Appendix A — Answer Key, Chapter 10: Properties of Quadrilaterals

SOL G.PC.1 (a, c, d) · Covers textbook Chapter 10 and the companion workbook. Item numbers match the textbook; workbook items are the same problems, so this key serves both. Item numbers run continuously from 1 to 132 across the chapter.

Conventions used in every answer below. A family's defining condition is the one thing you check; everything else it has is inherited from a family above it or is a theorem to be proved. A property is specific to a family when it separates that family from the one it sits inside — so diagonals bisect each other is never an answer about a rhombus. Virginia's exclusive trapezoid definition holds throughout: exactly one pair of parallel sides, so no parallelogram is a trapezoid. Bisect each other and congruent are different claims and are kept apart everywhere below. Constructions verify; rulers measure — and the arcs stay on the page.

The one line that prevents most errors in this chapter: rectangle → congruent diagonals; rhombus → perpendicular diagonals; both → square; and every one of them bisects, because every one of them is a parallelogram.

Lesson 10.1 — The Six Families

Guided practice

1. A rhombus is a parallelogram with four congruent sides.
2. A trapezoid has exactly one pair of parallel sides. Exactly is the whole of the word — it rules out a second pair.
3. A square is a rectangle and a rhombus at the same time, which is why the figure puts it where those two overlap.
4. Because the definitions cannot both hold. A trapezoid has exactly one pair of parallel sides and a parallelogram has two, so no figure is in both boxes and the boxes cannot touch.
5. The overlap of the rectangle and rhombus regions, inside the parallelogram box. It is there because a square satisfies both definitions, so it belongs to both regions at once.
6. The definition is the single condition you check to decide whether a figure is in the family. A theorem is anything else that is true of every member — it follows from the definition and has to be proved, not assumed.

Independent practice

7. A rectangle. Both pairs of opposite sides parallel makes it a parallelogram; the right angle makes it a rectangle.
8. No. A parallelogram has two pairs of parallel sides, and exactly one excludes a second pair. It is a trapezoid.
9. Yes. A square is a parallelogram with four congruent sides, which is exactly the definition of a rhombus.
10. No. A rhombus needs four congruent sides; a square needs a right angle as well, and most rhombi have none.
11. Yes. "Parallelogram with a right angle" is the definition of a rectangle, so every rectangle is a parallelogram by construction.
12. No. A parallelogram needs no right angle, so most are not rectangles. (This is the same asymmetry as 9 and 10, one level up.)
13. Square, rhombus, rectangle, and parallelogram — all four.
14. An isosceles trapezoid.
15. A rhombus. Four congruent sides is the definition; without a right angle it is not a square.
16. No. Under Virginia's exclusive definition, two pairs of parallel sides makes the figure a parallelogram, and a parallelogram is not a trapezoid in this course.
17. A rectangle. The two conditions checked were both pairs of opposite sides parallel (parallelogram) and a right corner (the right angle that upgrades it). Nothing was checked about the side lengths, so it cannot be called a square.
18. Equal sides do not disqualify anything. A rectangle is a parallelogram with a right angle, and a square is one — it just satisfies extra conditions besides. Extra conditions narrow which families a figure is in; they never remove one. Every square is a rectangle.
19. Every rhombus is a parallelogram, so anything proved for all parallelograms applies to it automatically. The reverse fails because not every parallelogram is a rhombus — one counterexample among parallelograms is enough to sink an upward claim.
20. Overlapping boxes would mean some figure is in both families — one with exactly one pair of parallel sides and also two pairs. No figure has both, so the intersection is empty and the boxes are drawn apart.

Exit ticket 10.1

21. A rectangle is a parallelogram with a right angle.
22. Parallelogram, rectangle, rhombus, square, trapezoid, and isosceles trapezoid.

Lesson 10.2 — Properties of Parallelograms

Guided practice

23. Opposite sides are parallel and congruent; opposite angles are congruent; consecutive angles are supplementary; the diagonals bisect each other.
24. The definition is only that both pairs of opposite sides are parallel. The congruent opposite sides, the congruent opposite angles, the supplementary consecutive angles, and the bisecting diagonals are all theorems.
25. $M(4, 2)$ — the point where the diagonals meet. It is the midpoint of both diagonals at once, which is what "bisect each other" means.
26. $\angle BAC \cong \angle DCA$, because $\overline{AB} \parallel \overline{DC}$; and $\angle DAC \cong \angle BCA$, because $\overline{AD} \parallel \overline{BC}$. Both pairs are alternate interior angles (Chapter 2).
27. ASA, using those two pairs of angles and the shared diagonal $\overline{AC} \cong \overline{AC}$ by the Reflexive Property. The sides then come from CPCTC.
28. Bisect each other means each diagonal cuts the other into two congruent halves at $M$ — a statement about the four halves. Congruent would mean the two whole diagonals have the same length — a statement about the two wholes. A parallelogram always does the first and generally not the second.

