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Building and Checking a Geometric Model: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 A carpet with a tapered end

The main body of a stage carpet has four straight sides and four square corners. Along one full side, a straight-sided extension tapers to a shorter far edge parallel to the joining seam. Decorative stitching leaves the outline and usable area unchanged.

The installer measures the seam at 2.65 m. For the calculation, the plan fixes the distance from the seam to the far edge at exactly 1.1 m. Both dimensions lie in the plane of the floor, and the distance between the parallel edges is perpendicular to them.

Which two shapes model the area? Classify the two input dimensions and name a detail that can be left out.

The dashed seam marks where the carpet starts to taper.

Text description of this figure

A carpet with a straight-sided body and a tapered end. A dashed seam separates the body from the tapered end. The long seam and the narrow far end are parallel. Decorative stitched marks appear inside the body. The sketch is schematic.
1. Hint 1
  Look for familiar shapes on either side of the seam.
2. Hint 2
  Ask how the two input values were obtained.
Answer

Use a rectangle and a trapezoid. The 2.65 m seam is measured, and the 1.1 m distance is stipulated. The decorative stitching can be left out.

Full solution

The body has a rectangular outline.
The straight-sided tapered part has parallel edges of different lengths, so use a trapezoid.

The seam length comes from the installer.
The distance between the parallel edges is fixed by the plan for this model.

The stitching changes neither the outline nor the usable area, so the area estimate can omit it.

Answer

Use a rectangle and a trapezoid. The 2.65 m seam is measured, and the 1.1 m distance is stipulated. The decorative stitching can be left out.

Key idea

Choose the shapes and inputs that describe the quantity being modeled.
Problem 2 A foam block for a display

A display block is modeled as a rectangular prism. Its reported dimensions are 1.44 m, 54 cm, and 26 cm.

Estimate its volume in cubic meters by using 1.5 m, 0.5 m, and 0.25 m as convenient lengths. Then calculate the model volume from the reported readings.
1. Hint 1
  Write all three lengths in the requested length unit.
2. Hint 2
  Use the prism volume formula on each set of dimensions.
Answer

The rough estimate is $0.1875\text{ m}^3$ . The model calculation gives $0.202176\text{ m}^3$ .

Full solution

Multiply the three convenient lengths.
$\begin{aligned}V_{\text{estimate}}&=1.5(0.5)(0.25)\text{ m}^3\\&=0.1875\text{ m}^3.\end{aligned}$
This is about 0.19 cubic meters.

The width and height readings convert to 0.54 m and 0.26 m.

Use the three lengths in meters.
$V=1.44(0.54)(0.26)$

The result is 0.202176 cubic meters, which is close to the rough estimate.
The digits describe the arithmetic with the reported readings.

Answer

The rough estimate is $0.1875\text{ m}^3$ . The model calculation gives $0.202176\text{ m}^3$ .

Key idea

Use the same length unit in all three directions before interpreting the product as cubic units.
Problem 3 Two ties on a dial

A dial reports a positive length as 14.2 mm to the nearest tenth of a millimeter. A halfway reading is sent to the larger report.

Find the measurement interval for the reading $x$ in millimeters. Explain what happens at each endpoint.
1. Hint 1
  Mark the previous and next reports, 14.1 mm and 14.3 mm.
2. Hint 2
  The endpoints are halfway to those neighboring reports.
Answer

$14.15\le x<14.25$ . The 14.15 mm reading is included, and the 14.25 mm reading is excluded.

Full solution

The half-step is 0.05 mm, so the endpoints are 14.15 mm and 14.25 mm.

Lower endpoint.
A tie between 14.1 and 14.2 goes to 14.2, so 14.15 mm is included.

Upper endpoint.
A tie between 14.2 and 14.3 goes to 14.3, so 14.25 mm is excluded.

The lower tie is included and the upper tie is excluded.

Answer

$14.15\le x<14.25$ . The 14.15 mm reading is included, and the 14.25 mm reading is excluded.

Key idea

The rounding rule decides which halfway readings are included.
Problem 4 Two surveys of the same clearing

An outline model based on tape measurements predicts a clearing area of 66.4 to 69.1 square meters, with both endpoints included. A grid survey that uses none of the tape readings gives one area estimate in each of two hypothetical cases.

