12 multiple-choice questions, progressively harder.
In convex trapezoid ABCDABCDABCD, ABABAB is parallel to CDCDCD. Let M,NM,NM,N be leg midpoints and PPP the midpoint of ACACAC. If MPMPMP is 111111 cm and PNPNPN is 666 cm, what is the length of base ABABAB?
Solution
Correct answer: A
PN is the midsegment in triangle ABC, so it has half the length of AB. Double 6 cm to obtain 12 cm. MP instead corresponds to base CD.
In convex trapezoid ABCDABCDABCD, bases AB,CDAB,CDAB,CD are parallel. MMM is the midpoint of ADADAD and PPP is the midpoint of ACACAC. Extend MPMPMP to meet BCBCBC at NNN. Which reason proves NNN is the midpoint of BCBCBC?
Correct answer: C
MP is parallel to CD by triangle ADC, hence to AB. In triangle CAB, the parallel through midpoint P cuts CB at the same one-half fraction by AA similarity.
Convex kite ABCDABCDABCD has equal pairs meeting at AAA and CCC, and diagonal intersection OOO. After SSS proves that ACACAC bisects angle AAA, which congruence proves BO=DOBO=DOBO=DO?
Correct answer: B
AB and AD are congruent; AO is common. The included angles BAO and DAO are congruent because AC bisects the angle at A. SAS therefore matches BO with DO.
Convex kite ABCDABCDABCD has equal pairs meeting at AAA and CCC. Its diagonals meet at OOO, and AO=OCAO=OCAO=OC. What is the strongest guaranteed classification among these choices?
Correct answer: D
AC already bisects BD by the kite proof. The extra midpoint makes both diagonals bisect each other, proving a parallelogram. Its adjacent equal-side pairs make all four sides equal, but no right angle at a vertex is given.
Two perpendicular segments ACACAC and BDBDBD meet internally at OOO. Which length data give a convex kite that is not a rhombus?
With the selected data, AC perpendicularly bisects BD, giving adjacent equal pairs by SAS. AO and OC differ, so BD does not bisect AC; it is not a rhombus. Each other option has mutual perpendicular bisection and gives four congruent right triangles.
In the equal-legs proof, convex trapezoid ABCDABCDABCD has exactly one parallel pair, longer base ABABAB, and congruent legs. Through CCC draw CECECE parallel to ADADAD, meeting ABABAB at EEE. Which triangle becomes isosceles, and why?
AECD is a parallelogram, so EC equals AD. The given equal legs then give equal sides EC and BC in triangle EBC. Its base-angle equality transfers to angles A and B.
Why does a non-parallelogram isosceles trapezoid fail to have a half-turn symmetry?
Its bases have different lengths: parallel and congruent opposite bases would make a parallelogram. A half-turn swaps opposite sides and preserves lengths, so it cannot swap these bases.
A convex trapezoid has chosen base lengths aaa and bbb. Its midsegment splits it into two smaller trapezoids. What is the midsegment length of the smaller trapezoid adjacent to the base of length aaa?
First find the original midsegment length m. Then average a with m for the adjacent trapezoid.
A convex kite has two different adjacent-pair side lengths. Why can it not have a second reflection axis?
At one vertex the two incident sides have the first pair length, and at the opposite vertex both have the other pair length. Each such vertex is unique because the two pair lengths differ. A reflection must therefore fix both individually. A reflection fixes exactly the points on its axis, so that axis must be the line through those two vertices.
In a convex kite, the given side pairs are AB,ADAB,ADAB,AD and BC,CDBC,CDBC,CD. Which triangle matching is justified by those pairs and one common diagonal, without further conditions?
SSS matches AB to AD, BC to DC, and AC to AC. That fixes the correspondence A to A, B to D, and C to C. The other matchings require extra equalities.
A parallelogram ABCDABCDABCD has congruent angles at chosen base ABABAB. Which conclusion is forced?
Consecutive parallelogram angles are supplementary. If those two angles are congruent, each is 90 degrees. Opposite-angle equality then makes all four right angles, so the shape is a rectangle.
Convex kite ABCDABCDABCD has equal pairs meeting at AAA and CCC. The side lengths ABABAB and BCBCBC are 101010 cm and 171717 cm. Diagonal BDBDBD is 161616 cm long. How long is ACACAC?
Let O be the diagonal intersection. AC perpendicularly bisects BD, so each half of BD is 8 cm. Apply Pythagoras separately in the two right triangles.
Convexity puts O inside AC, so add AO and OC.
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