Question

If $5,5r,5r^2$ are the side lengths of a triangle, then the possible value(s) of $r$ is/are

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

Answer 1

Answer: (A), (B)

$\frac{3}{4}$


As $5,5r,5r^2$ are the sides of a triangle.
So, $r>0$

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 equation $\left(1\right)$
$5r^2-5r-5<0$
$\Rightarrow r^2-r-1<0$
$r^2-r-1=0\Rightarrow r=\frac{1\pm \sqrt{5}}{2}$
So,
$r^2-r-1<0\Rightarrow \frac{1-\sqrt{5}}{2}<x<\frac{1+\sqrt{5}}{2}$
$\Rightarrow 0<r<\frac{1+\sqrt{5}}{2}\cdots \left(4\right)[\because r>0]$

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

From equation $\left(3\right)$
$5r^2-5r+5>0$
$\Rightarrow r^2-r+1>0D=1-4=-3<0\Rightarrow x\in R$

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

Hence, from the given options only $r=\frac{3}{2},\frac{3}{4}$ satisfied above condition.