Case A. The grid survey gives 67.6 square meters.

Case B. The grid survey gives 65.9 square meters.

For each case, decide whether the observation is consistent with the prediction under the stated assumptions. Explain what you would conclude.
1. Hint 1
  Compare each estimate with both ends of the prediction interval.
2. Hint 2
  Think about what could make a survey and a model disagree.
Answer

Case A is consistent with the prediction under the model assumptions. Case B is outside the prediction interval, so check the model and both sets of data.

Full solution

Case A.
The survey estimate lies between the bounds.
It agrees with this prediction under the stated assumptions, and other assumptions could give a similar area.

Case B.
The survey estimate lies below 66.4 square meters.
The prediction and observation disagree as reported.
Recheck the tape readings, shape approximation, and grid survey before deciding what to revise.

Answer

Case A is consistent with the prediction under the model assumptions. Case B is outside the prediction interval, so check the model and both sets of data.

Key idea

A separate observation checks the prediction, and a discrepancy can have several causes.
Problem 5 A square mold from one reading

The top of a mold is modeled as a square. Its measured side length was reported as 73 cm to the nearest centimeter, with ties sent to the larger report. Assume rounding is the only source of uncertainty in the model input.

Find the prediction interval from this input for the square's area in square centimeters. Explain which endpoint can occur.
1. Hint 1
  First recover the allowed side lengths from the report.
2. Hint 2
  For positive side lengths, a larger square side makes the area larger. Track which side length endpoints are allowed.
Answer

$5256.25\le A<5402.25$ . The lower endpoint is attained and the upper one is excluded.

Full solution

The side length $s$ in centimeters satisfies
$72.5\le s<73.5.$

For positive $s$ , squaring increases as $s$ increases.
Both dimensions of the square equal the same $s$ .

The lower endpoint is $72.5^2$ , or 5256.25 square centimeters.
The side 72.5 cm is allowed.

The area approaches $73.5^2$ , or 5402.25 square centimeters.
It cannot attain that bound because the side 73.5 cm is excluded.

Answer

$5256.25\le A<5402.25$ . The lower endpoint is attained and the upper one is excluded.

Key idea

When a positive side is used twice for a square, apply the endpoint rule to the shared side once.
Problem 6 The mat around an access panel

A mat is modeled by a rectangle of fixed length 5.5 m with one rectangular access opening strictly inside it. The outer width $w$ can range from 3.4 to 3.6 m, inclusive. The opening has fixed length 1.6 m and width $v$ from 0.7 to 0.9 m, inclusive. Any pair of allowed widths is possible, and the opening always fits.

Find the least and greatest mat areas. Explain why the two pairs of widths attain them.

The full rectangle and the opening use separate widths.

Text description of this figure

A rectangular mat has fixed length 5.5 meters and outer width w. A smaller rectangle strictly inside is an access opening of fixed length 1.6 meters and width v. The two widths are independent. The drawing is schematic.
1. Hint 1
  Think about changing one width while holding the other fixed.
2. Hint 2
  A larger opening removes more area. A larger outer width adds area.
Answer

The least area is 17.26 square meters at an outer width of 3.4 m and an opening width of 0.9 m. The greatest area is 18.68 square meters at an outer width of 3.6 m and an opening width of 0.7 m.

Full solution

The model area in square meters is
$A=5.5w-1.6v.$

At fixed $v$ , each increase in $w$ adds area.
At fixed $w$ , each increase in $v$ removes area.
These directions hold throughout the allowed ranges.

Least area.
Use the smallest outer width and largest opening width.
$A_{\min}=5.5(3.4)-1.6(0.9)$
The result is 17.26 square meters.

Greatest area.
Use the largest outer width and smallest opening width.
$A_{\max}=5.5(3.6)-1.6(0.7)$
The result is 18.68 square meters.

The width endpoints are included and both pairs are allowed, so both area endpoints are attained.

Answer

The least area is 17.26 square meters at an outer width of 3.4 m and an opening width of 0.9 m. The greatest area is 18.68 square meters at an outer width of 3.6 m and an opening width of 0.7 m.

