A circular fountain with a radius of 7 m is encircled by a grass bed that is 3.5 m wide. Determine the ratio of the area of the grass bed to the area of the fountain.

Correct — C. The bed is a ring, so its area is the outer circle minus the fountain.

Outer radius = 7 + 3.5 = 10.5 m
Bed = π(10.5² − 7²) = π(110.25 − 49) = 61.25π
Fountain = π × 7² = 49π

Divide, and π cancels: 61.25 ÷ 49 = 1.25
1.25 = 5⁄4, so bed : fountain = 5 : 4 → option (c)

Why the others are wrong

• (a)
  3:2 — 3 : 2 is 10.5 : 7, the outer radius against the fountain's radius. That compares two lengths. Areas go with the square of the radius, and the bed excludes the fountain.
• (b)
  2:1 — 2 : 1 is 7 : 3.5, the fountain's radius against the bed's width, two lengths rather than two areas. A bed twice the fountain's area would need an outer radius of 7√3 ≈ 12.1 m.
• (d)
  4:1 — 4 : 1 needs a total area of 5 × 49π, which means an outer radius of 7√5 ≈ 15.7 m and a bed about 8.7 m wide. This bed is only 3.5 m wide.

A path, border or bed of uniform width w around a circle of radius r is a ring (an annulus). Its area is the outer circle minus the inner one: π(R² − r²), with R = r + w.

In a ratio of two such areas π cancels, so only the squared radii matter.

Key facts

• Area of a ring with outer radius R and inner radius r = π(R² − r²) = π(R + r)(R − r).
• If two circles' radii are in the ratio a : b, their areas are in the ratio a² : b².
• For a ring of width w built outside a circle of radius r, the outer radius is r + w.

Common traps

• Adding the width twice to the radius: the bed adds 3.5 m on each side of the diameter, so the outer radius is 7 + 3.5 = 10.5 m, not 14 m.
• Comparing the whole outer circle with the fountain (9 : 4) and forgetting that the bed excludes the fountain.
• Answering with a ratio of lengths, such as 3 : 2 or 2 : 1, when the question compares areas.

No directly related past PYQ was found.