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

Answer: (D)

$\frac{7}{4}$

$5,5r,5r^2$ sides of triangle,

$5+5r>5r^2$ ...(1)

$5+5r^2>5r$ ...(2)

$5r+5r^2>5$ ...(3)

From (1) $r^2-r-1<0$,

$[r-\left(\frac{1+\sqrt{5}}{2}\right)][r-\left(\frac{1-\sqrt{5}}{2}\right)]<0$

$r\in (\frac{1-\sqrt{5}}{2},\frac{1+\sqrt{5}}{2})$ ...(4)

from (2),

$r^2-r+1>0\Rightarrow r\in R$ ...(5)

from (3),

$r^2+r-1>0$

So, $(r+\frac{1+\sqrt{5}}{2})(r+\frac{1-\sqrt{5}}{2})>0$

$r\in (-\infty ,-\frac{1+\sqrt{5}}{2})\cup (-\frac{1-\sqrt{5}}{2},\infty )$ ...(6)

from (4), (5), (6),

$r\in (\frac{-1+\sqrt{5}}{2},\frac{1+\sqrt{5}}{2})$