# VectorPaint right trapezoids: distinguish a length-five midsegment from areas nine and eleven
Language: en
Canonical: [VectorPaint right trapezoids: distinguish a length-five midsegment from areas nine and eleven]()
中文：[中文 HTML]()
English: [English HTML]()
Published: 2026-11-12T00:00:00+08:00
Modified: 2026-11-12T09:43:29+08:00
Author: Bison
Construct a trapezoid from vertices and leg midpoints, checking area twenty and why the midsegment does not bisect its area.
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.
## Related tutorials and sources
[VectorPaint parallelogram dissections: fix base and perpendicular height before area](), [VectorPaint triangle medians: check midpoints and the two-thirds centroid position]().
Checked on November 12, 2026: VectorPaint public version 1.0.8, requiring iOS / iPadOS 17.0 or later. Scope follows the [official App Store]() and [Apple public lookup](). Manual exercises use controls actually available in your installed version; external calculations, written records and diagrams imply no built-in automatic analysis. No step-by-step physical-device, private-file or all-export-environment testing is claimed.