Independent practice

29. $CD = 9$ and $DA = 14$ — opposite sides are congruent.

30. $m\angle C = 72°$ (opposite), and $m\angle B = m\angle D = 108°$ (consecutive, supplementary).

31. $m\angle D = 115°$ (opposite) and $m\angle C = 65°$ (consecutive, supplementary).

32. $MC = AM = 7$, so $AC = 14$.

33. $BM = 11$ — $M$ is the midpoint of $\overline{BD}$.

34. Opposite sides are congruent: $3x + 4 = 5x - 6 \Rightarrow 10 = 2x \Rightarrow x = 5$, and $AB = 19$. (Check: $CD = 5(5) - 6 = 19$.)

35. $2y + 7 = 4y - 9 \Rightarrow 16 = 2y \Rightarrow y = 8$, and $BC = 23$. (Check: $AD = 4(8) - 9 = 23$.)

36. Consecutive angles are supplementary: $(2y + 10) + 3y = 180 \Rightarrow 5y = 170 \Rightarrow y = 34$, so $m\angle A = 78°$ and $m\angle B = 102°$. (Check: $78 + 102 = 180$.)

37. Opposite angles are congruent: $4z - 20 = 2z + 30 \Rightarrow 2z = 50 \Rightarrow z = 25$, so $m\angle A = 80°$. (Check: $m\angle C = 2(25) + 30 = 80°$.)

38. The diagonals bisect each other, so $AM = MC$: $2x + 1 = 4x - 7 \Rightarrow 8 = 2x \Rightarrow x = 4$. Then $AM = MC = 9$ and $AC = 18$.

39. Opposite sides are congruent, so the perimeter is $2(AB) + 2(BC)$: $16 + 2(BC) = 46 \Rightarrow BC = 15$.

40. $M$ is the midpoint of each diagonal, so it is $\tfrac{7.2}{2} = 3.6$ ft from each endpoint — $3.6$ ft. (The $6$ ft and $4$ ft are not needed — bisection does not depend on them.)

41. Two different claims have been run together. Bisect each other says each diagonal is cut in half at $M$; congruent says the two diagonals have the same length. A parallelogram's diagonals always do the first and generally not the second — congruent diagonals are specific to a rectangle. The gate in the previous item is the counterexample: its diagonals bisect each other while measuring $7.2$ ft and something else entirely.

42. Because both conclusions are CPCTC on the same congruence. Once $\triangle ABC \cong \triangle CDA$ by ASA, the corresponding angles $\angle B$ and $\angle D$ are congruent immediately, and $\angle A \cong \angle C$ follows by adding the two pairs of congruent halves that the diagonal created. One congruence pays for the sides and the angles together.

43. Its opposite angle equals it, and each of the two consecutive angles is its supplement. So from $m\angle A = \theta$ you get $m\angle C = \theta$ and $m\angle B = m\angle D = 180° - \theta$ — all four from one.

44. Given: parallelogram $ABCD$ with diagonals meeting at $M$. Prove: $\overline{AM} \cong \overline{CM}$ and $\overline{BM} \cong \overline{DM}$.

    Statement	Reason
    1. $ABCD$ is a parallelogram	1. Given
    2. $\overline{AB} \parallel \overline{DC}$	2. Definition of a parallelogram
    3. $\overline{AB} \cong \overline{DC}$	3. Opposite sides of a parallelogram are congruent
    4. $\angle BAM \cong \angle DCM$	4. Alternate interior angles, $\overline{AB} \parallel \overline{DC}$
    5. $\angle ABM \cong \angle CDM$	5. Alternate interior angles, $\overline{AB} \parallel \overline{DC}$
    6. $\triangle AMB \cong \triangle CMD$	6. ASA (4, 3, 5)
    7. $\overline{AM} \cong \overline{CM}$, $\overline{BM} \cong \overline{DM}$	7. CPCTC
    8. The diagonals bisect each other	8. Definition of bisect (each is cut into two congruent halves at $M$)

Exit ticket 10.2

45. $m\angle A = 58°$ (opposite) and $m\angle B = 122°$ (consecutive, supplementary).
46. $AM = 9$ — the diagonals bisect each other.