Key idea

Justify a composite bound by tracking the direction of each input and checking that the chosen inputs can occur together.
Problem 7 A unit slips off the calculation

A banner is modeled as a rectangle. The reports for its sides are 128 cm and 0.85 m. Sam writes the following work to find its area in square meters.

Line 1. The rectangular area is the product of the two side lengths.

Line 2. The numerical product of 128 and 0.85 is 108.8.

Line 3. The area is 108.8 square meters.

Identify the first incorrect line and explain why. Correct the area and give a quick estimate to check it.
1. Hint 1
  The numerical multiplication can be correct even when the final unit is wrong.
2. Hint 2
  Find the unit of the product before converting either length.
Answer

Line 3 is the first incorrect line. The model area is $1.088\text{ m}^2$ , close to a rough estimate of about 1 square meter.

Full solution

Line 1 uses the correct area formula.
Line 2 states a true numerical product.

Line 3 treats centimeters times meters as square meters.
The two original lengths use different units.

The length 128 cm is 1.28 m.
Use the same length unit for both factors.
$A=(1.28\text{ m})(0.85\text{ m})$

The calculated model area is 1.088 square meters.
For a quick estimate, use 1.3 m and 0.8 m, whose product is 1.04 square meters.

Answer

Line 3 is the first incorrect line. The model area is $1.088\text{ m}^2$ , close to a rough estimate of about 1 square meter.

Key idea

Check the units carried by each factor before naming the unit of a product.
Problem 8 When two widths are one

A claim says that the smallest remaining area in a subtraction model always comes from the smallest possible whole area and the largest possible removed area.

Test the claim on an 11 m long rectangular platform with a 5 m long end strip removed across the whole width $w$ . Its width can range from 2.6 to 3.4 m, inclusive.

Find the number the claim would produce and the true least remaining model area. Explain why the results differ.

One width applies to the whole platform and to the strip.

Text description of this figure

An 11 meter long platform and an end strip 5 meters long. The strip reaches from the top edge to the bottom edge of the whole rectangle. The single vertical width w applies to both parts. The dashed divider separates the strip from the retained part. The drawing is schematic.
1. Hint 1
  Ask whether the whole and removed areas can use different values of the width.
2. Hint 2
  Calculate the length left after removing the strip. Its width is still $w$ .
Answer

The proposed rule gives $11.6\text{ m}^2$ . The true minimum is $15.6\text{ m}^2$ when the shared width is $2.6\text{ m}$ .

Full solution

The smallest whole uses width 2.6 m and has area 28.6 square meters.
The largest strip uses width 3.4 m and has area 17 square meters.

The proposed difference is 11.6 square meters.
These choices would give the strip and the platform different widths, so they cannot occur together.

The remaining rectangle has length 6 m, since the strip runs across the full width.
Its area in square meters is
$A=6w.$

The area increases with the positive width, so the allowed width of 2.6 m attains the true minimum of 15.6 square meters.
This contradicts the proposed rule.

Answer

The proposed rule gives $11.6\text{ m}^2$ . The true minimum is $15.6\text{ m}^2$ when the shared width is $2.6\text{ m}$ .

Key idea

Separate extrema from different terms can describe an impossible combination when the terms share an input.
Problem 9 Check a joined awning footprint

An awning footprint is modeled as a rectangle plus a nonoverlapping triangle. They share the rectangle's full width $w$ along the seam. The rectangle extends 2.8 m back from the seam. The triangle tip is a perpendicular distance of 0.8 m from the seam. Both distances are fixed.

The seam width was reported as 3.7 m to the nearest tenth, with ties rounded to the larger report. Assume rounding accounts for the width uncertainty in the model.

Part A. Find the predicted area bounds. Say which endpoint is attained.

Part B. An independent survey reports an area of 12.00 square meters to the nearest hundredth, with ties rounded to the larger report. Compare the two area intervals under these rounding assumptions.

The dashed seam is the width used by both pieces.

