Question

In a triangle $ABC$, right angled at $A$, on the leg $AC$ as diameter, a semicircle is described. If a chord joins $A$ with the point of intersection $D$ of the hypotenuse and the semicircle, then the length of $AC$ equals to

• A. $\frac{AB.AD}{AB + AD}$
• B. $\frac{AB \cdot AD}{\sqrt{AB^2 - AD^2}}$
• C. $\frac{AB. AD}{\sqrt{AB^2 + AD^2}}$
• D. $\sqrt{AB. AD}$

Answer 1

Answer: (B)

B $\frac{AB \cdot AD}{\sqrt{AB^2 - AD^2}}$

Answer 2

Explanation:
Place the right‐angled triangle with vertex A at the origin, letting AC = x (on the x–axis) and AB = b (on the y–axis). Then C = (x, 0) and B = (0, b). The semicircle with AC as diameter has center (x/2, 0) and radius x/2, so its equation is

$$\left(X-\frac{x}{2}\right)^2+Y^2=\left(\frac{x}{2}\right)^2.$$

The hypotenuse BC joining B(0, b) and C(x, 0) has equation

$$y=b-\frac{b}{x}X.$$

Let D = (d,, b-\frac{b}{x}d) be the other point (apart from C) where the hypotenuse meets the semicircle. Substituting the line’s coordinates into the circle’s equation and then factoring out the obvious solution $d=x$ (which corresponds to C), one finds the coordinate of D to be

$$d=\frac{x\,b^2}{x^2+b^2}\quad \text{and}\quad y_D=\frac{x^2\,b}{x^2+b^2}.$$

The length of the chord AD is:

$$AD=\sqrt{d^2+y_D^2}=\sqrt{\frac{x^2\,b^4+x^4\,b^2}{(x^2+b^2)^2}}=\frac{x\,b}{\sqrt{x^2+b^2}}.$$

Since $AB=b$ and the hypotenuse $BC=\sqrt{x^2+b^2}$, the relation becomes:

$$AD=\frac{x\cdot AB}{\sqrt{x^2+AB^2}}.$$

Solving for $x$ (which is AC), we get:

$$AD\,\sqrt{x^2+AB^2}=x\,AB.$$

Squaring both sides,

$$AD^2\,(x^2+AB^2)=x^2\,AB^2\quad\Longrightarrow\quad x^2(AB^2-AD^2)=AD^2\,AB^2.$$

Thus,

$$x^2=\frac{AD^2\,AB^2}{AB^2-AD^2}\quad\Longrightarrow\quad x=\frac{AB\cdot AD}{\sqrt{AB^2-AD^2}}.$$

Answer:
The length of AC is $\displaystyle \frac{AB\cdot AD}{\sqrt{AB^2-AD^2}}$, that is, Option B.