Question

$\ \textless \ font color="red"\ \textgreater \ \ \textless \ b\ \textgreater \ \ \textless \ i\ \textgreater$✨A pole has to be erected at a point on the boundary of a circular Park of diameter 13 metres in such a way that the differences of its distances from two diametrically opposite fixed points A and B on the boundary is 7 metres.At what distances from the two gates should the pole be erected ?✨

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Answer 1

$\huge{\bold{\underline{\underline{ANSWER:-}}}}$$<b><i>$

The poles should be erected at the distances 12 metres and 5 metres from the gates.

$\bold{\underline{\underline{Explanation:-}}}$

$\mathsf{Let\:P\:be\:the\:location\:of \:the\:pole\:on\:the}$

$\mathsf{circular\:boundary.\:Let \:the\:distance\:of\:the\:pole\:P\:from}$

$\mathsf{the\:gate\:A\:be\:x \:metres,i.e.,AP=x\:metres.}$

$\mathsf{As\:the\:difference\:of \:distances\:of\:the }$

$\mathsf{pole\:from\:the\:two \:gates = BP-AP}$

$\mathsf{(or\:AP-BP)is\:7 \:m, \:therefore, \:BP=(x+7)\:m.}$

$\mathsf{Given,AB=13\:m\:and\:AB\:is\:a\:diameter,}$

$\mathsf{\angle\:APB=90\degree......}$

$\mathsf{(angle\:in\:a\:semicircle\:is\:90\degree)}$

$\mathsf{In\:\triangle\:APB,\angle\:APB=90\degree.}$

$\mathsf{By\:Pythagoras\:theorem,\:we\:get}$

$\mathsf{{AP}^{2} + {BP}^{2} = {AB}^{2}}$

$\mathsf{\implies{x}^{2} + {(x + 7)}^{2} = {13}^{2}}$

$\mathsf{\implies{x}^{2} + {x}^{2} + 14x + 49 = 169}$

$\mathsf{\implies{2x}^{2} + 14x - 120 = 0}$

$\mathsf{\implies{x}^{2} + 7x - 60 = 0}$

$\mathsf{\implies{x}^{2} + 12x - 5x - 60 = 0}$

$\mathsf{\implies\:x(x + 12) - 5(x + 12) = 0}$

$\mathsf{\implies(x + 12)(x - 5) = 0}$

$\mathsf{\implies(x + 12) = 0 \: \: \:or \: \: \implies( x - 5) = 0}$

$\mathsf{\implies\:x = - 12 \: \: \: or \: \: \: \implies\:x = 5}$

$\mathsf{But\:x\:being\:distance\:AP\:can't\:be\:negative.}$

$\mathsf{\implies\:x = 5\:is\:the\:correct\:answer}$

$\mathsf{Therefore,x=5\:and\:x+7=5+7=12}$

$<b><u>$Hence, a pole can be erected on the boundary of the circular park at 5 m and 12 m from gates A and B or at 12 m and 5 m from gates A and B respectively.

Answer 2

Step-by-step explanation:

Let the required position of the pole be P.

We have, AB = 13 cm, PB = x cm.

Angle in a semi-circle, ∠APB = 90°.

By Pythagoras theorem, we get

⇒ AB² = AP² + x²

⇒ 13² = AP² + x²   ------ (i)

Given that fixed points A and B on the boundary is 7 metres.

⇒ AP - x = 7 m       ------------ (ii)

On Squaring both sides, we get

⇒ (AP - x² = 7²

⇒ AP² + x² - 2 * AP * x = 49

⇒ 13² - 2 * AP * x = 49

⇒ AP * x = 60

⇒ AP = (60/x)      ----------------- (iii)

Substitute (iii) in (ii), we get

⇒ (60/PB) - PB =7

⇒ 60 - x² = 7x

⇒ x² + 7x - 60 = 0

⇒ x² + 12x - 5x - 60 = 0

⇒ x(x + 12) - 5(x + 12) = 0

⇒ (x - 5)(x + 12) = 0

⇒ x = 5, 12{Rejected}

⇒ x = 5

So, PB = x = 5 m.

AP = 7 + x = 12 m.

∴ Poles can be erected at distances 5 m and 12 m from the gates.

Hope it helps!