4. On the shores of a circular island (viewed from above), there are cities $A, B, C$, and $D$. A straight asphalt road $A C$ divides the island into two equal halves. A straight asphalt road $B D$ is shorter than road $A C$ and intersects it. The speed of a cyclist on any asphalt road is 15 km/h. The island also has straight dirt roads $A B, B C, C D$, and $A D$, on which the cyclist's speed is the same. The cyclist reaches each of the points $A, C$, and $D$ from point $B$ via a straight road in 2 hours. Find the area enclosed by the quadrilateral $A B C D$.
Answer: 450 sq. km. Solution. The condition of the problem means that a quadrilateral $ABCD$ is given, in which angles $B$ and $D$ are right (they rest on the diameter), $AB=BC$ (both roads are dirt roads, and the cyclist travels them in the same amount of time), $BD=15 \frac{\text { km }}{\text { h }} 2$ hours $=30$ km. Drop two perpendiculars from point $B$: $BM$ - to the line $AD$, and $BN$ - to the line $CD$. Then $\triangle \mathrm{BMA}=\triangle \mathrm{BNC}$ (both are right triangles, the hypotenuses are equal, $\angle BNC=\angle BAM$ - each of these angles, together with $\angle BAD$, gives $180^{\circ}$). Therefore, the quadrilateral $MBND$ is equal in area to the quadrilateral $ABCD$. At the same time, $MBND$ is a square, the diagonal $BD$ of which is known. Therefore, its area is $\frac{30^{2}}{2}=450$ sq. km.