Question

$ABC$ is a right angled triangle right angled at $A$, with side $AC = 1$ and $AB = a$, a circle having $AC$ as diameter cuts the side $CB$ at $D$ if $CD = b$ then :

• A. ab>1
• B. ab<1
• C. $\frac{b}{a} > \frac{1}{a^2+\frac{1}{2}}$
• D. $\frac{b}{a} < \frac{1}{a^2+\frac{1}{2}}$

Answer 1

Answer: (B), (C)

(B) and (C)

Answer 2

1. Triangle and Circle Properties:

   • Given a right-angled triangle $ABC$ with $\angle A = 90^\circ$, $AC=1$, and $AB=a$.
   • The hypotenuse $BC = \sqrt{AC^2 + AB^2} = \sqrt{1^2 + a^2} = \sqrt{a^2+1}$.
   • A circle with diameter $AC$ passes through $A$ and $C$. Any point $D$ on this circle subtends a right angle at $D$, i.e., $\angle ADC = 90^\circ$.
   • The point $D$ lies on the hypotenuse $CB$.

2. Similar Triangles:

   • Consider $\triangle ABC$ and $\triangle DAC$.
   • $\angle BAC = 90^\circ$ (given).
   • $\angle ADC = 90^\circ$ (angle in a semicircle).
   • $\angle ACB$ is common to both triangles.
   • Therefore, $\triangle ABC \sim \triangle DAC$ by AA similarity.

3. Ratio of Sides:

   • From the similarity, the ratio of corresponding sides is equal: $\frac{AC}{CD} = \frac{BC}{AC}$
   • Substitute the given values: $AC=1$, $AB=a$, $CD=b$, and $BC=\sqrt{a^2+1}$. $\frac{1}{b} = \frac{\sqrt{a^2+1}}{1}$
   • Squaring both sides gives: $\frac{1}{b^2} = a^2+1$
   • This leads to the fundamental relationship: $b^2 = \frac{1}{a^2+1} \implies b = \frac{1}{\sqrt{a^2+1}} \quad (\text{since } b > 0)$

4. Evaluate Options:

   • (A) $ab>1$: Substitute $b$: $a \left(\frac{1}{\sqrt{a^2+1}}\right) = \frac{a}{\sqrt{a^2+1}}$. The inequality becomes $\frac{a}{\sqrt{a^2+1}} > 1$, which implies $a > \sqrt{a^2+1}$. Squaring both sides ($a>0$): $a^2 > a^2+1 \implies 0 > 1$, which is false. So, (A) is incorrect.

   • (B) $ab<1$: The inequality becomes $\frac{a}{\sqrt{a^2+1}} < 1$, which implies $a < \sqrt{a^2+1}$. Squaring both sides ($a>0$): $a^2 < a^2+1 \implies 0 < 1$, which is true. So, (B) is correct.

   • (C) $\frac{b}{a} > \frac{1}{a^2+\frac{1}{2}}$: Substitute $b$: $\frac{b}{a} = \frac{1}{a\sqrt{a^2+1}}$. The inequality becomes $\frac{1}{a\sqrt{a^2+1}} > \frac{1}{a^2+\frac{1}{2}}$. For $a>0$, both denominators are positive. This inequality is equivalent to: $a^2+\frac{1}{2} > a\sqrt{a^2+1}$ Squaring both sides: $\left(a^2+\frac{1}{2}\right)^2 > \left(a\sqrt{a^2+1}\right)^2$ $a^4 + a^2 + \frac{1}{4} > a^2(a^2+1)$ $a^4 + a^2 + \frac{1}{4} > a^4 + a^2$ $\frac{1}{4} > 0$ This is true for all $a>0$. So, (C) is correct.

   • (D) $\frac{b}{a} < \frac{1}{a^2+\frac{1}{2}}$: Since option (C) is true, its opposite (D) must be false.