Read · step-by-step explanationEntrance-exam arithmetic (moving figures)

How far can a dog on a rope move around its kennel?

A dog is tied by a 4 m rope to the corner of a 2 m by 4 m kennel. Where the rope catches on a corner, the centre and radius of the arc change. Noticing that switch is the judgment this problem turns on. Four steps, each with a figure and the equation.

2026-10-22　Domain: Arithmetic

Analysis data (cue → response texts, stage analyses, explanation steps) is shown in the original Japanese. English rendering of the data layer is pending; titles, summaries and observations are in English.

PROBLEM

The problem

SubjectA rectangular kennel measures 2 m by 4 m, as shown. A dog is tied by a rope to the corner A. The dog can move freely outside the kennel. When the rope is 4 m long, what is the area, in m², of the region the dog can reach? Take π to be 3.14.

CREATE BASE arithmetic materials (moving figures, Japanese junior-high entrance level). The original figure is omitted; redrawn figures used for the analysis are shown.

OUTPUT

Step-by-step explanation

Cues and operations one step at a time. Figures and equations come from this problem’s judgment data.

Step 1 · Right angle → mark 90°

The corners of a rectangle are 90°.

∠A＝∠B＝∠C＝∠D＝90°

Figure for step 1 (redrawn for the analysis)

Step 2 · Path of a point → find what is fixed

The dog moves on a circle of radius 4 m centred on A, over 270°: the full turn minus the 90° taken up by the kennel.

Centre A, radius 4 m, angle 270°

Figure for step 2 (redrawn for the analysis)

Step 3 · Path of a point → find what is fixed

When the rope catches on B, 2 m of it is used up; the remaining 2 m sweeps 90°.

Centre B, radius 4 − 2 = 2 m, angle 90°

Figure for step 3 (redrawn for the analysis)

Step 4 · Area of a sector → radius × radius × π × angle/360°

Add the areas of the two sectors.

4×4×π×270/360＝12π
2×2×π×90/360＝π
13×3.14＝40.82m²

Figure for step 4 (redrawn for the analysis)