Question

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

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Answer 1

$\huge \bf \red{ \mid{ \overline{ \underline{ANSWER}}} \mid}$

$\bf{step \: by \:step \: explanation}$

Let P be the required location of pole .

A and B are the two gates.

Let the distance of pole at p from gate be 'x' meters

i.e. PB = x m

•°• AP = (x + 7)m and AB = 13 m

$\sf{as \: \: \: \: \: \: \: \angle \: APB = 90 \degree}$

•°• AP² + BP² = AB² => (x +7)² + x =13²

$\sf \implies 2 {x}^{2} + 14x + 49 - 169 = 0 \\ \\ \sf \implies2 {x}^{2} + 14x - 120 = 0 \\ \\ \sf \implies {x}^{2} + 7x - 60 = 0 \\ \\ \sf \implies(x + 12)(x - 5)$

Rejecting x =-12, we get x = 5

•°• PB = 5m

and

AP = 12m

Thus, it is possible to erect a pole at a point p on the boundary of the park at distances 5m and 12m respectively from gates B and A.

Answer 2

${\huge {\mathfrak {Answer :-}}}$

Let P be the required location of pole .

A and B are the two gates.

Let the distance of pole at P from gate be 'y' meters

i.e. PB = y m

➡️ AP = (y + 7) m and AB = 13 m.

We have, angle APB = 90°.

${\mathfrak {\pink {- :Pythagoreas \:theorem :-}}}$

(AP)^2 + (BP)^2 = (AB)^2

Or, (y + 7)^2 + y^2 = 13^2

Or, y^2 + 14y + 49 + y^2 = 169

Or, 2y^2 + 14y + 49 - 169 = 0

Or, 2y^2 + 14y - 120 = 0

Or, y^2 + 7y - 60 = 0

Or, y^2 + 12y - 5y - 60 = 0

Or, y(y + 12) - 5(y + 12) = 0

Or, (y - 5)(y + 12) = 0

Or, y = 5,- 12.

Value of y can't be negative.

So, we get x = 5

➡️ PB = 5m and AP = (5 +7) = 12m

$<b>➡️ Hence , it is possible to erect a pole at a point p on the boundary of the park at distances 5m and 12m respectively from gates B and A. </b>$

$<i>$ That's it..

${\red {-:Arnab \: here :-}}$
$<marquee >$
${\huge {\blue {Thanks}}}$