Building and Checking a Geometric Model

Watch the video explainer

Learning goals

  • Choose geometric components and distinguish measured dimensions from dimensions fixed for the calculation
  • Convert inputs to consistent units and estimate the size of a result
  • Translate a rounding rule into a measurement interval with the correct endpoints
  • Bound a model prediction by tracking which dimensions raise or lower it
  • Compare a prediction with an independent observation and decide what to check or revise

Decide what the shapes need to capture

A geometric model uses familiar shapes and measurements to answer a question about a real object or space. For a first estimate of the garden’s plantable area, you could represent the outline by a rectangle and cut out a smaller rectangle around the tree. That choice straightens the bowed edge and replaces a rounded tree bed with a rectangular one. Those are simplifying assumptions, choices to leave out or idealize details for a stated purpose.

Garden outline and area modelLeft panel shows the bowed boundary and rounded tree bed. Right panel shows a containing rectangle with dashed edges at the removed corner. Outer dimensions are reported measurements and cutout dimensions are fixed in the model.Field sketchArea modelbowed edgetree bed9.7 m measured5.8 m measured2.4 m1.5 mfixed cutout
The field sketch records the bow and the rounded bed. The model uses straight sides and a rectangular exclusion. Both panels are schematic.

Worked example 1 Estimate the plantable area

Problem. A garden’s outer spans are reported as 9.79.7 m and 5.85.8 m. Model the outline by a rectangle with those dimensions and the rounded tree bed by a 2.42.4 m by 1.51.5 m corner cutout. Use the reported values as inputs. Find the model’s plantable area and name one detail that could change the actual area.

Solution. Multiply the outer dimensions and subtract the area excluded around the tree.

A=9.7(5.8)−2.4(1.5)A=52.66 m2.\begin{aligned} A&=9.7(5.8)-2.4(1.5)\\ A&=52.66\text{ m}^2. \end{aligned}

The inward bow could leave less plantable area than the model counts. The rounded bed could also change the result. The decimal is the result of the calculation with those inputs. For a rough estimate, report about 5353 square meters.

The outer spans were measured. A different reading could change them. The corner dimensions were stipulated for the model, meaning they were set for the calculation. A dimension given on a sketch could be a recorded measurement or a fixed input. Ask how the number was obtained before deciding whether it can vary.

Check your understanding

A museum measures a display base as 81 cm wide. Its planning model specifies an exact 12 cm border on each side. Which inputs are measured or stipulated?

Answer choices

Make the units agree and estimate first

A quick estimate gives you a size to expect. Use easy numbers close to the inputs after converting to consistent units. Then compare the calculated value with that estimate. If the answers are far apart, check the conversions and the formula before you check the physical model.

Worked example 2 Check the size of a floor area

Problem. A rectangular entry mat is measured as 265265 cm long and 1.321.32 m wide. Estimate its area in square meters using convenient numbers. Then calculate the model area from the reported readings.

Solution. Convert the length to meters.

265 cm=2.65 m265\text{ cm}=2.65\text{ m}

Use 2.72.7 m and 1.31.3 m for a rough estimate. Their product is 3.513.51 square meters, so expect about 3.53.5 square meters.

A=(2.65 m)(1.32 m)A=3.498 m2.\begin{aligned} A&=(2.65\text{ m})(1.32\text{ m})\\ A&=3.498\text{ m}^2. \end{aligned}

The sizes agree. The multiplication uses meters in both directions, so the result is in square meters.

Check your understanding

A rectangular practice strip is 380 cm long and 2.1 m wide. Which is a reasonable estimate of its area?

Answer choices

Recover the range behind a rounded reading

A report of 1.41.4 m to the nearest tenth stands for a range of readings. A reading of 1.361.36 m would round to it, and so would 1.441.44 m. The edges of the range are halfway from 1.41.4 to its neighbors 1.31.3 and 1.51.5.

In this lesson, for positive lengths reported to the nearest unit of precision, an exact halfway reading rounds to the larger report. The reading 1.351.35 m rounds to 1.41.4 m. The reading 1.451.45 m rounds to 1.51.5 m. The measurement interval is the range of readings allowed by that reporting rule.

Readings reported as 1.4 metersThe left midpoint, 1.35, rounds up to 1.4. The right midpoint, 1.45, rounds up to 1.5 and is outside the segment.1.351.41.45m
A filled dot includes the endpoint. An open dot excludes it. The segment marks readings that round to 1.4 m.

If rounding is the only uncertainty we are tracking, we use that interval for the input. A stretched tape or a poor choice of endpoints can affect the reading before it is rounded. Those effects would need separate information.

Worked example 3 Bound the area from one rounded width

Problem. A rectangular noticeboard model uses a fixed height of 0.720.72 m. Its measured width was reported as 1.41.4 m to the nearest tenth, with ties rounded to the larger report. Find the measurement interval for the width ww and the prediction interval for area AA from these inputs. Say which area endpoint can occur.

Solution. Half a tenth is 0.050.05 m. With widths in meters, the measurement interval is

1.35≤w<1.45.1.35\le w<1.45.

Multiply each side by the fixed positive height. Multiplication by a positive number preserves the order, so the area bounds are

0.972≤A<1.0440.972\le A<1.044

in square meters. The lower endpoint occurs at 1.351.35 m. The area can get arbitrarily close to 1.0441.044 square meters from below. That number is an upper bound, and the area never attains it with these inputs.