Lesson 10.3 — Rectangles and Rhombi

Guided practice

47. All four angles are right, and the diagonals are congruent.
48. The diagonals bisect each other — that comes from being a parallelogram, not from being a rectangle. (Congruent opposite sides, congruent opposite angles, and supplementary consecutive angles are also inherited.)
49. All four sides are congruent; the diagonals are perpendicular; each diagonal bisects a pair of opposite angles.
50. No. One diagonal in the figure is exactly twice the other ($AC = 4\sqrt{5}$ against $BD = 2\sqrt{5}$). Congruent diagonals are specific to a rectangle, and a rhombus has them only when it happens to be a square.
51. Rectangle → congruent diagonals. Rhombus → perpendicular diagonals.
52. Because both are parallelograms, and bisecting diagonals is a parallelogram theorem. It is inherited by everything below the parallelogram box, so it distinguishes nothing inside it.

Independent practice

53. $BD = 26$ (diagonals congruent) and $AM = 13$ (they bisect each other).
54. $AC = 15$ and $MC = 7.5$.
55. $m\angle AMB = 90°$ — a rhombus's diagonals are perpendicular.
56. $\overline{BD}$ bisects $\angle ABC$, so $m\angle ABD = \tfrac{118}{2} = 59°$.
57. $\overline{AC}$ bisects $\angle BAD$, so $m\angle BAC = 32°$. And $\angle ABC$ is consecutive to $\angle BAD$ in a parallelogram, so $m\angle ABC = 180 - 64 = 116°$.
58. Congruent diagonals: $4x + 3 = 6x - 11 \Rightarrow 14 = 2x \Rightarrow x = 7$, and $AC = 31$. (Check: $BD = 6(7) - 11 = 31$.)
59. Bisecting diagonals: $2y - 1 = y + 6 \Rightarrow y = 7$, so $AM = MC = 13$ and $AC = 26$.
60. All four sides congruent: $5w - 2 = 3w + 8 \Rightarrow 2w = 10 \Rightarrow w = 5$, so each side is $23$ and the perimeter is $4(23) = 92$.
61. A rhombus. Perpendicular diagonals are what distinguishes a rhombus among parallelograms; nothing here forces a right angle, so it need not be a square.
62. A rectangle, by the same reasoning with the other property.
63. A square — it is a rhombus and a rectangle at once.
64. The diagonals bisect each other and are perpendicular, so each side is the hypotenuse of a right triangle with legs $\tfrac{16}{2} = 8$ and $\tfrac{12}{2} = 6$. Side $= \sqrt{8^2 + 6^2} = \sqrt{100}$, so the side is $10$.
65. Each diagonal is $\sqrt{9^2 + 12^2} = \sqrt{225}$, so both are $15$, and both are that length because a rectangle's diagonals are congruent.
66. It has verified that the frame is a rectangle — provided it is already a parallelogram, since congruent diagonals in a parallelogram force the right angles. It has not verified that the frame is a parallelogram in the first place (an isosceles trapezoid also has congruent diagonals), and it has said nothing about the side lengths, so it does not establish a square.
67. Two errors. The conclusion: a rhombus's diagonals are perpendicular, not congruent — congruent is the rectangle's property. The reason: bisecting each other is true of every parallelogram, so it cannot establish anything specific to a rhombus, whichever conclusion is drawn from it.
68. Diagonal $\overline{AC}$ cuts the rhombus into $\triangle ABC$ and $\triangle ADC$ with $\overline{AB} \cong \overline{AD}$, $\overline{CB} \cong \overline{CD}$ (all four sides are congruent), and $\overline{AC} \cong \overline{AC}$ (Reflexive). That is SSS, so the triangles are congruent and CPCTC gives $\angle BAC \cong \angle DAC$ and $\angle BCA \cong \angle DCA$ — the diagonal bisects both angles it runs between. The same argument on $\overline{BD}$ gives the other pair.

