A person climbs a hill in a straight path from point ‘O’ on the ground in the direction of north-east and reaches a point ‘A’ after travelling a distance of 5 km. Then, from the point ‘A’ he moves to point ‘B’ in the direction of north-west. Let the distance AB be 12 km. Now, how far is the person away from the starting point ‘O‘ ?

The correct option is 13 km.

Let us break down the movement of the person step-by-step using a coordinate system or direction vectors to understand the geometry of the path:

1. Starting at point 'O':
Let point $O$ be the origin $(0,0)$ on a standard 2D map where North points vertically upward (+y direction) and East points horizontally to the right (+x direction).

2. First movement from 'O' to 'A':
The person travels in the North-East direction. The North-East direction makes an angle of $45°$ with both the North and East directions (bearing of $45°$). The distance travelled is $OA=5\text{km}$.

3. Second movement from 'A' to 'B':
From point $A$, the person turns and moves in the North-West direction to reach point $B$. The North-West direction makes an angle of $45°$ with both the North and West directions.

4. Finding the angle between the two paths:
Since the North-East direction points at an angle of $45°$ East of North, and the North-West direction points at an angle of $45°$ West of North, the angle between the North-East path ($OA$) and the North-West path ($AB$) is:
$\angle OAB=45°+45°=90°$.

5. Applying the Pythagorean Theorem:
Since $\angle OAB=90°$, the triangle $OAB$ formed by the starting point $O$, the intermediate point $A$, and the final destination $B$ is a right-angled triangle at vertex $A$.

Using the Pythagorean theorem to find the straight-line distance $OB$ from the starting point:
$O{B}^2=O{A}^2+A{B}^2$

Substitute the given distances ($OA=5\text{km}$ and $AB=12\text{km}$):
$O{B}^2=5^2+{12}^2$
$O{B}^2=25+14=169$
$OB=\sqrt{169}=13\text{km}$.

Thus, the person is 13 km away from the starting point 'O'.