Question

If $5,5r,5r^2$ are the lengths of the sides of a triangle, then $r$ cannot be equal to :

• A. $\frac{3}{2}$
• B. $\frac{3}{4}$
• C. $\frac{5}{4}$
• D. $\frac{7}{4}$

Answer 1

The correct option is D $\frac{7}{4}$


The triangle is possible if,
$5+5r>5r^2\cdots \left(1\right)$​​​​​
$5r+5r^2>5\cdots \left(2\right)$​​​​​
$5r^2+5>5r\cdots \left(3\right)$​​​​​

From eqn $\left(1\right)$
$5r^2-5r-5<0$
$\Rightarrow r^2-r-1<0$
$\Rightarrow [r-\left(\frac{1-\sqrt{5}}{2}\right)][r-\left(\frac{1+\sqrt{5}}{2}\right)]<0$
$\Rightarrow 0<r<\frac{1+\sqrt{5}}{2}\cdots \left(4\right)$
$\because r>0$

From eqn $\left(2\right)$
$5r^2+5r-5>0$
$r^2+r-1>0$
$(r-\left(\frac{-1-\sqrt{5}}{2}\right))(r-\left(\frac{-1+\sqrt{5}}{2}\right))>0$
$r>\frac{-1+\sqrt{5}}{2}\cdots \left(5\right)$
$\because r>0$

From eqn $\left(3\right)$
$5r^2-5r+5>0$
$\Rightarrow r^2-r+1>0\Rightarrow r\in R^{+}\cdots \left(6\right)(\because D=1-4=-3<0)$

Therefore, from $\left(4\right),\left(5\right)$ and $\left(6\right)$, we get
$\left(\frac{-1+\sqrt{5}}{2}\right)<r<\left(\frac{1+\sqrt{5}}{2}\right)$
$r\in (0.618,1.618)$
Hence, $r\ne \frac{7}{4}$