Given a semicircle with O as the centre, as shown in the figure, the ratio$\frac{\overline{AC}+\overline{CB}}{\overline{AB}}$ is _____, where$\overline{AC},\overline{CB}and\overline{AB}$ are chords.

The correct answer is:

$\sqrt{2}$

1. Understanding the Geometric Properties from the Figure:
From the provided image, we observe a semicircle with centre O and a horizontal diameter chord AB. A point C lies on the arc of the semicircle. The line segment CO represents a radius connecting the centre O to point C on the boundary.
A right-angle symbol (∟) is shown at the point O, which indicates that the radius segment CO is perpendicular to the diameter chord AB:

$CO\perp AB$

2. Defining Lengths in terms of Radius (R):
Let the radius of the semicircle be:

$R$

This gives us the following lengths for the segments:
• $OA=R$
• $OB=R$
• $OC=R$
The diameter AB (which is also a chord) is equal to:

$AB=OA+OB=2R$

3. Calculating the Chord Lengths AC and CB:
Since CO is perpendicular to AB, the triangles AOC and BOC are right-angled triangles at O.
Applying the Pythagorean theorem to right-angled triangle AOC:

$A{C}^2=O{A}^2+O{C}^2$

Substituting the radius values:

$A{C}^2=R^2+R^2=2R^2$

Taking the square root on both sides:

$AC=R\sqrt{2}$

Applying the Pythagorean theorem similarly to right-angled triangle BOC:

$C{B}^2=O{B}^2+O{C}^2=R^2+R^2=2R^2$

Which simplifies to:

$CB=R\sqrt{2}$

4. Evaluating the Ratio:
We need to find the ratio:

$\frac{AC+CB}{AB}$

Substituting the values of AC, CB, and AB in terms of $R$:

$\frac{AC+CB}{AB}=\frac{R\sqrt{2}+R\sqrt{2}}{2R}$

Simplifying the numerator:

$\frac{AC+CB}{AB}=\frac{2R\sqrt{2}}{2R}$

Canceling $2R$ from the numerator and denominator yields:

$\frac{AC+CB}{AB}=\sqrt{2}$