Exit ticket 10.3

69. A rectangle's diagonals are congruent; a rhombus's are perpendicular. (Both bisect each other, because both are parallelograms.)
70. Legs $\tfrac{10}{2} = 5$ and $\tfrac{24}{2} = 12$, so the side is $\sqrt{5^2 + 12^2} = \sqrt{169}$, which is $13$.

Lesson 10.4 — Squares, and What "Specific To" Means

Guided practice

71. The diagonals are congruent — from the rectangle; perpendicular — from the rhombus; and bisect each other — from the parallelogram.
72. Because a square is defined as a rectangle and a rhombus at once, its property list is exactly the union of two lists that already existed. There is no third condition left to impose.
73. Opposite sides parallel and congruent · opposite angles congruent · diagonals bisect each other · all sides congruent · diagonals perpendicular. The last two are the specific ones; the first three are inherited.
74. Rectangles, squares, and isosceles trapezoids.
75. Any of opposite sides parallel and congruent, opposite angles congruent, or diagonals bisect each other — each is checked in all four parallelogram columns, so knowing it tells you nothing about which of the four you are in.
76. Specific to means the property is true of that family and not of the larger family it sits inside — so it is a property that actually identifies the figure rather than one it inherited.

Independent practice

77. Diagonals perpendicular. The other two belong to every parallelogram.
78. Diagonals congruent. The other two belong to every parallelogram.
79. A square — four congruent sides makes it a rhombus, the right angle makes it a rectangle, and both at once is a square.
80. No. An isosceles trapezoid has congruent diagonals and is not a rectangle. The correct statement adds a hypothesis: a parallelogram with congruent diagonals must be a rectangle.
81. Rhombus.
82. Rectangle.
83. Square.
84. Isosceles trapezoid. (Exactly one pair of parallel sides rules out every parallelogram, so the congruent diagonals point here instead of to a rectangle.)
85. Rhombi and squares.
86. Parallelograms, rectangles, rhombi, and squares — every family in the parallelogram box, and no others.
87. Rhombus is confirmed, and the perpendicular diagonals add nothing new — that is a theorem which already follows from four congruent sides. To call it a square you would still need one right angle (equivalently, congruent diagonals).
88. Diagonals bisecting each other establishes that the figure is a parallelogram — nothing more, since every parallelogram has it. A square needs the diagonals to be congruent (rectangle) and perpendicular (rhombus), and neither was checked.
89. Because a square is by definition the intersection of those two families, so its property set is exactly the union of their property sets. Anything true of a square is true because it is a rectangle, or because it is a rhombus, or because both are parallelograms — there is no fourth source.
90. Because every parallelogram has congruent opposite sides, so the property is satisfied by rhombi and by parallelograms that are neither. It is inherited, so it cannot separate rectangles from anything. A rectangle answer is all four angles right or diagonals congruent.

Lesson 10.5 — Trapezoids and Isosceles Trapezoids

Guided practice

91. Each has exactly one pair of parallel sides — the bases. That single condition is the definition of a trapezoid.
92. Its legs are congruent (tick-marked in the figure). That one extra condition is the whole definition of an isosceles trapezoid.
93. $AD = 2\sqrt{5}$ and $BC = 5$. They are different, so the legs are not congruent and the figure is a plain trapezoid.
94. The legs are congruent (the definition); each pair of base angles is congruent; the diagonals are congruent.
95. Yes, the diagonals are congruent — $AC = BD = \sqrt{89}$ in the figure. No, they do not bisect each other; bisecting is a parallelogram property, and this figure is not a parallelogram.
96. Because it has exactly one pair of parallel sides and a parallelogram has two. Congruent legs do not make a second pair parallel — they only make the two non-parallel sides the same length.

