Entrance Ramp Compliance and a Right-Triangle Planter
A primary school is replacing a step at its rear entrance with a concrete ramp. The finished floor is 24 inches above the pavement. Local code adopts the 1:12 maximum slope rule - for every 12 inches of horizontal run, the ramp may rise at most 1 inch - and requires a 5-foot flat landing at the top of the ramp within the same footprint. The only space between the door and the fence line is 22 feet 6 inches of horizontal distance. Separately, the garden committee is building a triangular raised planter whose two shorter sides meet at a right angle and measure 1.2 m and 1.6 m. It will be edged with timber strip costing $8.50 per metre along all three sides and filled with soil to a uniform depth of 0.30 m. The committee needs the edging cost and the soil volume before ordering, and wants the right angle confirmed rather than assumed.
1. Show whether a 24-inch rise can be achieved in the available 22 ft 6 in, given the landing requirement. State the shortfall and one workable remedy.
Read the worked answer
Required sloped run at 1:12: rise × 12 = 24 × 12 = 288 inches of run = 24 ft.
Available run: 22 ft 6 in = 22 × 12 + 6 = 270 inches.
Deduct the landing (the landing is flat, so it consumes footprint but provides no rise): 270 − 60 = 210 inches of sloped run.
Maximum rise achievable: 210 ÷ 12 = 17.5 inches, against a required 24 inches → shortfall 6.5 inches.
Even ignoring the landing the ramp fails: 270 ÷ 12 = 22.5 inches < 24 inches, short by 1.5 inches. So the landing is not the only obstruction; the footprint itself is too short.
Remedy: lower the finished threshold by at least 6.5 inches (to 17.5 inches) so the ramp fits with the landing; alternatively extend the footprint by at least 78 inches (6.5 × 12) to reach 288 + 60 = 348 inches total. A switchback does not rescue the design: two 12-inch runs would need 288 inches of sloped run plus a mid landing, exceeding 270 inches by more.
2. Express the 1:12 slope as a percent grade and as an angle of inclination, and interpret both against the 5% guideline commonly used for level walkways.
Read the worked answer
Percent grade: rise/run × 100 = 1/12 × 100 = 8.33%.
Angle: θ = arctan(1/12) = arctan(0.08333) ≈ 4.76°.
Interpretation: the slope is a small angle but a substantial grade. Compared with a 5% walkway guideline (1:20), 1:12 is 8.33 ÷ 5 = 1.67 times as steep, which is why 1:12 is a maximum for wheelchair users and not a design target. Two cautions for the school:
- The 1:12 figure applies to the sloped section only. If the 5-foot landing is mistakenly built with any fall, its rise must not be counted toward the 24 inches, or the effective slope of the remaining run increases.
- Percent grade is rise over horizontal run, not over the sloped surface length. Using the 24-inch rise and the sloped hypotenuse instead of 288 inches of run understates the grade and can make a non-compliant ramp appear compliant.
3. For the planter, confirm the right angle, find the third side, and compute the edging cost and soil volume.
Read the worked answer
Confirm the right angle via the converse of the Pythagorean theorem. If the 1.2 m and 1.6 m sides enclose the right angle, the third side must be √(1.2² + 1.6²) = √(1.44 + 2.56) = √4.00 = 2.00 m. Because 1.2² + 1.6² = 2.00² exactly, the triangle with those three sides is right-angled, so the quoted corner is genuinely 90°. (The sides are a 3-4-5 triangle scaled by 0.4.)
Edging: perimeter = 1.2 + 1.6 + 2.0 = 4.80 m. Cost = 4.80 × $8.50 = $40.80. Allow a waste/offcut margin of about 5–10%, i.e. order 5.1–5.3 m (about $43–$45).
Soil volume: area = ½ × 1.2 × 1.6 = 0.96 m²; volume = 0.96 × 0.30 = 0.288 m³. Ordering by the bag, that is 288 litres, so 6 bags of 50 L gives 300 L - enough, but bedding settlement means ordering 7 is prudent.
Estimation check: the planter is roughly a 1 m × 1 m half-square, so ~0.5 m²… wait - ½ × 1.2 × 1.6 is 0.96 m², close to a full square metre because the legs are both near 1.4 m; the figure is consistent with a footprint about 1.4 m on a side. Volume near 0.3 m³ is therefore believable.