A rectangular park is 60m by 40m. A semicircular pond of diameter 40m is inside. What percent of the park's area does the pond occupy?
- (a)33.3%
- (b)26.18%
- (c)28.35%
- (d)31.42%
Answer
Why
Correct — B. Compare the pond's area with the park's.
Park area = 60 × 40 = 2,400 m²
Pond radius = 40 ÷ 2 = 20 m
Pond area = ½ × π × 20² = 200π ≈ 628.32 m²
Percentage = 628.32 ÷ 2,400 × 100 ≈ 26.18% → option (b)
Why the others are wrong
- (a)33.3% — 33.3% is about one-third of the park, 800 m². The semicircle covers 200π ≈ 628 m², about 172 m² less.
- (c)28.35% — 28.35% of 2,400 m² is about 680 m², some 52 m² more than the semicircle's 200π ≈ 628 m².
- (d)31.42% — 31.42% of the park is about 754 m², some 126 m² more than the semicircle's 200π ≈ 628 m².
Concept
A semicircle's area is half a circle's, πr²⁄2, with r half the given diameter.
The share of a region covered is part ÷ whole × 100, so here it is 200π ÷ 2,400 = π⁄12 ≈ 0.2618.
The pond fits: its 40 m diameter can lie along a side, and its 20 m height is less than both 40 m and 60 m.
26.18% comes from π ≈ 3.1416. With π = 22⁄7 the pond is 628.57 m² and the share is 26.19%, so the nearest option is still 26.18%.
Key facts
- Area of a semicircle = πr²⁄2.
- Radius = diameter ÷ 2, so a 40 m diameter gives r = 20 m.
- π⁄12 ≈ 0.2618.
Study next
Common traps
- Using the diameter as the radius: π × 40² ÷ 2 = 800π ≈ 2,513 m², more than the whole park.
- Using the full circle: 400π ≈ 1,257 m² gives about 52.4%, which is not an option.
A part-of-the-whole area percentage is also asked at 21 Sep 2025, 16:00, Quant Q.8, where a triangle of 600 m² is 37.5% of a 1,600 m² triangle-plus-rectangle field.
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