Independent practice

97. $m\angle B = 68°$ — base angles at the same base are congruent.
98. $\angle A$ and $\angle D$ are same-side interior angles across the parallel bases, so they are supplementary: $m\angle D = 112°$, and $m\angle C = 112°$ as the other base angle there.
99. $m\angle C = 105°$ (the other base angle at that base), and $m\angle A = m\angle B = 75°$ (supplementary across the parallel bases).
100. $BD = 17$ — the diagonals are congruent.
101. Congruent legs: $3x - 5 = x + 7 \Rightarrow 2x = 12 \Rightarrow x = 6$, and each leg is $13$. (Check: $6 + 7 = 13$.)
102. Congruent diagonals: $5y + 2 = 8y - 13 \Rightarrow 15 = 3y \Rightarrow y = 5$, and $AC = 27$. (Check: $8(5) - 13 = 27$.)
103. No. The diagonals are congruent to each other but do not bisect each other. Bisection is a parallelogram property, and an isosceles trapezoid is not a parallelogram — so $M$ is not the midpoint of either diagonal.
104. Not isosceles — the legs are $7$ and $9$, and congruent legs is the definition. It follows that neither pair of base angles is congruent, and the diagonals are not congruent either. All three properties stand or fall together.
105. An isosceles trapezoid.
106. Rectangles, squares, and isosceles trapezoids.
107. Perimeter $= 12 + 8 + 5 + 5 = 30$, so $30$ ft. The second leg is $5$ ft, and that is known without measuring because congruent legs is the definition of an isosceles trapezoid — the panel would not be one otherwise.
108. Congruent and bisecting are different claims. Congruent says the two diagonals have the same length, which this figure does have. Bisecting each other says each cuts the other in half, which requires the figure to be a parallelogram — and an isosceles trapezoid, with exactly one pair of parallel sides, is not one.
109. Drop a perpendicular from each end of the short base to the long base. The two right triangles that result have congruent hypotenuses (the congruent legs) and congruent vertical legs (the distance between the parallel bases), so they are congruent by HL — and CPCTC gives the congruent base angles. The angles at the other base are same-side interior angles with those, across the two parallel bases and along a leg as transversal, so they are supplementary to them (Chapter 2).
110. Check whether the diagonals bisect each other — a rectangle's do, an isosceles trapezoid's do not. Equivalently, check whether both pairs of opposite sides are parallel, or whether the angles are right. Congruent diagonals alone cannot tell them apart, which is the whole point of item 80.

Exit ticket 10.5

111. $m\angle B = 74°$ (base angles at the same base) and $m\angle D = 106°$ (same-side interior across the parallel bases).
112. $AC = 21$ — the diagonals are congruent. No, $AM \neq MC$: they do not bisect each other, because this figure is not a parallelogram.

Lesson 10.6 — Constructions That Verify Properties

Guided practice

113. The parallel line construction.
114. The angle bisector construction, and it points to a rhombus — each diagonal of a rhombus bisects a pair of opposite angles.
115. The perpendicular line construction.
116. The compass was set to the length of $\overline{AB}$, and swung from $A$ and from $C$ — two opposite vertices, one opening.
117. That all four sides are congruent, which makes the figure a rhombus. The perpendicular diagonals then follow as a theorem, with nothing more to check.
118. Because the arcs are the verification. They show that a length was carried from one side to another rather than judged by eye, and G.PC.1d assesses the construction itself. Erase them and what is left is a drawing, which verifies nothing.

Independent practice

119. The parallel line construction, used twice — once for each pair of opposite sides. That is the definition of a parallelogram, so both pairs have to be checked.
120. The congruent segment construction.
121. The perpendicular line construction.
122. The angle bisector construction.
123. Set the compass to $\overline{AB}$ and do not change it. Swing an arc from $A$ and check that $D$ lies on it. Swing an arc from $C$ and check that both $B$ and $D$ lie on it. If all four vertices land on the arcs, then $AB = AD = CB = CD$, all four sides are congruent, and the figure is a rhombus. Leave the arcs.
124. Use the congruent segment construction on the two legs: set the compass to one leg, then from the endpoint of the other leg swing an arc of that radius. If the arc passes through the other leg's far endpoint, the legs are congruent and the trapezoid is isosceles.
125. Use the perpendicular line construction at one vertex: construct the perpendicular to one side through that vertex and check that it falls along the adjacent side. A parallelogram with one right angle is a rectangle, so one corner is enough.
126. A rectangle — both pairs of opposite sides parallel makes it a parallelogram, and the right angle upgrades it. Nothing has been verified about the sides, so it is not a square.
127. Set the compass to one side of the panel and, without changing the opening, swing arcs from two opposite corners. If all four corners lie on those arcs, the four sides are congruent and the panel is a true rhombus. What is left as evidence: the scribed arcs themselves, still on the panel, showing that one length was carried onto the others rather than eyeballed or read off a tape.
128. Every parallelogram's diagonals bisect each other, so the check confirmed only what was already given — it rules out nothing and cannot distinguish a rhombus from a rectangle or from a plain parallelogram. To settle it, use the perpendicular line construction on the diagonals (perpendicular ⇒ rhombus), or the congruent segment construction across the sides (four congruent sides ⇒ rhombus).
129. A ruler reports that two sides look equal to whatever precision its markings allow, and the comparison passes through a number that was read twice. A construction carries one length directly onto the other with the compass and shows whether they coincide — no number, no reading, no scale. That is why G.PC.1d asks for the construction and why the arcs are the answer.
130. Because "diagonals are perpendicular" is a theorem that follows from four congruent sides, and a quadrilateral with four congruent sides is a rhombus. Once the sides are verified, the perpendicularity is proved — constructing the diagonals would only re-illustrate something already established.

