Question 14 Marks
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 difference 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$?$
Answer
View full question & answer→Let $P$ be the position of the pole and $A\ B$ be the opposite fixed gates. Let, $BP = x$ metres.
$\therefore AP = x + 7$
In right triangle $APB,$
$A P^2+B P^2=A B^2$
$\Rightarrow(x+7)^2+x^2=13^2$
$\Rightarrow x^2+49+14 x+x^2=169$
$\Rightarrow 2 x^2+14 x+49-169=0$
$\Rightarrow 2 x^2+14 x-120=0$
$\Rightarrow 2\left(x^2+7 x-60\right)=0$
$\Rightarrow x^2+7 x-60=0$
$\Rightarrow x^2+12 x-5 x-60=0$
$\Rightarrow x(x+12)-5(x+12)=0$
$\Rightarrow(x+12)(x-5)=0$
Either $x+12 = 0$, then $x = -12$ which is not possible being negative or $x - 5 = 0,$ then $x = 5.$
Thus $P$ is at a distance of $5m$ from $B$ and $5+7 = 12m$ from $A.$
$\therefore AP = x + 7$
In right triangle $APB,$
$A P^2+B P^2=A B^2$
$\Rightarrow(x+7)^2+x^2=13^2$
$\Rightarrow x^2+49+14 x+x^2=169$
$\Rightarrow 2 x^2+14 x+49-169=0$
$\Rightarrow 2 x^2+14 x-120=0$
$\Rightarrow 2\left(x^2+7 x-60\right)=0$
$\Rightarrow x^2+7 x-60=0$
$\Rightarrow x^2+12 x-5 x-60=0$
$\Rightarrow x(x+12)-5(x+12)=0$
$\Rightarrow(x+12)(x-5)=0$
Either $x+12 = 0$, then $x = -12$ which is not possible being negative or $x - 5 = 0,$ then $x = 5.$
Thus $P$ is at a distance of $5m$ from $B$ and $5+7 = 12m$ from $A.$