JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Properties of Triangle

For $5,5r,5r^2$ to be a triangle's sides the triangle inequalities must all hold. Dividing out the $5$, the binding ones are $1+r>r^2$ and $r+r^2>1$, which together confine $r$ to between $\frac{\sqrt{5}-1}{2}$ and $\frac{\sqrt{5}+1}{2}$, about $(0.618,1.618)$. Of the offered values $\frac{3}{4}$, $\frac{5}{4}$ and $\frac{3}{2}$ all fall inside, but $\frac{7}{4}=1.75$ lies beyond the upper limit, so $r$ cannot equal $\frac{7}{4}$.