Question

Instead of turning right at the end if he took left and walked 20 km, what is the shortest distance to his starting point?

• A. \(3\sqrt{7}\) km
• B. \(2\sqrt{5}\) km
• C. \(7\sqrt{2}\) km
• D. \(5\sqrt{2}\) km

Hint:

Translate every turn into a direction on the \(xy\)-plane and keep running coordinates. Then use the Pythagorean theorem for the straight-line distance.

Answer 1

The Correct Option is D

Path with coordinates:
Start at \(O(0,0)\). Take \(+x\) east, \(+y\) north.
- Walk 25 km west \(\Rightarrow (-25,0)\).
- Left (to south) 15 km \(\Rightarrow (-25,-15)\).
- Left (to east) 30 km \(\Rightarrow (5,-15)\).
- Now take left (to north) and walk 20 km \(\Rightarrow (5,5)\).
Distance from start:
\[ d=\sqrt{(5-0)^2+(5-0)^2}=\sqrt{25+25}=\sqrt{50}=5\sqrt{2}\ \text{km}. \] \[ \boxed{5\sqrt{2}\ \text{km}} \]

Answer 2

Instead of plotting coordinates on axes, let's keep a running tally of net east-west and net north-south displacement across the four legs of the walk.

Leg 1: 25 km west \(\Rightarrow\) East-West change \(-25\). Leg 2: 15 km south \(\Rightarrow\) North-South change \(-15\). Leg 3: 30 km east \(\Rightarrow\) East-West change \(+30\). Leg 4: 20 km north (the left turn instead of right) \(\Rightarrow\) North-South change \(+20\). Net East-West \(=-25+30=+5\) km (i.e. \(5\) km east of the start), and net North-South \(=-15+20=+5\) km (i.e. \(5\) km north of the start).

1. Option \(3\sqrt{7}\) km: Squaring gives \(9\times7=63\), but the actual sum of squares here is \(5^2+5^2=50\), so this value is too large.
2. Option \(2\sqrt{5}\) km: Squaring gives \(4\times5=20\), far short of the required \(50\), so this is too small.
3. Option \(7\sqrt{2}\) km: Squaring gives \(49\times2=98\), noticeably more than \(50\), so this overshoots.
4. Option \(5\sqrt{2}\) km: Squaring gives \(25\times2=50\), exactly matching \(5^2+5^2=50\) from the tally above.

The east-west/north-south tally gives a straight-line distance of \(\sqrt{50}=5\sqrt{2}\) km.

Hence, the correct answer is \(5\sqrt{2}\) km.