Exit ticket 10.6

131. Congruent segment, congruent angle, angle bisector, perpendicular line, and parallel line.
132. The congruent segment construction — one compass opening carried onto all four sides. The arcs stay on the page, because they are the evidence that the sides were compared rather than assumed.

Chapter 10 Review — answers

Review 1 (G.PC.1 a, c).

• $\angle A$ and $\angle B$ are consecutive, so they are supplementary: $(3x + 15) + (5x + 5) = 180 \Rightarrow 8x + 20 = 180 \Rightarrow x = 20$. Then $m\angle A = 75°$, $m\angle B = 105°$, $m\angle C = 75°$ (opposite $\angle A$), and $m\angle D = 105°$ (opposite $\angle B$). Check: $75 + 105 + 75 + 105 = 360$.
• The diagonals bisect each other, so $AM = MC$: $2y - 3 = y + 4 \Rightarrow y = 7$. Then $AM = MC = 11$ and $AC = 22$.
• The first used consecutive angles are supplementary; the second used the diagonals bisect each other. Both are theorems. The definition — both pairs of opposite sides parallel — was used in neither computation directly; it is what both theorems were proved from, back in Lesson 10.2.
• A parallelogram with congruent diagonals must be a rectangle. To call it a square you would still need four congruent sides (equivalently, perpendicular diagonals). And $ABCD$ cannot be a rectangle: its angles are $75°$ and $105°$, not right — so its diagonals are not congruent, even though they bisect each other. That is the distinction this chapter keeps making.

Review 2 (G.PC.1a).

• Rhombus. Specific to it: the diagonals are perpendicular (or: all four sides congruent, or each diagonal bisects a pair of opposite angles).
• Rectangle. Specific to it: the diagonals are congruent (or: all four angles right).
• Square. Specific to it: it is a rectangle and a rhombus at once — the diagonals are congruent and perpendicular, which no other family manages.
• Isosceles trapezoid. Specific to it: the legs are congruent (or: each pair of base angles congruent). Its congruent diagonals are shared with rectangles, so among trapezoids they identify it, but they are not enough on their own to name a family — see item 80.
• Trapezoid. Specific to it: exactly one pair of parallel sides, and nothing more — different-length legs is exactly the absence of the isosceles condition, so none of the three isosceles properties holds.

In every case, an inherited property — diagonals bisect each other, opposite sides congruent, opposite angles congruent — would be a wrong answer to the second half of the question, even though it is a true statement about the figure.

Review 3 (G.PC.1 a, c, d).

• Perimeter $= 14 + 10 + 6 + 6 = 36$, so $36$ in. The two legs are both $6$ in because the panel is an isosceles trapezoid.
• $m\angle B = 71°$ — the base angles at the long base are congruent. The two angles at the short base are each $180 - 71 = 109$, so $109°$ — they are same-side interior angles with the base angles, across the parallel bases along a leg (Chapter 2), so supplementary. Check: $71 + 71 + 109 + 109 = 360$.
• Use the congruent segment construction. Set the compass to one leg, and from the endpoint of the other leg swing an arc of that radius; if it passes through that leg's far endpoint, the two legs are congruent. The scribed arc stays on the panel — it is the evidence that the $6$ in was carried across rather than measured twice.
• Diagonals that bisect each other make the panel a parallelogram, and under Virginia's exclusive definition no parallelogram is a trapezoid — so it cannot be the intended isosceles trapezoid. Add congruent diagonals to a parallelogram and you get a rectangle. The panel is a rectangle, which is a genuine quality failure and not a mislabelling: the bases would have to be equal, so a $14$ in and a $10$ in base is impossible.