A prediction interval from these inputs gives the model’s range using the stated input bounds and assumptions. Its endpoints come from those inputs. They carry no probability or confidence level.

Check your understanding

A length in centimeters is reported as 86 to the nearest centimeter. Ties go to the larger report. Which inequality describes the possible reading xx?

Answer choices

Track which way each input moves the result

For an outer rectangle minus a cutout, a larger outer width adds area and a larger cutout width takes area away. The input choices for a lower bound can therefore go in opposite directions.

Worked example 4 Find both bounds on a paved area

Problem. A courtyard model subtracts a rectangular drain opening from a containing rectangle. The outer length is fixed at 66 m, and its width can be anywhere from 3.83.8 to 4.24.2 m, inclusive. The opening has fixed length 22 m, and its width can be anywhere from 1.11.1 to 1.31.3 m, inclusive. Assume it fits inside for every allowed pair and the two widths can vary independently. Find the least and greatest paved area and justify both choices.

Solution. At fixed lengths, increasing the outer width increases the area of the whole rectangle. Increasing the opening width increases the area taken away.

Least area. Pair the smallest outer width with the largest opening width.

Aleast=6(3.8)−2(1.3)Aleast=20.2 m2.\begin{aligned} A_{\mathrm{least}}&=6(3.8)-2(1.3)\\ A_{\mathrm{least}}&=20.2\text{ m}^2. \end{aligned}

Greatest area. Pair the largest outer width with the smallest opening width.

Agreatest=6(4.2)−2(1.1)Agreatest=23 m2.\begin{aligned} A_{\mathrm{greatest}}&=6(4.2)-2(1.1)\\ A_{\mathrm{greatest}}&=23\text{ m}^2. \end{aligned}

Both pairs are allowed, so the two endpoints can occur.

If two components share a measurement, choose one value for it. For instance, suppose a 66 m long rectangle has a 22 m long strip removed across its full width ww. The model’s paved area is

A=(6−2)w.A=(6-2)w.

The remaining length is 44 m. Increasing ww makes both the whole rectangle and the removed strip larger, and the net effect is to increase the paved area. Giving the two widths different values would describe a different layout.

Check your understanding

A cover is modeled by a fixed outer area minus a rectangular opening. Only the opening width can vary. Its length is fixed. Which opening gives the lowest cover area?

Answer choices

Compare with an observation from another method

An independent observation measures or estimates the target quantity without using the same inputs to compute it in the same way. For an area predicted from tape measurements, a separate estimate from a grid placed over the actual outline can help. Repeating the original multiplication is an arithmetic check.

Worked example 5 Recheck the curved garden edge

Problem. A rectangular-outline garden model with a tree-bed cutout predicts between 52.352.3 and 53.053.0 square meters of plantable area from the input bounds. A team that did not use those inputs estimates the same plantable area by counting grid squares. Its estimate is between 50.850.8 and 51.251.2 square meters. Another field measurement suggests the model includes an extra patch roughly shaped like a triangle with a 6.26.2 m base and 0.500.50 m perpendicular height. Treat its estimated area as fixed in the revised model. Compare the original prediction with the grid result, revise the prediction interval, and interpret the comparison.

Solution. The original prediction interval lies above the grid estimate. Under the stated bounds they disagree. Check the tape measurements, grid estimate, and the model’s treatment of the bowed edge.

The extra patch has estimated area

Apatch=12(6.2)(0.50)Apatch=1.55 m2.\begin{aligned} A_{\mathrm{patch}}&=\tfrac12(6.2)(0.50)\\ A_{\mathrm{patch}}&=1.55\text{ m}^2. \end{aligned}

Subtract 1.551.55 square meters from both original bounds. The revised prediction runs from 50.7550.75 to 51.4551.45 square meters. The grid estimate lies inside it, so the two are consistent under the revised assumptions. The triangle is still an approximation. A fresh observation could check how well the revised model works beyond this comparison.

When the observation is outside the predicted range, the discrepancy gives you a place to investigate. It could come from the inputs, the model, or the observation. When the observation falls inside, the result is consistent with the prediction under the assumptions. Other models could give the same agreement.

Check your understanding

A flat-roof model predicts 32 to 35 square meters. A separate survey reports 34 square meters for the same surface. What does the comparison establish?

Answer choices

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Progressively harder sets of questions. Each opens on its own page.

Core practice

Practice problems at the level of the course, to be worked out on paper. Hints one at a time, then the answer or the full worked solution, with your progress kept in this browser.

Core practice Work it out on paper 10 problems Start →
More resources (optional)

Other explanations of this lesson, if you want a second take.

A bit of history (optional)

In Measurement of a Circle, Archimedes approached a curved boundary from two sides. He worked with a many-sided polygon fitted inside the circle and another wrapped around it. Each time he halved the angles used in his construction, he could work with shorter sides and more of them. His calculation reached polygons with 96 sides.

That effort gave two different fractions for a single quantity. Archimedes placed the ratio of the circumference to the diameter above 3+10713+\tfrac{10}{71} and below 3+173+\tfrac17. The familiar fraction 227\tfrac{22}{7} is the upper of those two bounds.

Archimedes, Measurement of a Circle, Proposition 3, in T. L. Heath’s translation

Your progress

See all your progress →