A child is looking for his father. He went 90 meters in the East before…

Concept: In a direction-and-distance problem, split the path onto two perpendicular axes, east-west and north-south. Each "turn right" rotates the walking direction by 90 degrees. To find the straight-line distance from the starting point, add up the movement along each axis separately (legs in the same direction add, legs in opposite directions cancel), then combine the two net displacements as the two legs of a right triangle using the Pythagorean theorem.

Application:

1. Starting at the point A and facing East, the child walks 90 m East to point B. Net east-west movement so far: 90 m East; net north-south movement: 0.

2. At B he turns right (now facing South) and walks 20 m South to point C. Net north-south movement so far: 20 m South.

3. At C he turns right again (now facing West) and walks 30 m West to point D, his uncle's place. This is opposite to the first leg, so the net east-west movement reduces to 90 - 30 = 60 m East.

4. From D he walks 100 m North to point E, where he meets his father. This is opposite to the second leg, so the net north-south movement becomes 100 - 20 = 80 m North.

5. The child's position relative to the start is now a net 60 m East and 80 m North -- two perpendicular displacements that form the legs of a right triangle whose hypotenuse is the straight-line distance from the start.

6. By the Pythagorean theorem, the distance is √(602 + 802) = √(3600 + 6400) = √10000 = 100 m.

Cross-check: 60, 80, 100 is a 3-4-5 right triangle scaled by 20 (3x20=60, 4x20=80, 5x20=100), confirming the hypotenuse without needing a calculator.