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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.
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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 ($\frac{1}{2}$ mark). In right $\Delta APB$, $(x+35)^2 + x^2 = (65)^2$ (1 mark). $x^2 + 35x - 1500 = 0$ (2 marks). $(x+60)(x-25) = 0$, $x=25$ ($\frac{1}{2}$ mark). Hence, $x+35=60$. Distance of $P$ from $A = 60$ m, Distance of $P$ from $B = 25$ m ($\frac{1}{2} + \frac{1}{2}$ marks).