The diagram represents a plot of land, ABCD. The police is looking for a robber hiding in the land. Scale: 1 cm represents 1 km Here are the accounts from three witnesses:
Witness 1: The robber is located more than 6 km away from A.
Witness 2: The robber is located nearer to B than A.
Witness 3: The robber is located nearer to AD than AB. Shade the region where the robber is hiding.

1. Represent the condition “more than 6 km away from A”.

Since the scale is $1\text{ cm}:1\text{ km}$, $6\text{ km}$ is represented by $6\text{ cm}$.

Draw a circle with centre $A$ and radius $6\text{ cm}$.

The robber must be in the region outside this circle.

2. Represent the condition “nearer to B than A”.

Construct the perpendicular bisector of $AB$.

Points on this perpendicular bisector are equally far from $A$ and $B$.

Since the robber is nearer to $B$ than $A$, choose the side of the perpendicular bisector that contains $B$.

3. Represent the condition “nearer to AD than AB”.

Construct the angle bisector of $\angle DAB$.

Points on the angle bisector are equally far from the lines $AD$ and $AB$.

Since the robber is nearer to $AD$ than $AB$, choose the side of the angle bisector closer to $AD$.

4. Shade the required intersection inside the plot.

Within the boundary of plot $ABCD$, shade only the region satisfying all three conditions:

$\text{outside circle centre } A \text{ radius } 6\text{ cm}$

$\cap\ \text{side of perpendicular bisector of } AB \text{ containing } B$

$\cap\ \text{side of angle bisector of } \angle DAB \text{ nearer to } AD.$