Question

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

Answer 1

Let $P$ be the pole and $A\&B$ are opposite fixed gates.

Given$PA-PB=Y$

Let$PA=a,PB=b$

hence$a-b=7\therefore a=7+b-\left(i\right)$

In right$△PAB,$

${AB}^2={AP}^2+{BP}^2${ By pythagoras thm}

$132={(7+b)}^2+b^2$

$169=49+14b+2b^2$

$⟹b^2+7b-60=0$

$\therefore b=5$ or$-12$

$\therefore b=5$and$a=7+5=12$

$PA=12mandPB=5m$

Sum of the gates$=12+5=17m$