45
There is a circular park of diameter $65$ m as shown in the following figure, where AB is a diameter. An entry gate is to be constructed at a point P on the boundary of the park such that distance of P from A is $35$ m more than the distance of P from B. Find distance of point P from A and B respectively.
_q_1.png)
Show SolutionHide Solution↓
Let distance of gate at P from point B is $x$ m. Then distance of gate at P from point A is $(35 + x)$ m.
In right $\Delta APB$, $(x + 35)^2 + x^2 = (65)^2 \implies x^2 + 35x - 1500 = 0 \implies (x + 60)(x - 25) = 0 \implies x = 25$.
Hence, $x + 35 = 60$. Distance of P from $A = 60$ m, Distance of P from $B = 25$ m.
In right $\Delta APB$, $(x + 35)^2 + x^2 = (65)^2 \implies x^2 + 35x - 1500 = 0 \implies (x + 60)(x - 25) = 0 \implies x = 25$.
Hence, $x + 35 = 60$. Distance of P from $A = 60$ m, Distance of P from $B = 25$ m.