Edging, Soil, and the Scale Factor
A neighborhood association is building a free-standing flower bed shaped like a rectangle 12 feet long and 8 feet wide, with a semicircle attached to one of the 12-foot sides. The bed will be edged with flexible border along its entire outer boundary and filled with soil to a depth of 3 inches. Volunteers will buy edging in 50-foot rolls and soil in 2-cubic-foot bags. In the planning meeting, one volunteer computes the area as 96 + 36π ≈ 209.1 square feet and the edging as 8 + 12 + 8 + 12 + 6π ≈ 58.8 feet. A second volunteer, hearing 58.8 feet, concludes that two rolls of edging are required. The group wants a defensible order before Saturday's work day, and the treasurer asks how the quantities would change if the plan were enlarged by a scale factor of 1.5 so the bed reads better from the street.
1. Find the correct area and length of edging, and identify both errors in the volunteer's work.
Read the worked answer
Area. Rectangle: 12 × 8 = 96 ft². The semicircle has diameter 12, so radius 6, and a full circle would be π(6²) = 36π ≈ 113.10 ft². A semicircle is half of that: 18π ≈ 56.55 ft². Total area ≈ 152.55 ft².
The volunteer used the full circle (36π) instead of half, inflating the area by about 56.5 ft², roughly 37%.
Edging. The outer boundary consists of the two 8-foot sides, the one 12-foot side not shared with the semicircle, and the arc. The other 12-foot side is the diameter where the semicircle meets the rectangle, so it lies inside the figure, not on the boundary. Boundary = 8 + 12 + 8 + (half the circumference) = 28 + π(6) ≈ 28 + 18.85 = 46.85 ft.
The volunteer's expression 8 + 12 + 8 + 12 + 6π counts the shared diameter as if it were exposed and also then fails to halve the arc consistently. Note that 6π is correctly the arc length (half of 12π), so the arc is right and the diameter is the error. A quick sanity check: the boundary must be shorter than the perimeter of the enclosing 12-by-20 rectangle, and the figure shows why the interior segment cannot be edged.
2. How many rolls of edging and bags of soil should the group order? Show the rounding reasoning.
Read the worked answer
Edging. One 50-foot roll covers 46.85 ft, leaving about 3.15 ft, or roughly 6.7%, of slack. Mathematically, one roll suffices. Practically, 3 feet is thin for a project with corners, arc curvature, and cutting waste, and border material usually cannot be invisibly spliced. A defensible recommendation is one roll plus a plan for the arc (measure the arc with a string, cut once), and a second roll only if the material is brittle or the crew expects waste above 7%. The judgement to make explicit is that "one roll is enough" is a mathematical conclusion, while "buy one roll" is a procurement decision with a stated risk.
Soil. Volume = area × depth = 152.55 ft² × 0.25 ft = 38.14 ft³. At 2 ft³ per bag: 38.14 ÷ 2 = 19.07, so 19 bags fall short by 0.14 ft³. Round up to 20 bags. If the group follows the usual practice of adding 10% for settling and spillage, 38.14 × 1.1 = 41.95 ft³, requiring 21 bags (20.98). Either way, the rule is round up after dividing, never round to nearest, and check the depth unit first: 3 inches is 0.25 foot, and forgetting that conversion yields a volume four times too large.
3. The group enlarges the design by a scale factor of 1.5. Determine the new edging length and soil volume, and explain why they do not both scale by 1.5.
Read the worked answer
Edging (a length). 46.85 × 1.5 ≈ 70.28 ft. Two 50-foot rolls (100 ft) are now needed, with about 29.7 ft left over - enough that the group might reconsider the scale or ask whether leftover border has another use. One roll no longer suffices, so the scale change crosses a purchasing threshold.
Area. 152.55 × 1.5² = 152.55 × 2.25 ≈ 343.24 ft².
Soil volume. 343.24 × 0.25 ≈ 85.81 ft³, so 85.81 ÷ 2 = 42.9, requiring 43 bags.
Why the difference: every length on the boundary is multiplied by 1.5, so the boundary total is multiplied by 1.5. But area is a product of two lengths, so each is scaled by 1.5 and the product by 1.5 × 1.5 = 2.25. The depth of 3 inches is not scaled at all (a vertical dimension of the material, not of the plot), so volume scales exactly as area does. A student who scales the soil by 1.5 would order 57 ft³ and be about 29 ft³ short - the sort of error the treasurer is right to ask about before the order is placed.