP² = AB²

   (x + 7)² + x² = (13)²

   x² + 14x + 49 + x² = 169

   2x² + 14x + 49 – 169 = 0

   2x² + 14x – 120 = 0

   2(x² + 7x – 60) = 0

   x² + 7x – 60 = 0

   x² + 12x – 5x – 60 = 0

   x(x + 12) – 5(x – 12) = 0

   (x + 12)(x – 5) = 0

   x + 12 = 0

   x = -12

   Or

   x – 5 = 0

   x = 5

   But the side of right triangle can never be negative

   Therefore, x = 5

   Hence, P is at a distance of 5 metres from the gate B.

   ⇒ BP = 5m

   Now, AP = (BP + 7)m = (5 + 7)m = 12 m

   ∴ The pole has to be erected at a distance 5 mtrs from the gate B and 12 m from the gate A.

   [tex]\\\\\\[/tex]

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