On a large ground, there is a straight tall vertical wall of length 28 m. A goat is tied to a point on the ground which is at the middle of the wall, using a rope. If the length of the rope is 21 m, what is the area of the region (in sq. m) around the wall that the goat can access?

Correct — A, 847. The peg is at the middle of a 28 m wall, so the wall runs 14 m each way from it and the 21 m rope is 7 m longer than each half. On the goat's own side of the wall the rope sweeps a clean half-circle of radius 21 m, which is ½ × (22/7) × 21² = 693 sq. m. At each end of the wall the rope has 21 − 14 = 7 m to spare, and it bends round the corner onto the far side, sweeping a half-circle of radius 7 m there: ½ × (22/7) × 7² = 77 sq. m at each end, so 154 sq. m for the two ends together. The total is 693 + 154 = 847 sq. m.

• (c)693 — This is the half-circle on the goat's own side and nothing else. It ignores the 7 m of rope left over at each end of the wall, which lets the goat reach round both corners.
• (b)851 — No consistent reading of the figure produces 851. It sits four square metres above the right answer, which is close enough to look plausible to anyone who has the method right and slips in the arithmetic.
• (d)654 — Smaller than the half-circle of radius 21 m, which is 693 sq. m and is entirely accessible. Any answer below 693 can be rejected before you compute anything else.

A tethered-animal question is a sum of circular sectors. Wherever the rope can swing freely it sweeps a sector of radius equal to the rope; wherever it catches on a corner, the leftover length becomes the radius of a new sector pivoting about that corner. A long straight wall blocks half the plane, so the free sweep is a half-circle, and each end of the wall contributes a further half-circle of the leftover rope on the other side.

Two decisions settle these items. First, how much of the plane the obstacle blocks: a wall much longer than the rope would block a full half-plane and the answer would be the semicircle alone, which is why 693 is offered. Second, whether the rope reaches the ends of the obstacle at all — here 21 m against a 14 m half-wall, so it does, with 7 m to spare. Round a wall's end the leftover rope sweeps a half turn, because the goat can move from lying flat along the wall on one side, round the end, to lying flat along the wall on the other. Keep π as 22/7 throughout: with 21 and 7 in the radii, every multiplication cancels cleanly and no decimals appear.

• Free sweep against a long straight wall = half-circle of radius equal to the rope length.
• Leftover rope at a wall's end = rope length − distance from the peg to that end; here 21 − 14 = 7 m.
• The leftover sweeps a half-circle about that end, on the far side of the wall.
• With π = 22/7, ½π(21)² = 693 sq. m and ½π(7)² = 77 sq. m.
• Total 693 + 77 + 77 = 847 sq. m, and any offered value below 693 can be rejected on sight.

• Stopping at the semicircle and forgetting the two ends.
• Treating the sweep round a corner as a quarter-circle when the obstacle is a thin wall and the swing is a half turn.
• Using the full rope length as the radius round the corner instead of what is left after reaching the end.

Asked with numbers chosen so that 22/7 cancels exactly, so the examiner is testing the decomposition of the region rather than arithmetic stamina.

No directly related past PYQ was found.

• practice — not a real PYQ

  A goat is tied by a 14 m rope to a peg in open ground with no obstacle anywhere near. What area can it graze?

  • (a)308 sq. m
  • (b)616 sq. m
  • (c)44 sq. m
  • (d)154 sq. m

  Answer(b) 616 sq. m — with nothing in the way the sweep is a full circle, (22/7) × 14² = 616.

• practice — not a real PYQ

  A cow is tied to a corner of a square shed of side 10 m by a rope 7 m long, and grazes only outside the shed. What area can it reach?

  • (a)38.5 sq. m
  • (b)77 sq. m
  • (c)115.5 sq. m
  • (d)154 sq. m

  Answer(c) 115.5 sq. m — the shed's corner leaves three quarters of the circle free, that is (3/4) × (22/7) × 49.