Text description of this figure

A rectangular awning footprint extends into a triangle at one edge. The triangle shares the full edge of width w with the rectangle. The rectangle extends 2.8 meters back from this edge. The tip has perpendicular distance 0.8 meters from the shared edge. This plan is schematic.
1. Hint 1
  Recover the interval for the width before doing any area calculations.
2. Hint 2
  Use the same width in both parts. The triangle contributes half its perpendicular height times that width.
3. Hint 3
  Find the range of unrounded survey readings that would give 12.00. Look for common values with the model interval.
Answer

A. The model allows 11.68 up to 12.00 square meters, including the lower endpoint only. B. The survey allows 11.995 up to 12.005 square meters, also including the lower endpoint only. The ranges overlap.

Full solution

Part A.
The rounding rule allows widths $3.65\le w<3.75$ in meters.

The two parts share the same width.
$\begin{aligned}A&=2.8w+\tfrac12(0.8)w\\&=3.2w.\end{aligned}$

The area increases with the width.
Multiply each endpoint by 3.2 to obtain 11.68 and 12.00 square meters.
The lower endpoint is attained and the upper one is excluded.

Part B.
With the survey rounding rule, the unrounded reading can be at least 11.995 and less than 12.005 square meters.

The two intervals share all values from 11.995 up to, but excluding, 12.00 square meters.
This overlap makes the survey and the model consistent under the stated assumptions.

Answer

A. The model allows 11.68 up to 12.00 square meters, including the lower endpoint only. B. The survey allows 11.995 up to 12.005 square meters, also including the lower endpoint only. The ranges overlap.

Key idea

When both predictions and observations come from rounded data, compare the ranges their precision allows.
Problem 10 Revising a platform at the cove

An initial straight-edged model gives 45 to 47 square meters for the top area of a platform, including both endpoints. The interval tracks only the uncertainty in its measured spans. A team using separate survey data estimates the same area to be 40.9 to 41.8 square meters, including both endpoints.

Part A. Compare the original prediction with the survey. What does the comparison justify checking?

Part B. Field inspection shows the model crosses a gap where the platform edge curves inward. A revised model treats the gap as a triangle with base 7.5 m and perpendicular height 1.4 m. Use its calculated area as a fixed correction. Find the new prediction interval and compare it with the survey.

Part C. Say what this revised comparison can establish and name a fresh independent check.

The survey team found a gap between the straight model edge and the curved platform edge.

Text description of this figure

The straight model edge crosses a gap where the actual platform edge curves inward. The curve joins the straight edge at two points 7.5 meters apart. The perpendicular offset shown between the straight and curved edges is 1.4 meters. The drawing is schematic.
1. Hint 1
  Start by looking for any common values in the two original intervals.
2. Hint 2
  Decide whether the gap adds to the platform or was counted by mistake.
3. Hint 3
  Use a check that looks at the actual platform without reusing the input spans or the triangle calculation.
Answer

A. The intervals are disjoint. Recheck inputs and model assumptions. B. The revised interval is 39.75 to 41.75 square meters, inclusive, and overlaps the survey. C. They are consistent. Check with a fresh overhead survey.

Full solution

Part A.
The survey stops at 41.8 square meters, below the original lower bound of 45.
The difference justifies checking the span measurements, shape approximation, and separate survey.

Part B.
The triangular gap is outside the real platform but inside the initial model.
Its fixed model area is
$A_{\text{gap}}=\tfrac12(7.5)(1.4)$

The gap calculation gives 5.25 square meters.
Subtract that value from both original endpoints.
They remain included because the correction is fixed.

The revised interval is 39.75 to 41.75 square meters.
The survey interval is 40.9 to 41.8 square meters.
Their overlap is 40.9 to 41.75 square meters, with both ends included.

Part C.
The overlap shows that the revised range can agree with the observation under the assumptions.
The triangle may miss parts of the curved gap, and the revised interval bounds model outputs from these inputs.
Another team could estimate the area from a new overhead survey and count its covered grid squares, including partial squares near the curve.

Answer

A. The intervals are disjoint. Recheck inputs and model assumptions. B. The revised interval is 39.75 to 41.75 square meters, inclusive, and overlaps the survey. C. They are consistent. Check with a fresh overhead survey.

Key idea

After an independent check, revise the geometry with field evidence and keep track of the assumptions that still affect the result.

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