If 5, 5r, 5r² are the lengths of the sides of a triangle, then r cannot be equal to
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: B
Solution
For three sides to form a triangle, the sum of any two sides must exceed the third. The critical inequality for sides 5, 5r, 5r² (assuming r > 0) is that the sum of the two smaller sides must exceed the largest.
For r > 1: largest side is 5r², so need 5 + 5r > 5r² → r² - r - 1 < 0 → r < (1+√5)/2 ≈ 1.618. Since 7/4 = 1.75 > 1.618, it fails this inequality.
For r < 1: largest side is 5, so need 5r + 5r² > 5 → r + r² > 1 → r > (−1+√5)/2 ≈ 0.618. Options 3/4 = 0.75 and 3/2 = 1.5 both satisfy their respective conditions, and 5/4 = 1.25 < 1.618 also works.
Therefore r = 7/4 cannot be a valid side ratio.
For r > 1: largest side is 5r², so need 5 + 5r > 5r² → r² - r - 1 < 0 → r < (1+√5)/2 ≈ 1.618. Since 7/4 = 1.75 > 1.618, it fails this inequality.
For r < 1: largest side is 5, so need 5r + 5r² > 5 → r + r² > 1 → r > (−1+√5)/2 ≈ 0.618. Options 3/4 = 0.75 and 3/2 = 1.5 both satisfy their respective conditions, and 5/4 = 1.25 < 1.618 also works.
Therefore r = 7/4 cannot be a valid side ratio.
[JEE Main 2019 · 10 Jan · Shift I]