VectorPaint right trapezoids: distinguish a length-five midsegment from areas nine and eleven

English translation of the Chinese original, translated on 2026-11-12. Original publication dates, attribution, code and version notes are retained.

A trapezoid’s midsegment bisects its legs but need not bisect its area. Draw a right trapezoid, its midsegment and the resulting regions, checking lengths and areas separately. VectorPaint supplies paths, colours and labels; averages, perpendicular height and area are calculated externally, without assuming built-in geometric proof.

List vertices in one coordinate system

In mathematical coordinates use A=(0,0), B=(6,0), C=(4,4), D=(0,4), joined ABCDA. The horizontal parallel bases have lengths 6 and 4; left leg AD is vertical with length 4, while BC slopes. Label invented drawing units, not unmeasured centimetres. OpenStax’s trapezoid material supports bases and area formula. Independently calculate this example rather than presenting its textbook figure as your own.

Map to screen without changing vertices

Place the origin at screen (300,900) with 100 pixels per unit: screen x=300+100x and screen y=900−100y. Vertices become (300,900), (900,900), (700,500), (300,500). Downward screen increase does not reverse mathematical y. Close the four-point path in order; joining A to B to D to C would cross it. If placement is approximate, retain the coordinate table and label that limitation, rather than using imperfect pixels to challenge the mathematical formula.

Join the leg midpoints

The left-leg midpoint M=(0,2) and right-leg midpoint N=(5,2), averaging B and C coordinatewise. MN is horizontal with length 5=(6+4)/2. It joins leg midpoints, not base midpoints, and need not pass through diagonal intersections. Small endpoint marks identify midpoints without making MN a third outer boundary. Check N lies on BC before checking parallelism. A length-five line placed near the centre without joining the legs does not complete the construction. At heights 1 and 3, cross-sections end at x=5.5 and 4.5 with left endpoints x=0. These check that the sloping leg is straight. Neither is the midsegment; hide them after checking rather than labelling all three cross-sections midsegments.

Use perpendicular height for total area

The bases are 4 apart, giving area (6+4)×4/2=20 square units. A perpendicular from C to (4,0) splits off a 4×4 rectangle of area 16 and a base-2, height-4 right triangle of area 4, again totalling 20. The sloping leg has squared length 20, not perpendicular height 4. Keep both calculations to detect substituting the wrong quantity. Length 5 and area 20 have different units and cannot be compared as the same kind of measure.

Equal heights do not imply equal region areas

Above MN, bases 4 and 5 with height 2 give area 9; below it, bases 5 and 6 with height 2 give 11, totalling 20. Equal heights do not compensate for different average widths, so neither is 10. Use pale region fills labelled 9 and 11, retaining one outer boundary and avoiding overlapping darker false regions. Appearance alone does not establish 9: write the calculation. If rotating the figure, preserve the original base-side identities rather than treating screen-top as automatically the original upper region.

Keep a sloping-height error separate

In one erroneous copy, substitute BC=√20 for height, obtaining 5√20≈22.36, labelled wrong sloping-leg substitution rather than 20. Another copy changing both region labels to 10 wrongly asserts area bisection; that is a different failure. Change one condition at a time, restoring height 4 and regions 9/11 without moving vertices to accommodate wrong numbers. A base-perpendicular should identify the error. No area plug-in is invoked; explicit arithmetic, not automatic marking, checks both errors.

Separate labels and checking references for delivery

Retain vertex/midpoint names and parallel/right-angle cues without covering M or N. Hide ordinary drawn grids, perpendiculars or error labels when outside delivery scope; inspect native-grid behaviour in actual exports. Open a PNG to check bases 4/6, height 4, midsegment 5 and regions 9/11. Monochrome labels must distinguish regions: faint colour loss is not lost area. Open vector output separately to inspect nodes and labels. Scaling or font substitution changes no values in the coordinate table.

Accept length and area conditions separately

Retain coordinates, both total-area decompositions, midpoint evidence, regions 9/11 and both errors. Nonparallel MN requires midpoint repair; 22.36 requires checking perpendicular height; two tens require checking each region’s bases. Recalculate after repair rather than recolouring alone. This particular trapezoid has midsegment 5, total area 20 and unequal split; do not claim every midsegment fails to bisect area or treat raster pixel counts as proof. No part is fabricated or geometry automatically solved: coordinates and manual calculations support the result.